Combinations

Q6...If point (p,q) lies on the perimeter of the rectangle above, what is the minimum possible value of (q+p)? −8 −5 −2 −1 0

A Correct. The minimum value of (q+p) will occur when p and q are both at their minimal value. The minimum point will occur in the third quadrant, at the bottom left corner of the rectangle. Both p and q will be at -4 at this point. Therefore, the minimum value of (q+p) is (-4)+(-4)=-8.

Warning:

A common mistake here is to focus only on the forbidden combinations - the jackets and pants. Students often make the mistake of calculating the number of forbidden choices as just 1×2=2. However, a combination is made of 1 jacket, 1 pair of shoes, and 1 pair of pants, and one piece of luggage (suitcase or carry bag). The "forbidden combinations" do not include just the one pants with the two jackets - it's the 24 combinations of these "forbidden items" with the rest of the sources (shoes, suitcases and carry bag).

If the volume of a certain cube is twice its surface area, what is the width of that cube? 3 6 12 24 36

C Correct Remember, the surface area of a cube is 6(Side²). Also, the volume of a cube is Side³. Equate the volume with twice the surface area to find the answer. Since the volume is twice the surface area, Side³ = 2 × 6 × Side². Divide by Side² to get Side³/Side² = 12 --> Side = 12. Hence, this is the correct answer.

To sum up:Choosing from a single source

Choosing from a single source - work the problem in a step-by-step mode. Consider whether problem include repetition or not - whether the items chosen can be chosen more than once or not. Without repetition, the SeBoxes's size changes from step to step. set up the problem as: n × (n-1) × (n-2) × (n-3)... Where n is the number of items to choose from and k is the number of choices to be made (the number of SeBoxes).

To sum up: Combinations "At least" questions

Combinations "At least" questions are considered one of the tougher question types in this already tough subject, but the correct approach will quickly unravel them. The major first step is identifying them by the use of "at least" in the question. "At least" questions are usually constructed in a way that the "Forbidden" scenarios will be fewer in number and easier to calculate than the "good" scenarios. Therefore, The presence of the phrase "at least" is a surefire trigger to breaking down the question using the "forbidden" combinations approach: [Total combinations - "Forbidden" combinations] = Good combinations

To sum up the basics of our method so far:Identifying Combinations problems:

Identifying Combinations problems: Problem requires counting the number of ways to choose k items out of a single source of n possible items, under the following conditions: 1) without repetition 2) the order of choice DOESN'T matter. The question doesn't introduce "titles" - it just asks how many ways to choose k items out of n possible items, regardless of the order of choosing. The question deals with groups, rather than with orderings. Solving method: 1) Set up the problem: write k SeBoxes, one for each item chosen. 2) Find the number of possible options for each SeBox, one step at a time. No repetition means n, n-1, n-2,... 3) Multiply the numbers to find the total number of options. 4) Order of choosing doesn't matter - Divide the result by k!. Alternative method - plug in n and k into the Combinations formula:

To sum up:

When a problem introduces forbidden choices, the calculation method is as follows: [Total combinations - "Forbidden" combinations] = Good combinations 1) Calculate the total # of combinations using the regular Step-By-Step method. 2) Calculate the total # of forbidden combinations using the same method. Remember to include the other sources (the ones without forbidden choices in them) as well. 3) Subtract the # of forbidden combinations from the total # of combinations to find the # of good combinations.

Alice has chosen A, B, and C to be her gold, silver and bronze rank BFFs, and now wants to take a picture of them to post in her blog. She picks up her camera, tells them to stand in a row, and waits impatiently while the three friends decide in what order they wish to stand. In how many different ways can the three friends arrange themselves? We have three places in our row, so three SeBoxes: 1st Place( ) 2nd Place( ) 3rd Place( ) Work the problem step by step, as before. How many friends can you choose from for the first place in the row?

3 Good. Either A, or B, or C - that's three possible friends in the first place. Write 3 in the first SeBox: 1st Place(3) 2nd Place( ) 3rd Place( ) Move on. Once we've placed a friend in the first place, how many friends can you choose from for the 2nd place? 2 Correct. Since one friend is already standing in the 1st place, you only have 2 choices for friends to stand in the 2nd place. Write "2" in the second SeBox: 1st Place(3) 2nd Place(2) 3rd Place( ) Since by now two friends have already been chosen, there's only one possible friend to place in the 3rd place: 1st Place(3) 2nd Place(2) 3rd Place(1) Multiply the numbers to get 3 × 2 × 1 = 6 ways of arranging 3 friends.

An electrical supplies store sells 140 power stabilizers every week. If each power stabilizer costs the store $16, what is the minimum selling price per unit that will ensure a weekly profit of at least $5600 from sales of power stabilizers? 24 26 40 56 60

A Incorrect. Remember: profit = revenues - expenses Find the minimum profit per unit. Since the question asked for the minimum selling price (revenue), don't forget to add the cost (expenses) of the stabilizer to the store. B Incorrect. D Correct. Divide the required profit by the number of units to find the minimum dollar profit-per-unit. 5600 / 140 = 560 / 14 = 40. Therefore, in order to make a $40 profit per unit, the store must charge 40+16=$56 per unit. Alternative method: numbers in the answer choice and a specific question call for reverse Plugging in. Assume that the selling price is $56. Take away the cost of $16 to get a profit of $40 per unit. Multiply by 140 to reach the minimum profit of $5600.

Q7...In the right triangle ABC above, AB=BC=4. If arc BD is centered at C, what is the area of the shaded region? 4-∏ 8-2∏ 4√2-∏ 8-∏ 8√2-∏ Now in this order eliminate answer choices that are: Negative - identify the expression that suggest negative area and POE them quickly. Not in the expected format: we expect subtraction with pi. Not in the ballpark: What portion of the triangle is occupied by the shaded area? divide the triangle in two, so you can assess more accurately.

B Correct. In dealing with shaded areas you deal with areas that are created by subtracting one geometric shape from another. The curved contour of the shaded region tells you that this subtraction involves a part of a circle and therefore you expect to find ∏ there. The principles of ballparking remain unchanged: Identify the given area - it's the area of triangle ABC= AB*BC/2 = 4*4/2=8. The target area you seek is the shaded region of course.

"Ice-Cold" Ice-cream factory produces only tricolor ice-cream products, where each ice-cream has three stripes of different colors. The factory uses the colors pink, purple, orange, silver, blue and red. How many different ice-cream products have at least one stripe out of the following colors: pink, purple or orange (assume that the order of the stripes in a single ice-cream does not matter)? 12 14 18 19 20

B Incorrect. In "at least" questions subtract the number of forbidden combinations from the total number of combinations to receive the good (wanted) combinations: [Total combinations - "Forbidden" combinations] = Good combinations In this case, the Total combinations is the total number of ways of choosing 3 colors out of 6, regardless of restrictions of colors. Now define the "Forbidden" combinations": groups of 3 colors in which there are no pink, purple or orange. The "Good" combinations are the remainder: all those trios that contain at least one of the three mentioned colors. D Correct. Total number of different ice-creams: Pick 3 colors out of 6, no order, no repetition. C(6,3) = 20 OK. Now, for the forbidden choices - Those are ice-creams that do not contain the colors pink, purple or orange, and thus they contain only the three other colors - silver, blue, and red. There is only one ice-cream with those colors (it's actually picking 3 colors out of 3). And thus, 20-1 = 19 ice-creams have at least one of the wanted colors.

Martha has 5 dogs and 4 cats. She takes each animal separately for a daily walk. She can take the animals for a walk to the park or to the barber shop. How many different options for a walk with a single animal does Martha have? 8 9 10 18 20

B Incorrect. Use SeBoxes to calculate separately the number of dog walks and cat walks. Then add the two numbers, because Martha can take either a dog OR a cat on a walk. D Correct. Martha can take a dog OR a cat for a walk. She can take each type of animal to 2 places. Let's start with the dogs: D(5) ₓ Places (2) = 10 So Martha can have 10 different trips with her dogs. Now the cats: C(4) ₓ Places (2) = 8 Martha can have 8 different trips with her cats. Since she can take a dog OR a cat on a walk, add the results: D(5) ₓ Places (2) + C(4) ₓ Places (2) = 18 Alternative method: since Martha has 9 animals in total and she can take each animal on 2 different trips, the total number of options that she has is: Animals (9) ₓ Places (2) = 18

In how many ways can the letters D, I, G, I, T be arranged so that the two I's are not next to each other? 36 48 72 96 128

C Close enough - you took 2 minutes to answer this question. Incorrect. Whenever you have forbidden options, calculate them and subtract them from the total number of options to receive only the wanted options. [Total combinations - "Forbidden" combinations] = Good combinations Notice that since you have two identical I's, their internal order doesn't matter - divide by the number of internal orderings of two items=2!. D Incorrect. A Correct. First, find the number of total choices: the number of ways to arrange 5 letters of which 2 are identical equals the number of arrangements for 5 different letters divided by the number of internal arrangements of the 2 identical letters (which we don't want to count, since they do not yield distinct arrangements): 5! / 2! = 60 Next, find the number of Forbidden choices (in which the 2 I's are next to each other) - treat the 2 I's as one, so you have to arrange only 4 terms (3 different digits plus one "big" digit of two Is): that's 4! options. Note: since the two I's are identical, you need NOT multiply that by the number of internal arrangements of the I's (2!) Finally, find the number of Good choices: 60 - 4! = 60-24 = 36

Consider the following question: A restaurant offers a business lunch menu composed of a main course and a glass of wine. The main course menu presents a choice of 2 beef dishes (which can go with a glass of either red or white wine) and 2 fish dishes (which only go with white wine). How many combinations of a business lunch menu of a single main course and a single glass of wine can be ordered? We identify this as a combinations question because it asks for the number of ways of combining or arranging things. Ask yourself: how many sources of selections are there? There are actually several possible ways to answer this, but don't get confused - focus on what you need to select: 1 main course and 1 glass of wine. Therefore, you need to select two items, or two SeBoxes, from two different sources. Draw the two SeBoxes: Main Course( ) Wine( ) Now, consider the Main Course SeBox: How many Main Courses can you choose from? And Herein lies the problem. The question presents four choices of main courses - two beef, two fish. However, the two subcategories are different in their relation with the other source of choices - Some go with both red and white, others only go with white. The solution: split the problem into two separate scenarios: Beef OR Fish. Deal with each scenario separately: Calculate the two number of combinations using the same Step-By-step, SeBox by SeBox method we've seen so far. Once you figure out the number of combinations in each separate scenario, Add the two together - each scenario presents a different way to reach the desired conclusion of a business lunch menu, and the OR relation between the scenarios dictates adding their separate results.

Begin with the Beef scenario: Draw the two SeBoxes: Beef Course ( ) Wine ( ) How many options of Beef course can we choose from? That's right, 2. Write "2" in the Beef SeBox. Beef Course (2) Wine ( ) Now, consider the wine SeBox: If we chose Beef, we have 2 possible choices of wine - red or white. Therefore, in this the Beef scenario, we write "2" in the Wine SeBox: Beef Course (2) Wine (2) Since the business lunch consists of Beef AND Wine, multiply the results to get: Beef Course (2) × Wine (2) = 4 That's 4 possible combinations for the Beef Scenario. Now, consider the Fish Scenario, which is similar to the Beef scenario, with one main difference: the 2 possible choices of fish only go with 1 choice of wine: white wine. Therefore, in THIS scenario, the SeBoxes will look like the following: Fish Course (2) × Wine (1) = 2 That's 2 possible total combinations for the Fish scenario. Now that we know there are 4 combinations with Beef, and 2 combinations with Fish, what do we do with these nice numbers? We add the two together - 4+2 = 6 possible combinations for a lunch menu We multiply the two together - 4×2=8 possible combinations for a lunch menu A.... Absolutely right. Remember that the two scenarios present an OR relation - a guest may choose either Beef OR Fish. When we discussed seperate sources of selections and their relations, we've seen that an OR relation means "add" the SeBoxes. The same goes between scenarios - if there are several seperate scenarios, presenting a choice of one OR the other, add the number of combinations of the individual scenarios to get the final number of combinations. Bottom line: I If the problem presents several scenarios, or several separate ways to reach the desired combinations: 1) Calculate the number of combinations for each scenario separately using the regular step-by-step method 2) Since the scenarios present an OR relationship, add the result.

How many three digit numbers contain the digit 5 at least once? 52 128 252 648 900

C Correct. In "at least" questions subtract the number of forbidden combinations from the total number of combinations to receive the good (wanted) combinations.: [Total combinations - "Forbidden" combinations] = Good combinations In this case, the number of Total combinations is the total number of 3 digit numbers. Now define the the "Forbidden" combinations: Containing the digit "5" at least once means that all the scenarios where the digit 5 appears once, twice or three times are good combinations. The "Forbidden combinations" are the remainder: all those numbers that do not include the digit 5 at all. First find the total number of 3 digit numbers. Remember that a number cannot begin with a "0", and thus the first box has only 9 options. Since the number of options changes between boxes, it is preferable to use SeBoxes instead of formulas for this question: 1st(9) 2nd(10) 3rd(10) = 900 Now, find the forbidden options, meaning numbers that do not have the digit 5 in them at all. Again, the first digit cannot be 0, but now it cannot be 5 as well: 1st(8) 2nd(9) 3rd(9) = 648 Lastly, subtract the number of "no 5" numbers from the total number of three digit numbers to receive the number of good options - numbers that have 5 in them at least once: 900-648 = 252.

As x increases from 141 to 142, which of the following increases? I) x−1/x II) 17−1/x III) (17+x)/x I only II only III only I and II only I and III only

D Correct. This question is about trends. When x grows, expressions I-III change. It seems that you have to deal with tricky algebra and ugly numbers, but actually you can avoid those awkward numbers. Instead of checking what happens to these expressions when x grows from 141 to 142, try checking the trends for friendlier numbers. If that sounds to you like plugging in, well, you are not mistaken. For example, check what happens to I-III when x grows from 1 to 2. X=1 X=2 result I) x−1/x 1-1/1=0 2-1/2=1.5 expression increases II) 17−1/x 17-1/1=16 17-1/2=16.5 expression increases III) (17+x)/x (17+1)/1=18 (17+2)/2=19/2≈10 expression decreases Only I and II have increased, so this is the right answer.

How many dressing combinations of one pair of shoes, one shirt, and one pair of trousers are possible if a person has 12 pairs of shoes, 5 shirts, and 2 pairs of trousers? 19 36 120 128 136

C Correct. This problem presents a case of Multiple distinct sources - shoes, shirts and trousers. Break the problem in a step-by-step method using SeBoxes. There are 3 items, so you need 3 SeBoxes. Find the number of choices for each SeBox, then multiply or add according to the AND / OR relationship between them. Since our person needs Shoes AND shirts AND Trousers, multiply the number of choices in each SeBox: Shoes(12) ₓ Shirts(5) ₓ Trousers(2)=120

How many combinations options are there for a code consisting of 2 even digits followed by 2 odd digits, if no digit can be repeated? 200 250 400 625 1000

C Correct. This problem presents a case of Single source - DIGITS. Break the problem in a step-by-step method using SeBoxes. There are 4 digits, so you need 4 SeBoxes. Since no digit can be repeated, there's no repetition and the SeBoxes' size changes from one digit to the next. There are 5 even digits and 5 odd digits. Once you used an even digit, you have only 4 options for the second even digit. The same goes for the odd digits, therefore the number of combinations for such a code is: Even1(5)×Even2(4)×Odd1(5) ×Odd2(4)=400

A class consists of 5 boys and 4 girls. Given that one kid can only hold one title, in how many ways can you pick 2 boys to be the class clown and the teacher's pet or 2 girls to be the most beautiful girl in class and the smartest kid on the block? 9 18 32 60 240

C Correct. This problem presents a case of choosing from multiple sources - boys AND girls. Since you need 2 boys OR 2 girls, work out the number of choices for each source separately, then ADD the number of choices together. Within each source (boys or girls), the problem presents a case of choosing several items from a single source. Since you cannot pick the same kid twice, there's no repetition. Finally, since you have to give the kids different roles, the order of choice matters. Now that we have reduced the problem to a simple case of choosing k kids out of n, without repetition, order matters, use the Permutations formula: For the boys, you have to pick k=2 out of n=5, where order matters (since there are two disparate titles): Same goes for the girls, but here we choose k=2 out of only n=4: Now, since you have to pick 2 boys OR 2 girls, ADD the two results to get: 20+12=32

Q2...If O is the center of the circle shown above, what is the value of x? 120º 130º 140º 150º 160º

C Correct. This question is involves arcs and angles. Use the method you have learned: 1. Copy the figure to your noteboard. 2. Indicate the measure of EVERY arc on the circle. 3. Then solve for x. After you copy the figure to your noteboard, indicate the measure of every arc on the circle. This is what your noteboard should look like: The inscribed 110° angle defines an arc of 220°, marked in Red. The measure of the remaining part of the circumference, i.e., the Blue arc, is 360°−220°=140°. Now, solve for x: The 140° arc is equal to the central angle that defines it, therefore x=140°.

How many 5 person committees chosen at random from a group consisting of 3 men, 5 women, and 2 children contain at least 1 woman? 250 251 252 275 300

C Very good! You took 1 minutes and 41 seconds to answer this question. Incorrect. In "at least" questions subtract the number of forbidden combinations from the total number of combinations to receive the good (wanted) combinations: [Total combinations - "Forbidden" combinations] = Good combinations In this case, the Total combinations is the total number of ways of choosing 5 members out of 10 people, regardless of restrictions. Now define the "Forbidden" combinations": at least 1 woman means that 1, 2 ,3, 4 or 5 women are all good combinations. The only forbidden combinations are the groups that do not contain any women, or 0 women. Calculate each number separately, then subtract the "Forbidden" combinations" from the Total combinations. Note: This is the number of total choices, and not only choices with women. B Correct. Total possible committees - picking 5 persons from 10 (5 women, 3 men, 2 children), no order, no repetition: 3×2×7×6 This is a total of 252 options. Now, the forbidden options - committees with no women at all - picking 5 persons out of 5 (just the men and the children): This is only 1 option - picking all men and all children. Thus, the good options are 252 - 1 = 251 options.

Q5...What is the area of the shaded region, formed by the intersection of an equilateral triangle with area √3, and an inscribed circle, as shown above? √3-3∏ √3-4/3∏ 3√3/4 √3-∏/3 3√3-∏

D Correct. Don't go over your head to calculate your way to the wrong answer. In shaded regions always ballpark. It's shorter and safer. Remember the POE criteria: negative area, subtraction with ∏. In order to better guesstimate the area of the shaded regions, divide the triangle into three identical parts: Within each part, the shaded region is about half the total area, hence, the total shaded area should be about 0.5·√3≈0.8. Now, eliminate any answer choice that is not in the right ballpark.

How many three digit numbers that do not contain the digit 2 are there? 100 200 512 648 729

D Correct. This problem presents a case of choosing from a single source - DIGITS. work the problem in a step-by-step mode. Consider whether problem includes repetition or not - whether the items chosen can be chosen more than once or not. Also, note the digit(s) that cannot appear. A three digit number requires three SeBoxes. Since the problem doesn't state that the digits cannot repeat themselves, this is a problems with repetition, and the SeBoxes' size should remain the same. Now, consider the constraints presented by the problem: Since you cannot use the digit 2, you have only 9 options for each box (the digits 0,1,3,4,5,6,7,8 or 9). Also, since the problem asks for NUMBERS, the first digit cannot be 0, and therefore has only 8 options. Multiply the numbers in the boxes, because you need all three digits to form the number. The number of three digits numbers without the digit "2" is therefore: 1st(8)×2nd(9)×3rd(9)=648

Arnie's closet contains three shirts and two pairs of pants. How many different dressing combinations (of shirt & pants) does Arnie have? 2 3 5 6 8

D Correct. Work the problem step-by-step using SeBoxes. How many boxes do you need? Is this an AND or an OR relationship? Arnie's outfit should include two items of clothing - a shirt and a pair of pants. Therefore, you need two SeBoxes. The shirts Sebox has 3 options and the pants SeBox has 2 options: S(3) AND P(2) Since The outfit includes a shirt AND pants, multiply the numbers in the boxes: S(3) × P(2) = 6

A "Sven" number is defined as a five-digit number which goes according to the following rules: the leftmost digit is even, any digit to the right of an even digit must be an odd digit, and any digit to the right of an odd digit can only be one of the digits 1 or 7. How many different 5-digit numbers are "Sven" numbers? 20 80 160 220 250

D Incorrect. Calculate the number of possible choices for each of the digits separately, using SeBoxes. Then multiply accordingly. Remember that a "Sven" number is still a 5 digit number. This means that the first digit cannot be zero - for example, 04352 is not a 5-digit number. C Correct. First digit - even: 4 choices (2, 4, 6, 8. Since you need 5 digit numbers, the first digit cannot be zero 0). Second digit - odd: 5 choices (1, 3, 5, 7, 9). Third (to the right of an odd digit): 2 options (1 or 7) Since 1 and 7 are both odd, the fourth and fifth digits will be to the right of an odd digit, and therefore also have 2 choices each - 1 or 7. Use SeBoxes to mark the number of choices for each digit, and multiply the boxes, since you need ALL digits to form each number. That's a total of 160 options: 1st(4) ₓ 2nd(5) ₓ 3rd(2) ₓ 4th(2) ₓ 5th(2) = 160

A committee of 6 is chosen from 8 men and 5 women so as to contain at least 2 men and 3 women. How many different committees could be formed? 635 700 1404 2620 3510

D You grossly underestimated the time this question took you. You actually solved it in 5 minutes and 45 seconds. Incorrect. This is not a typical "at least" combinations questions. It does not ask you about "at least one," but "at least two" or "at least three". In this case, it does not matter if you directly calculate the number of "good" choices, or if you subtract the number of "forbidden" choices from the total. In general, there are two "good' groups of committees of 6 people: the committees which contain 2 men and 4 women, and the committees which contain 3 men and 3 women. Any other numbers will not satisfy the "at least 2 men and 3 women" rule. Note that for each of the two "good' committee types there is no repetition and no importance to order of selection (since there are no titles for committee members). Use the Combinations formula to Calculate each type separately and add the results. B Correct. Start with the first group: 2 men and 4 women. You have to pick 2 men out of 8 and 4 women out of 5. Order does not matter, and since you are picking humans, there is no repetition. C(8,2) = 28 C(5,4) = 5 Since you have to pick 2 men AND 3 women, multiply the results and receive a total of - 28·5 = 140 committees of 2 men and 4 women. OK. Now, for the committees with 3 men and 3 women. This means picking 3 men out of 8 and 3 women out of 5, no order, no repetition: C(8,3) = 56 C(5,3) = 10 So there are 56·10 = 560 committees of the second type. This gives you a grand total of 140 + 560 = 700 committees with at least 2 men and 3 women.

The three competitors on a race have to be randomly chosen from a group of five men and three women. How many different such trios contain at least one woman? 10 15 16 30 46

E Correct. In "at least" questions subtract the number of forbidden combinations from the total number of combinations to receive the good (wanted) combinations.: [Total combinations - "Forbidden" combinations] = Good combinations In this case, the Total combinations is the total number of ways of choosing 3 competitors out of 8 people, regardless of restrictions of man or woman. Now define the "Forbidden" combinations": a trio with at least one woman means that all the scenarios in which 1, 2 or 3 women are chosen as competitors are good combinations. The "Forbidden" combinations are the remainder: all those trios that do not include a woman at all. First calculate the number of total arrangements. You need three SeBoxes, and remember that the number decreases because there is no repetition (you cannot choose the same person twice) and that the order does not matter, because there aren't any different roles: 56 Now, calculate the number of forbidden options - those in which there is no women, only three men. This is actually picking 3 competitors out of only 5 men, not ordered: 10 And lastly, subtract the number of "only men" trios from the "total number of trios" to find the number of "at least one woman" trios: 56-10 = 46

If the coordinates of points A and B are (0,−6) and (5,0) respectively, what is the slope of the line that passes through points A and B? −1.2 −5/6 5/6 0.9 1.2

E Correct. The slope is given in the formula Given the coordinates of the two points, the slope of the line is -6 - 0 / 0 - 5 = 6/5 = 1.2. hence, this is the correct answer.

If the length of a rectangular box B is 100, the width of B is 10, and the height of B is 1, what is the surface area B? 222 666 1000 1110 2220

E Correct. The surface area of a rectangular box is given in the formula: Surface Area=2wl+2hw+2lh. If the length of a rectangular box B is 100, the width of B is 10, and the height of B is 1, then: --> 2wl+2hw+2lh = 2·100·10+2·100·1+2·10·1 =2000 + 200 + 20 = 2220

10 different biology books and 8 different chemistry books lie on a shelf. In how many ways can a student pick 2 books of each type? 80 160 720 1100 1260

E Correct. This problem presents a case of choosing from a Multiple distinct sources - biology AND chemistry Books. Since you need 2 biology books AND 2 chemistry books, Work out the number of choices for each source separately, then MULTIPLY the number of choices together. Within each source (biology and chemistry), the problem presents a case of choosing several items from a Single source. Since you cannot pick the same book twice, there's no repetition. Finally, since you only want to know which books were chosen, rather than the order in which they're chosen, the order of choice doesn't matter. You have to pick 2 out of 10 AND 2 out of 8, and order of picking doesn't matter. You can use the formula twice (for each kind of book) and multiply the results, since this is an "AND" relationship: C(n,k) = n! / (n-k)! k! or you can use the SeBoxes. You need 2 SeBoxes for each kind of book, and remember to decrease the number by 1, since you cannot repeat a selection. Finally, for each type divide by 2! since the order does not matter, then multiply the results to get (10)×(9) / 2×1 × (8)×(7) / 2×1 =45×28=1260

How many 4 digit numbers can be formed if the tens and thousands digits are even, and the hundreds and units digits are odd? 100 200 300 400 500

E Correct. This problem presents a case of choosing from a Single source - DIGITS. Break the problem in a step-by-step method using SeBoxes. There are 4 items, so you need 4 SeBoxes. For each SeBOX, ask yourself "How many choices do I have for this digit?". Since the problem doesn't state otherwise, the digits can repeat and the SeBoxes size remains unchanged. Notice that the problem asks for 4 digit numbers, so the first digit cannot be zero - 0134 is not a four digit number. Fill the boxes step by step: 1st digit - even, no 0 (since 0 cannot be at the beginning a number) --> 4 options: 2,4,6,8 2nd digit - odd --> 5 options: 1,3,5,7,9 3rd digit - even --> 5 options: 0,2,4,6,8 (keep in mind that zero is even). 4th digit - odd --> 5 options: 1,3,5,7,9 4×5×5×5 = 500

Q4...(Office Excel)Which of the following represents the coordinates of a point that lies on the perimeter of the rectangle above? (−6,3) (3,−2) (1,1) (1,−5) (4,−4)

E Correct. Use POE - eliminate answer choices which cannot lie on the perimeter of the rectangle. The rectangle lies in the first and fourth quadrants, with x coordinate ranging from 2 to 4, and y coordinates ranging from 2 to -6. POE A, as it has a negative x value. POE answer choices C and D - the smallest x coordinate of the rectangle must be 2 (the coordinates of the leftmost line). Down to B and E. Take a look at B - (3,-2) is the center of the rectangle, not a point on its perimeter. Therefore, POE B, and you're left with E.

How many options are there for license plate numbers if each license plate can include 2 digits and 3 letters (in that order), or 3 digits and 2 letters (in that order)? (Note: there are 26 letters in the alphabet) 37×260² 36×260² 36×270² 10²×26^3 10²×26²

E Incorrect. This problem presents a case of Multiple distinct sources. Break the problem in a step-by-step method using SeBoxes. There are 5 places in each license plate, so you need 5 SeBoxes. There is no repetition limit, so you can repeat and the SeBOxes size remains unchanged. Finally, The order of choice matters. For each place in a license plate mark the number of choices. Multiply or add accordingly. Correct. Let's start with 2 digits followed by 3 letters. There are 10 digits AND 26 letters, so the number of combinations will look like this: B Now for 3 digits followed by 2 letters. There are 10 digits AND 26 letters, so the number of combinations will look like this: Since a license plate can be of the first type OR of the second type, add the combinations: 10²×26^3 + 10^3×26² = 10²×26²(26+10)=36×260²

In how many ways can the letters {U, U, U, R, R} be arranged? This deceptively short question represents a harder type of Combinations questions. It is difficult because it doesn't immediately resemble any known C & P formulas we've learned.

Instead of dealing with 5 letters, deal with the question in a "1st place, 2nd place, 3rd place.." step-by-step method. Each place will have a choice of letters, and therefore deserves its own SeBox. 1st Place() 2nd Place() 3rd Place() 4th Place() 5th Place() How many letters can you choose from for the first SeBox? A...2 - U or R B...5 - Any one of the five letters {U, U, U, R, R} B Correct. Disregard the fact that there are only Us and Rs for now, and treat them all as LETTERS. The first SeBox has a choice of 5 letters to choose from: U, U, U, R, R. Write that number in the 1st SeBox: 1st Place(5) 2nd Place() 3rd Place() 4th Place() 5th Place() Since the first letter is already in place, the second SeBox only has a choice of 4 letters to choose from. The next SeBox only 3, then 2, then 1. 1st Place(5) 2nd Place(4) 3rd Place(3) 4th Place(2) 5th Place(1) The next logical step would be to multiply the numbers together to find the number of possible arrangements of five letters. Indeed, if the letters were all distinct letters (such as the set {A, B, C, D, E}, the answer would be 5×4×3×2×1 = 5! = 120. However, we still need to bring into account the fact that we actually have two groups of similar letters, rather than 5 distinct letters. The 120 arrangements counted include redundant arrangements which simply displace similar Us or similar Rs; This stems from the fact that within each group (Us or Rs), the order of choosing doesn't matter! We've seen that when the order of choosing doesn't matter, we must divide by the number of internal arrangements, which is k!. Because the internal order of 3 Us doesn't matter, divide by 3!. Because the internal order of 2 Rs doesn't matter, divide by 2!. Our final number of arrangements is: 5!/3!×2! = 10

What if John had to choose one pair of pants from a selection of pants, as well as the jacket? If John needed to also choose a pair of pants out of, for example, 4 possible pairs, that selection would merit its own SeBox, with a different label of "Pants". Remember: Every item deserves its own SeBox.

We'll discuss the use of multiple SeBoxes later on. For now, just remember this: A SeBox is a box with the number of possible options to choose a single item from. Every item deserves its own SeBox. Every time a C&P question asks you to choose an item out of many possible items: 1) Draw a box with a label on top 2) Ask yourself - "how many items can I choose from in this box?" Write the total number of available options inside the box.

Okay, now for the good news: 1) Although these problems are difficult, they can be dealt with using the right method. In our Combinations and Permutations lessons we'll be looking at the concept of the Selection Box, or SeBox, and how it can help you deal with each problem in a structured, step-by-step method. Stronger students who already feel at ease with Combinations and Permutations (and remember some high-school formulas regarding C & P) are still advised to follow our method - we'll show how it relates to your existing body of knowledge later on. 2) Combinations and Permutations problems are not that common on the GMAT - you'll see only one or two in a quantitative section. 3) Remember that the GMAT test is adaptive - it adapts to your level. Although the subject of Combinations and Permutations is considered difficult as a whole, the GMAT does have some easier questions that even weaker test-takers can tackle using the right approach. By no means do we recommend skipping over Combinations questions just because you fear them - Stick with the method, and you might be able to steer your way out of a tight spot in the test.

We'll get back to our business exec's problem above later. For now, just remember this: Recognizing Combinations and Permutations questions: The question asks for the number of ways of choosing or arranging items chosen from one or more sources.

For any positive n, f(n) is defined by the following equation, f(n)=(−1)n+2. What is the value of the product f(1)·f(2)·f(3)·...·f(9)? 16 17 80 81 243

D Correct. To solve this question efficiently, you should find a pattern in the sequence. Calculate the first terms of the sequence by plugging in n=1,2,3... From these calculations it is evident that whenever n is odd, the whole term is 1, and when n is even, the whole term is 3. Since the problem asks for the product of these 9 terms, the odd n don't count - multiplying by 1 changes nothing. Thus, the real issue of this question is how many times is n even for n=1,2,3,...9. Between 1 and 9, n is even four times (2,4,6,8.) Therefore, the product of the nine terms, f(1)·f(2)·f(3)·...·f(9), is 34=3×3×3×3=9×9=81.

The overall approach is therefore:

1) Define the "Total combinations" scenario in terms of the question - determine k and n when in a scenario that ignores any limitations imposed by the question, and find the total number of combinations. In the above case, choosing 4 people out of 9, regardless of boy or girl. Calculate using the relevant method, paying attention to number of sources, repetition, and order of choosing. 2) Break down the "at least" into the distinct scenarios covered by it. In our case, at least 1 girl allows any number of girls between 1 and the maximum of 4. Use the list of "good" scenarios" to define the list of "Forbidden" scenarios. In the above case, our "forbidden" scenario was 4 boys, no girls, so our scenario was defined as choosing 4 people out of 5 boys, so that there are no girls chosen. Calculate using the relevant method, paying attention to number of sources, repetition, and order of choosing. 3) Subtract the number of "Forbidden" combinations from the Total number of combinations to find the number of "Good combinations.

In how many different 10 letter combinations can can the letters B, E, N, E, F, I, C, I, A, L be ordered? 10×9×8×7×6×5×3×2 10×9×8×7×6×5×4×3 8! 9! 10!

A Close enough - you took 1 minutes and 10 seconds to answer this question. Correct. Notice that our group of letters contains two identical pairs of letters: 2 E's and 2 I's. Therefore, this is an Internal ordering Combinations questions: the question asks how many ways are there to arrange a large group composed of several sub groups in which the order doesn't matter. The solution will be of the form of: Nos of Arrangements = Nos of ways o arrange large group / Nos of internal arrangements of all small groups You have 10! options to arrange 10 different items. Since 2 of the items are E's and two others are I's, you have to divide 10! twice by 2!: 10! / 2!×2! = 10×9×8×7×6×5×3×2

How many 6 letter combinations can be made using the letters A,B,C or the letters D,E,F ? 2×3^6 3×2^6 3×3^6 2×2^6 6^6

A Correct. There are two scenarios presented by this problem - a 6 letter combination made of the letters A, B, C, and a separate 6 letter combination for the letters D, E, F. Calculate each scenario separately, then add the results, as they have an OR relationship. Within each scenario: The problem presents a case of choosing from a Single source - LETTERS. Since you need a 6 letter combination, the three letters can repeat and the SeBoxes size remains unchanged. Set up the problem as: n × n × n × n... = nk Where n is the number of items to choose from (e.g. three letters A, B, C) and k is the number of choices to be made (number of letter places) Let's start with the A,B,C words. For each of the six places in the combination you can use all 3 letters. When the number of choices for each place remains the same, use nk where n is the number of choices and k is the number of places. Thus you have 36 options. The same goes for the D,E,F words. Since it's an "OR" relationship, add the two results to get to 36+36=2×36.

Back to our flying executive: A business executive is packing for a conference. He has 5 jackets, 4 pairs of shoes, 3 pairs of pants, 2 suitcases and a carry bag. He needs to choose 1 jacket, 1 pair of shoes, and 1 pair of pants to wear on the flight, and one piece of luggage (suitcase or carry bag) to carry the rest of his clothes. How many combinations of clothing and luggage can he choose from? Let's complicate the problem a bit: our fashion-minded exec realizes that one of the pants clashes with two of jackets, and there's no way he can wear these items together. How many combinations of clothing and luggage does he have now? How would you go about calculating this? We already know he has 180 combinations total. Let's subtract the "forbidden choices" - the number of choices involving the offending pants with the two jackets. Split into seperate scenarios: Calculate the number of "good" choices without the offending articles - first jackets without the pants, then the pants without the jackets, then add the numbers.

A Very good. What we want to calculate is [Total combinations - "Forbidden" combinations] = Good combinations Calculate the number of combinations involving "forbidden choices" (the clashing pants AND 2 jackets) using the same method for choosing from different sources. (Here's how): 1) Break down the question {number of forbidden choices} into a series of SeBoxes, one SeBox at a time for each "item". Forbidden choices: J( ) Sh( ) P( ) Su( ) C( ) 2) For each SeBox, find the number of choices in the box. Remember that we're calculating the number of "forbidden choices", so the number of choices for the pants and jackets will include ONLY the forbidden articles (2 J, 1 P): Forbidden choices: J(2) Sh(4) P(1) Su(2) C(1) 3) Multiply/add the number of choices from each source according to its relationship (AND/OR): Forbidden choices: J(2) AND Sh(4) AND P(1) AND {Su(2) OR C(1)} 2×4×1×(2+1) = 24 Remember that these are 24 "forbidden combinations" - combinations containing the clashing pants AND the clashing jackets. To find out how many "good combinations" we have, subtract 24 from the total of 180 to get: [total combinations - "forbidden" combinations] = 180 - 24 = 156 combinations.

Remember our paranoid security officer and his changing passkey? Well, the company has decided to install a new combinations lock, slightly different from the earlier model: A company has installed a combination lock on its front door. The lock requires a 4-digit passkey, where no digit may be used more than once. The company's paranoid security officer has decided to change the passkey every day. In how many days will he run out of new passkeys? Once again, the first question we ask ourselves is the same question: How many SeBoxes does the problem present? Still 4- 1st digit, 2nd digit, 3rd digit, and 4th digit. Every digit merits its own SeBox, so four digits = four SeBoxes. Your notebook should look like this: 1st Digit( ) 2nd Digit( ) 3rd Digit( ) 4th Digit( )

Again, work the problem step-by-step. We already know we have 10 options for the first digit - write that in the first box. Now, how many digits can he choose from for the 2nd digit? 1st Digit(10 ) 2nd Digit( ) 3rd Digit( ) 4th Digit( ) 9 Correct! Let's suppose the officer chose "8" for the first place. Since 8 cannot be used more than once, the second SeBox will only have 9 possible digits to choose from - all digits except "8". Write that in the SeBox: 1st Digit(10 ) 2nd Digit(9) 3rd Digit( ) 4th Digit( ) If, for example, he chose "5" for the 2nd digit, the 3rd SeBox will only have 8 possible digits (not "8", not "5"), and the 4th SeBox will only have 7 possible digits. 1st Digit(10 ) 2nd Digit(9) 3rd Digit(8) 4th Digit(7) What do we do with all these nice numbers? Think about the relations between the SeBoxes: since the officer has 10 possible digits for the 1st place AND 9 digits for the 2nd place AND 8 digits for the 3rd place AND 7 digits for the fourth, we multiply the numbers to reach 10 × 9 × 8 × 7 = 90×56 = 5,040 possible codes. The above is an example of choosing from a single source without repetition. We've seen that Without repetition, the SeBoxes's size changes from step to step. The question can be set up as: n × (n-1) × (n-2) × (n-3)..., where n is the number of items to choose from. The number of n's in the above setup is determined by the number of SeBoxes (which we call k). In the case of our 4 digit code, for example, we had 4 SeBoxes - one for each digit. Since n=10 digits, we ended up with 10×9×8×7. This question also demonstrates the importance of working Combinations and Permutations problems step-by-step: Instead of working out a 4 digit code without repetition, we work out 1st digit, then 2nd, then 3rd then 4th digit, and the number of possible choices in each SeBox depends and what has been changed by the choices in the preceding one.

X is an even integer greater than 300,000 and smaller than 1,000,000. How many numbers can X be? 300,000 349,999 350,000 399,999 400,000

B Correct. Break down X into its digits (units digit, tens digits, etc.) and find the number of choices for each digit using SeBoxes. The problem doesn't state otherwise, so digits may appear more than once, and the problem presents a case with repetition. Think about the limitations presented by the problem - X must be an Even number Within the specified range. There are 7 choices for the left-most digit (3, 4, 5, 6, 7, 8 and 9), 5 choices for the units digit (0, 2, 4, 6 and 8), and 10 choices for each of the remaining digits. Thus, the total number of choices is: (7)×(10)×(10)×(10)×(10)×(5)= 350,000 Don't forget to subtract 1, because you were asked how many numbers are greater than 300000, and thus the number 300000 itself should not be counted: 350,000-1= 349,999

In a phase III clinical trial of Dosaxin, a new drug, patients received a progressively growing dosage of Dosaxin for a few days. On the first day, each patient received 15 milligrams of Dosaxin. On each of the following days, the daily dosage was m milligrams greater than the dosage received the day before, reaching a dosage of 43 milligram on the last day of the trial. For how many days did the trial last, if each patient received a total amount of 145 milligrams of Dosaxin during the whole trial ? 4 5 6 7 9

B Correct. Sum of a list of numbers with constant increments can be calculated using: Sum = Average x Number of numbers, where average = (largest value + smallest value) / 2. Use the above to calculate the number of days i.e. Number of numbers. Given that Sum = Average x Number of days: 145 = [(43 + 15) / 2] x Number of days 145 = (58 / 2) x Number of days 145 = 29 x Number of days 145 / 29 = Number of days 5 = Number of days. Hence, this is the correct answer.

If m is a positive integer, which of the following must be a multiple of 6? (2m+3)(2m+4) (2m+2)(2m)(2m+1) (m+2)(m+3) (6m+1)(m+6) m(m+2)(m+3)

B Correct. This is a MUST be true Q. Plug in numbers and try to eliminate four answer choices that are not always true. An answer that yields a value that is NOT a multiple of 6 for a single value of m cannot qualify as MUST be true for all possible values of m - eliminate it. Since the question must have a right answer, the last answer standing (the last one that has not been eliminated) must be the right answer. Plug in a good number such as m=2 into all answer choices: (2·2+2)(2·2)(2·2+1) = 6·4·5 is a multiple of 6, so this answer choice cannot be eliminated for m=2. Since all other answer choices are eliminated for the same plug-plug in, this is the right answer choice. Alternative explanation: Rearrange answer choice B as (2m)(2m+1)(2m+2) and see that it actually consists 3 consecutive integers. Recall that the product of n consecutive integers must be divisible by n!.Therefore, this expression must be divisble by 3!=3·2·1=6.

In a certain province in France there are 15 cities. If a single road segment connects only two cities, how many road segments are required in order to connect the cities so that each city is connected to all other cities with a single road segment? 30 105 330 14! 15!

B Correct. When choosing from the same source with no order, set up the SeBoxes and divide by the number of possible internal arrangement. The question is, how many Seboxes are needed here? The question wants the number of roads to connect each city to all others. If the cities are A, B, C, D, etc. a road segment would be a combination of two letters: A-B, A-C, C-D, E-F, etc. The road A-B is the road leaving from A, going to B Thus, we need only two Seboxes -the first is the letter the road leaves FROM, the second is the letter the road arrives at. Set up the SeBoxes - there are 15 possible cities for the first SeBox. The number decreases by 1, as you cannot repeat a city in the same connection. Multiply the boxes, since this is an "AND" relationship (each of the 15 cities has 14 other cities it can connect to): 1st(15)×2nd(14) Last question: Does the order of choosing matter? When paving roads in France (particularly), it doesn't matter if you connect Marseilles to Lyon or Lyon to Marseilles - it's the same road! Therefore, the order of selection of each pair doesn't matter, and to compensate for that you need to divide by the number of possible internal arrangements - in this case for two cities - 2!: 1st(15)×2nd(14)/2×1 = 105

How many 4 digit numbers are there, if it is known that the first digit is even, the second is odd, the third is prime, the fourth (units digit) is divisible by 3, and the digit 2 can be used only once? 20 150 225 300 320

C You grossly underestimated the time this question took you. You actually solved it in 5 minutes and 6 seconds. Incorrect. Make sure you assigned for each SeBox the correct number of items that meet the terms. Think again - how many numbers between 0-9 are divisible by 3? B Incorrect. Because of the limitation on the usage of "2", use the "forbidden combinations" method - calculate all combinations and then subtract those you cannot use. [Total combinations - "Forbidden" combinations] = Good combinations For the first digit, remember that question asks for numbers (not codes), so zero is not an option: zero is indeed even, but a number cannot begin with zero. A Incorrect. D Correct. Find the total number of combinations, disregarding the restrictions on "2", by breaking down the problem set-by step using SeBoxes: The options for the digits are: First digit: 4 options ( 2, 4, 6, 8). Second digit: 5 options (1, 3, 5, 7, 9). Third digit: 4 options ( 2, 3, 5, 7). Fourth digit: 4 options (0, 3, 6, 9). That gives us a total of 320 options: 1st(4)×2nd(5)×3rd(4)×4th(4) = 320 Now, define the "Forbidden" combinations: the phrasing "the digit 2 can be used only once" means that the Good combinations include those numbers where the digit "2" does not appear at all, or appears once. Therefore, the "Forbidden" combinations include the remainder: those scenarios in which the digit "2" appears more than once. Since "2" can only appear in the two even digits, the "Forbidden" combinations include those numbers where both of the even digits are "2". We have to reduce the number of options in which 2 appears twice. To do that we can assume that 2 does appear twice and check the other places: First digit: 1 option (2) Second digit: 5 options (1, 3, 5, 7, 9). Third digit: 1 option (2). Fourth digit: 4 options (0, 3, 6, 9). That's a total of 20 "forbidden options", in which 2 appears twice: 1st(1)×2nd(5)×3rd(1)×4th(4) = 20 Now we subtract the "Forbidden" choices from the total choices to find the good options - numbers in which the digit 2 appears only once, or does not appear at all: 320-20=300 options.

How many 5 person committees chosen at random from a group consisting of 5 men, 5 women, and 5 children contain at least 1 woman? 700 1221 1434 2751 3011

C You slightly underestimated the time this question took you. You actually solved it in 2 minutes and 39 seconds. Incorrect. In "at least" questions subtract the number of forbidden combinations from the total number of combinations to receive the good (wanted) combinations: [Total combinations - "Forbidden" combinations] = Good combinations In this case, the Total combinations is the total number of ways of choosing 5 members out of 15 people, regardless of restrictions. Now define the "Forbidden" combinations": at least 1 woman means that 1, 2 ,3, 4 or 5 women are all good combinations. The only forbidden combinations are the groups that do not contain any women, or 0 women. Calculate each number separately, then subtract the "Forbidden" combinations" from the Total combinations. D Correct. Total possible committees - picking 5 people from 15 (5 women, 5 men, 5 children), no order, no repetition: 7×3×13×11 Now, the Forbidden options - committees with no women at all - picking 5 persons out of 10 (just the men and the children): 3×2×7×6 Thus, the number of good combinations is 7×3×13×11 - 3×2×7×6. Wait before you go and figure out each number separately. Extract the common factors 3×7: 3×7×(13×11 - 2×6) = 21×(143-12) = 21×131 = (20+1)×131 = 2751

However, the obvious choice is not the best one here. It will take too long, and leave much room for careless errors. Notice that our scenario list is missing a single instance: 0 girls, 4 boys. This is the only "forbidden" scenario according the terms of the question, requiring at least one girl on the team. What do we do with Forbidden scenarios?

Calculate the number of combinations for the forbidden scenario, and subtract them from the total number of combinations. Exactly. When a problem introduces forbidden choices, the calculation method is as follows: [Total combinations - "Forbidden" combinations] = Good combinations In terms of our question this reads: [Total number of ways of choosing team of four - number of ways of choosing a team with no girls ] = Number of ways of choosing a team with at least one girl This is now our master plan for this question: we will calculate each scenario (Total and Forbidden) separately, then subtract them to find the number of good combinations - combinations including at least one girl. Let's do this. Figure out the Total number of ways of choosing team of four first. We want the general case: choose 4 monitors out of the available population, without any limitations as to boys or girls. Therefore, k=4 and n= total number of boys and girls = 5+4 = 9. Repetition: Since we need a team of 4, there's no repetition - a person can't be chosen twice for the team, because then he or she will take up two spots and the team will have less than 4 people. So we're choosing k=4 from a single source of n=9 people, without repetition. This leaves us with one final consideration, before we calculate: Does the order of choice matter? Do we need to divide by k!=4!? A...Order of choice matters: leave as is, do not divide by 4!. B...Order of choice doesn't matter: divide by 4!. B Correct. The question only cares about which people are on the team. Since there are no "titles" (head of team, first boy, etc), it doesn't matter in which order we choose them: if we chose person A, then person B, it would mean exactly the same as choosing B first, then A - the bottom line in both cases is that A and B are on the team. Since the order doesn't matter, Use the Combinations formula : 126 OK, now focus on the Forbidden scenario: no girls, 4 boys. We want to choose 4 monitors, so k is still 4. As to n, in this limited scenario the group you're choosing these monitors from is limited to the group of 5 boys, as you do not want the girls to be a choice. Repetition: Still no repetition. Order matters / doesn't matter?: Same as before - order doesn't matter. Use the Combinations formula: 5 As a final step, go back to our original master plan: [Total combinations - "Forbidden" combinations] = Good combinations. So we know that there are 126 ways of choosing 4 people out of 9, and 5 "forbidden" choices of choosing 4 people out of 5 boys (no girls). The number of good combinations is therefore 126-5 = 121.

A photographer is to take group photographs of a class of students for the school magazine such that each photograph should have five students. If there are four girls and four boys in the class and each photograph must not have two girls or two boys standing together, how many different photographs can the photographer take? 80 288 4^4 576 288^2

D Correct This fearsome looking question requires some thought. First, how can we make sure that we won't have two boys or two girls standing together? If we start with a boy on the left side of the row, then we cannot place another boy next to him - we must place a girl. Since that girl cannot have another girl standing next to her, we must alternate the third place with a boy again, etc. It follows that each group photo of five must alternate boy/girl/boy. However, there are two scenarios here, depending on whether we start with a boy (B-G-B-G-B) OR a girl (G-B-G-B-G). Calculate each scenario separately, then add the results, as they have an OR relationship. Within each scenario: The problem presents a case of choosing from Two sources - Boys and Girls. Keep in mind that the options from each source cannot be repeated, as no boy can logically appear twice in the same photo. Let's start with the photographs with girls and boys standing in the order GBGBG. Since there's no repetition, the number of options for each successive child from the same gender decreases: for example, if there are 4 choices for the first girl, there are only 3 choices for the next girl (third place), and two choices for last girl (5th place). The same goes for boys (2nd and 4th places). G(4)B(4)G(3)B(3)G(2) = 288 Now, do the same for photographs with girls and boys standing in the order BGBGB. Note that intuitively, this scenario should have the same number of permutations, as we have an identical 4 boys and 4 girls to choose from. B(4)G(4)B(3)G(3)B(2) = 288 Since there is an OR relationship here between either of the two types of photographs that can be taken, add the combinations. 288 + 288 = 576. Hence, this is the correct answer.

Four representatives are to be randomly chosen from a committee of six men and six women. How many different 4-representative combinations contain at least one woman? 400 415 455 480 495

D Correct. In "at least" questions subtract the number of forbidden combinations from the total number of combinations to receive the good (wanted) combinations: [Total combinations - "Forbidden" combinations] = Good combinations In this case, the Total combinations is the total number of ways of choosing 4 members out of 12 people, regardless of restrictions of man or woman. Now define the "Forbidden" combinations": a quartet with at least one woman means that all the scenarios in which 1, 2, 3, or 4 women are chosen as representatives are good combinations. The "Forbidden" combinations are the remainder: all those trios that do not include a woman at all. First calculate the number of total arrangements. You need four SeBoxes, and remember that the number decreases because there is no repetition (you cannot choose the same person twice) and that the order does not matter, because there aren't any different roles: 495 Now, calculate the number of forbidden options - those in which there are no women, only four men. This is actually picking 4 people out of a limited selection of 6 men (again, not ordered because there are no roles): 15 Lastly, subtract the number of "only men" quartets from the "total number of quartets" to find the number of "at least one woman" quartets: 495 - 15 = 480

(Office Excel) Q1... Arcs DE, EF, FD are centered at C, B and A, in equilateral triangle ABC, as shown above. If the area of the triangle is 6, what is the area of the shaded region formed by the intersection of the arcs and the triangle? 6-√3∏/3 6-√3∏/2 ∏/√3 6-√3∏ ∏/6

D Correct. Leave the details of strange new regions to NASA. While you stay on the land of GMAT, use Ballparking in order to quickly eliminate irrelevant answer choices. You start by identifying the given value, which in our case is the area of triangle ABC=6. The target area is a shaded region so you expect that you'd be able to POE some of the answer without an actual calculation - just on the basis of either the answer represents a negative area or it is not in the expected format- subtraction with a pi. Triangle ABC can be divided to 4 similar triangles: The shaded area is about a half of a quarter of ΔABC, i.e. ≈0.75. Eliminate answer choices that are far off the right ballpark, either in terms of size or because they do not apply to the form Shaded region = Total area−Unshaded

How many 5 letter combinations can be made from the letters of the word VERMONT if the first letter has to be a vowel and the last letter has to be a consonant, and each letter can be used only once? 21 42 120 600 720

D Correct. This problem presents a case of choosing from a Single source - letter Break the problem in a step-by-step method using SeBoxes. There are 5 items in the required combinations, so you need 5 SeBoxes. Since you can use each letter only once, there's no repetition and the SeBoxes' size changes. Finally, since changing the order of the chosen letters yields a different word, the order of choice matters. When there are restrictions on certain terms, start the filling the SeBoxes from those "problematic" terms: The first letter can only be a vowel - 2 choices The fifth letter can only be a consonant - 5 choices Then move to the other terms: The second through fourth letters can be anything, but since we already used two letters, there are only 5 choices left for the second letter (2 out of 7 have already been used) and then 4 and 3 choices respectively for the third and fourth letters (the number decreases with each consecutive letter, because there is no repetition). The SeBoxes should therefore look like this: 1st Digit(2)×2nd Digit(5)× 3rd Digit(4)×4th Digit(3)× 5th Digit(5) 2×5×4×3×5 = 600 combinations

(House Excel)\Q3...An equilateral triangle is inscribed in a circle, as shown above. What is the area of the shaded region, if the area of the circle is 2? 2 - 3√3 2 - 3√3/∏ 2 - 3√3/4 2 - 3√3/2∏ 2∏ - 3√3/2∏

D Correct. Use Ballparking and avoid unnecessary calculations. The shaded area is the target value that you seek. In your ballparking calculations you use the given value - in our case it's 2, the area of the circle. Run on each answer using elimination criteria, from the faster elimination to the slower. Start by checking if the expression is negative - An area can't be negative. Next ballpark for estimated size and POE answers that are not in the ballpark. To better guesstimate, divide the circle in three symmetrical sections. This is how the diagram looks like after cutting it in three: In each section, the shaded region is slightly larger than the area of the adjacent sub-triangle. Therefore, the area of the shaded regions altogether is a little larger than half the area of the circle, i.e., a little larger than 1. Click on the small arrow on the box above to scroll back other answers and check POE considerations.

Ed throws a blue die and a red die. How many combinations are there in which the two dice do not yield the same result and the sum of the results is not 9? 10 24 26 30 36

D Incorrect. In this problem, the limitations are phrased very clearly, leading to a "Forbidden combinations" approach: [Total combinations - "Forbidden" combinations] = Good combinations Calculate the "Forbidden" combinations, then subtract them from the Total combinations to get the Good combinations (the wanted choices). C Correct. There aren't that many Forbidden choice, so you can just count them: Start with the results in which the dice are the same: 1-1, 2-2, 3-3, 4-4, 5-5 and 6-6 - total of 6 "Forbidden combinations" where the dice yield the same result. None of these equals 9, so count the "Forbidden combinations": Results that sum up to 9 are 3-6, 6-3, 4-5, and 5-4 - a total of 4 Combinations. So there are 10 "Forbidden" choices. Calculate the total number of possible results for rolling two dice using SeBoxes. With six choices for the Die I AND six choices for Die II, the total number of combinations is: Die I(6) × Die II(6) = 36 So the number of good choices are 36-10 = 26 results.

The Coen family consists of a father, a mother, two children and a dog. A photographer is about to take the family's picture. How many different arrangements (of standing in a row) does the photographer have, if it is known that the father insists on standing by his woman, as the song says? 12 24 48 60 120

D Incorrect. This problem presents a case of choosing from a Single source - Family members. It also presents an added twist - the father and the mother have to stay together. When two or more objects have to be adjacent to each other, begin by considering them as one object at first. Then break the problem in a step-by-step method using SeBoxes. There are 4 items (father+mother object, two children, 1 dog), so you need 4 SeBoxes. Since we want everyone in the picture, there's no repetition and the SeBoxes' size changes. Finally, since we're asked for the different ways to arrange the family in a row, the order of choice matters. After calculating the number of arrangements, remember to multiply by the number of internal arrangements in the big object. B Incorrect. Since the big "father+mother" object actually has 2! arrangements possible (father to the right of mother, or to her left), multiply the result by 2!. C Correct. Calculating the number of different arrangements while treating the father and mother as one object yields 4! choices: 1st(4) ₓ 2nd(3) ₓ 3rd(2) ₓ 4th(1) = 24 But, since the big "father+mother" object actually has 2! arrangements possible (father to the right of mother, or to her left), multiply the result by 2!: 4!×2! = 24×2 = 48 different family arrangements.

Father John forms a choir from the church attendants. 30 people attend John's church, and the choir has 28 spots available, with one person as the lead singer. How many different combinations does John have? 3005 4412 6544 12180 24366

D You overestimated the time this question took you. You actually solved it in 2 minutes and 20 seconds. Correct. Pick the lead first - there are 30 choices. Now John has to pick 27 more choir members out of the 29 still available. The order of picking the other members does not matter, since they are not assigned different roles in the choir. Since John can't pick the same person twice, there is no repetition. Use the combinations formula for selecting 27 out of 29, no repetition, not ordered. Now, there are 30 choices for the lead vocal - multiply the number of choices for the choir by 30, because you need a lead AND 27 other members: 30×29×14 choices Hmmm. This is painful to calculate, but the solution is a number that ends with 0 (why? look at the unit's digit for each of the numbers that are multiplied: 0×9×4 = something that ends with 0). 12180 is the only answer that ends with 0. Hallelujah John!

M and N are among the 5 runners in a race, and there can be no tie. How many possible results are there where M is ahead of N? 10 25 35 60 80

D You slightly overestimated the time this question took you. You actually solved it in 3 minutes and 49 seconds. Correct. This problem requires common sense much more than strict formulas. Calculate the total number of options to arrange 5 terms in a row. Order matters here, since each order yields a different result in the race. There are 5! = 120 ways of ordering 5 runners. Now think - if there is no possibility for a tie, in how many of these 120 arrangements is M ahead of N? Considering the fact that there is no possibility of a tie, M will be either ahead or behind N. Intuitively, there is no real reason why there would be more results with M ahead of N than behind. Thus, the number of combinations where M is ahead of N should be a straightforward half of the total number of combinations - the total number of ways of ordering 5 runners in a race. There are 5! = 120 ways of ordering 5 runners, and half of that is simply 120/2=60. If you want to analyze this by numbers, do the following analysis: If N is last, then there are 4 places where M is ahead of N (1st, 2nd, 3rd and 4th places), and none where he is behind. If N is in fourth place, there are 3 places where M is ahead (1st, 2nd, 3rd) and one where he is behind. If N is in third place, there are exactly two places ahead and two behind. If N is in 2nd place, there is only one place ahead and 3 places behind. Note that this situation is the exact symmetrical opposite of the situation where n is in fourth place. If N is first, there are no places where M is ahead, and four places behind - the exact symmetrical opposite of the situation where N is last. If you go down the above list and sum the number of cases either way (ahead/behind), you will find that the number of places where M is ahead (4+3+2+1+0) is equal to the number of cases where M is behind N (0+1+2+3+4). This proves that there is no real bias either way, and thus the number of possible results where M is ahead of N should be exactly half of the total number of results, or 120/2=60.

There are 10 points on a circle. A hexagon can be formed by linking 6 of the 10 points. How many such hexagons are possible? 60 120 200 210 600

D You slightly overestimated the time this question took you. You actually solved it in 57 seconds. Correct. This problem presents a case of choosing from a Single source - points on a circle. You need to pick 6 out of 10 points. Does the order of picking them matter here? For a certain hexagon it doesn't matter if you chose vertex x before or after vertex y - in both cases, x and y will be vertexes of the hexagon. Therefore, the order of choosing the vertexes doesn't matter. You can either use the formula, in which case n=10 and k=6: C(n,k) = 210 or you can break down the problem using SeBoxes. There are 10 possible points to choose from for the first Vertex. You cannot choose the same vertex twice (no repetition), thus the number in each box decreases by 1. Since the order of choosing doesn't matter, divide the product of the boxes by the number of possible arrangements of the 6 chosen vertexes = 6!:

In the Land of Oz only one or two-letter words are used. The local language has 66 different letters. The parliament decided to forbid the use of the seventh letter. How many words have the people of Oz lost because of the prohibition? 65 66 67 131 132

D You slightly underestimated the time this question took you. You actually solved it in 3 minutes and 57 seconds. Incorrect. The number of lost words is the difference between the number of words originally and the number of words after the prohibition. Remember to consider the two scenarios presented by the question: 1 letter words, and 2 letter words. Calculate the total number of options for each scenario before and after the prohibition, then subtract the two scenarios before from the two scenarios after. Within the two letter scenario: break the problem into two SeBOxes (1st letter, 2nd letter). Remember that the problem does not state that each word has to contain two different letters - thus, this is a case with repetition, and the SeBox size remains the same. E Correct. Before the prohibition there were 66 one-letter words. There were also 66² two-letter words: 1st Letter (66)×2nd Letter(66) =66² After the prohibition there are 65 letters, so there are 65 one-letter words, and 65² two-letter words: 1st Letter (65)×2nd Letter(65) =65² The lost words are the difference between the number of words before and after the prohibition: (66+66²)-(65+65²) = 66-65+(66²-65²) = 1+(66²-65²) Expand the second part using the third recycled quad: a2-b2 = (a-b)(a+b): 66²-65² = (66-65)×(66+65) = 1×131 = 131. So the total number of forbidden words is 1+131=132.

A company that ships boxes to a total of 12 distribution centers uses color coding to identify each center. If either a single color or a pair of two different colors is chosen to represent each center and if each center is uniquely represented by that choice of one or two colors, what is the minimum number of colors needed for the coding? (Assume that the order of the colors in a pair does not matter) 4 5 6 12 24

D You slightly underestimated the time this question took you. You actually solved it in 3 minutes and 7 seconds. Incorrect. This questions presents a tougher nut to crack, but don't worry: use the answer choices and Reverse PI in such cases! Each answer gives you the number of 1 color codes (for example, 4 colors such as red, blue, green, black means 4 single-color codes with the same). For the 2-color codes, this is a simple case of choosing 2 colors out of the n colors, not ordered, where n is the number of colors you plug in from the answer choice. Use the Combinations formula: C Incorrect. B Correct. Since the question asks for the minimum number of colors needed, start by plugging Answer choice A back into the question: If there are 4 colors, there are 4 codes that have only 1 color. As for 2 color codes, pick 2 out of 4, not ordered: C(4,2) = 6 So together there are 10 possible codes, which is not enough, since you need at least 12. Eliminate A, and move on to B: If there are 5 colors, there are 5 codes that have only 1 color. As for 2 color codes, pick 2 out of 5, not ordered: C(5,2) = 10 So together there are 15 possible codes, which is more than the 12 needed.

How many boy and girl couples can be made from 7 girls and 8 boys? 14 15 32 48 56

E Correct. This problem presents a case of choosing from Multiple distinct sources - girls and boys. Break the problem in a step-by-step method using SeBoxes. There are 2 kinds of items, so you need 2 SeBoxes. Find the number of choices for each SeBox, then multiply or add according to the AND / OR relationship between them. There are 7 options for girls and 8 options for boys. Multiply the option to get a couple of a boy AND a girl: Girls(7)×Boys(8) = 56

If x and y are positive integers, and 3x=10y, then x·y CANNOT be which of the following? 30 120 240 480 3000

E Incorrect. This is an integers-question. Obviously, the equation 3x=10y poses some constraints on the possible values of x and y. To find out what this equation really means, copy it to your noteboard and find simple numbers that can be plugged into x and y to satisfy the equation. 3x = 10y 3 x 10 = 10 x 3 with x=10 and y=3 3 x 10 x 2 = 10 x 3 x 2 with x=10x2 and y=3 x 2 If you want to add a factor to x, you will have to add it also to y to satisfy the equation. This means that any factor other than the 10 or 3, is shared by the two variables. Therefore, xy will contain a factor of 3, a factor of 10, and other factors that come in pairs (one from each variable.) For example, x may by 10x2, and y 3x2, so xy is 3x10x2x2. The factor that was added after the 3 and 10 appears in both variables and therefore appears twice in the product. Now go down the answer choices and see if their factors besides 3 and 10 come in pairs. 3000 = xy=30·10·10. Thus, x=10·10 and y=3·10 satisfy the equation, and x·y could equal 3000. C Correct. A 30 - that's easy to POE: x=10 and y=3 are possible values, so x·y could equal 30. B 120 = 30×4 = 30 × 2·2. Thus, x=10·2 and y=3·2 satisfy the equation, and x·y could equal 120. C 240 = 30×8 = 30 × 2·2·2. The factors other than 30 do not come in pairs, so there are no possible integer values for x and y. Therefore, xy cannot equal 240. Since you've an answer choice xy CANNOT equal, you can stop right there, choose the answer choice and move on. We provide the POE for the other two answer choice follows: D 480 = 30 × 16 = 30 × 4·4. Thus, x=10·4 and y=3·4 satisfy the equation, and x·y could equal 480. E 3000 = xy=30·10·10. Thus, x=10·10 and y=3·10 satisfy the equation, and x·y could equal 3000.

Four people are to be selected to receive a modeling contract from a group of 10 men, 5 women and 2 children. How many ways are possible in which 2 men, one woman and one child will receive a contract? 90 180 225 450 900

E Incorrect. This problem presents a case of Multiple sources - men, women and children. Since you can't choose the same person twice, there's no repetition . Finally, for contract purposes the order of the chosen ones doesn't matter. For each selection, use the combinations formula: C(n,k) = n! / (n-k)! k! for picking k terms out of n, without repetition and without order. Note that the order of choosing the two men doesn't matter, so you need to divide the number of ways of choosing men by k!=2! E Correct. You have to pick 2 men out of 10 (order of picking doesn't matter here): C(10,2) = 45 Now you have to pick 1 woman out of 5 - that's 5 options, and 1 child out of 2 - that's two options. Since you have to pick 2 men AND a woman AND a child, multiply the three results to get the total possible options: 45×5×2 = 45×10 = 450

How many different arrangements are possible to place seven different books on a shelf if all three math books must be placed next to each other? 120 240 360 540 720

E You grossly underestimated the time this question took you. You actually solved it in 5 minutes. Correct. In questions in which objects must be adjacent, count them as one object at first. Here, treat the 3 math books as one - It's as if you have 4 books plus one "big" book. The question is then reduced to a relatively simple scenario of arranging 5 books (4 books plus 1 "big" math book) so you have 5! options to arrange them. However, there's another step: the three math books are not REALLY one book, and they can be rearranged between themselves. So now, since you can arrange the three math books in k!=3! different arrangements, multiply by the number of internal arrangements within the "big" book (3!): --> 5!×3! = 120×6 = 720

Alice has 8 friends, but only three can be awarded a gold, silver and bronze star "best friends forever" ranking in her Facebook profile. In how many ways can she choose and order three BFFs? Okay, a combinations problem, asking to count the number of ways to choose and order items - in this case, choose friends". How many sources to choose from? That's right, one source - 8 friends. Since Alice has to choose 3 BFFs, there's no repetition - each friend can only appear once on the list (She can't give one friend two awards, because then she'd have only 2 friends on the list). How many SeBoxes? 3 friends = 3 SeBoxes. Set up the problem: Gold BFF( ) Silver BFF( ) Bronze BFF( )

Fill them out step-by step: How many possible friends to choose from for Gold BFF? That's right, 8 friends. Let's say we choose friend A. How many possible friends to choose from for Silver BFF? only 7 left. Let's say we choose friend B. How many possible friends to choose from for Bronze BFF? only 6 left. Let's say we choose friend C. Multiply the number of options (And relationships) to get 8 × 7 × 6 = 336 This problem is quite similar to the 4-digit combination lock without repetition problem we've seen. We bring this problem up to discuss the issue of ORDER OF CHOICE. In this problem, the Order of choice MATTERS. We chose specific friends, but the details aren't important. There are 336 options to order 3 BFFs out of 8 friends. ABC is just one such ordering; BAC would be a different ordering (B as gold BFF and A as silver BFF is a different choice than A as gold and B as silver - just ask friend A!). All of these different orderings involving A, B, and C, as well as other friends, are counted within our 336 options. Remember this bit when we later discuss a similar kind of problem where the order of choice DOESN'T matter.

To sum up the basics of our method so far:

Identifying Permutations problems: Problem requires counting the number of ways to choose k items out of a single source of n possible items, under the following conditions: 1) without repetition 2) the order of choice matters. The question will introduce "titles": 1st, 2nd 3rd,; Gold, Silver, Bronze; President, VP, CFO. ABC and BAC are considered two different orderings. Solving method: 1) Set up the problem: draw k SeBoxes, one for each "title". 2) Find the number of possible items to choose from for each "title", one step at a time. No repetition means n, n-1, n-2,... 3) Multiply the numbers to find the total number of options. Alternative method - plug in n and k into the Permutations formula: P(n,k) = n! / (n-k)!

Got it. I now know how to solve THIS question. How do I handle similar questions? Is there a general rule?

Internal arrangements Questions represent the harder combinations questions encountered by high-scoring students only, and are considered very rare. It is perfectly possible to get a score of even 50 or 51 on the GMAT quantitative section without encountering a question of this sort. Here's the general case for handling this type of question: Identify Internal ordering Combinations questions by the following structure: the question asks how many ways are there to arrange a large group composed of several sub groups in which the order doesn't matter. This is opposed to most Combinations questions, which are variations on the theme of "how many ways of choosing a single small group from a large group". Once you've identified a Combinations question as an Internal ordering question, the solution will be of the form of: Nos of Arrangements = Nos of ways o arrange large group / Nos of internal arrangements of all small groups Another way of putting the above expression is Nos of Arrangements = n!/k!×k!×k!... Where n is the number of members of the large group, and the ks are the number of identical members in each of the subgroups where the order doesn't matter.

Remember Alice and her gold, silver and bronze star BFFs on Facebook? Alice has 8 friends, but only three can be awarded a gold, silver and bronze star "best friends forever" ranking in her Facebook profile. In how many ways can she choose and order three BFFs? That problem presented a Permutations question - choosing from one source, ordering k=3 BFFs out of n=8 possible friends, no repetition, order of choosing matters.

Let's look at a slightly different version of the same problem: Alice has 8 friends, but only three can be listed as "best friends forever" in her Facebook profile. In how many ways can she choose her three BFFs? We approach this problem the same way we approached it before. Again, a combinations problem, choosing from one source - 8 friends. Since Alice has to choose 3 BFFs, there's no repetition - each friend can only appear once on the list. Set up the problem (3 friends = 3 SeBoxes): BFF( ) BFF( ) BFF( ) How many possible options for BFF on the left? That's right, 8 friends. Let's say we choose friend A. How many possible options for middle BFF? only 7 left. Let's say we choose friend B. How many possible options for BFF on the right? only 6 left. Let's say we choose friend C. Our SeBoxes look like this: BFF(8) BFF(7) BFF(6) Multiply the number of options to get So far, we've done the exact same thing for both versions of the question. Indeed, It LOOKS like the answer is the same for both. However, if that were indeed the case, then why are we bothering you with this new version? What's the crucial difference between the two versions? BFF(8) BFF(7) BFF(6) =56*6 = 336 In this new version there's no order of rank (Gold, silver, bronze) - we just choose 3 BFFS out of 8. That's exactly right. It's a subtle difference, but it makes a huge difference in the answer. In the older version, Alice had to choose AND order her 3 BFFs - Gold BFF, Silver BFF, Bronze BFF. ABC and BAC were two DIFFERENT choices - In ABC, A is the Gold star Best Friend Forever, in BAC B is the Gold star Best Friend Forever. That's a huge difference, especially for what A and B think of Alice! However, in the new version, the order DOESN'T MATTER. Alice just needs to choose 3 BFFs - no need to rank them. Under these condition, ABC and BAC are the SAME CHOICE - in both cases, A, B, and C retain the honor of being Alice's BFFs, and the order of their choosing is irrelevant. So we think we have 8×7×6 = 336 different ways of choosing 3 BFFs, but in fact we have LESS options: those 336 ways of choosing 3 BFFs actually include several redundancies - several seemingly different choices that are actually the SAME CHOICE.

We begin our journey into Combinations land by introducing the concept of the Selection Box, or SeBox. A SeBox, as its name implies, is a box of selections. Every time a Combinations & Permutations question asks you to select one item out of many possible items (e.g. one jacket out of 5 possible jackets), you will model this Selection in a SeBox. This is the basis of our step-by-step method to solve C & P problems.

Let's look at this (very) simple example: John has 5 jackets. How many ways are there for John to choose one jacket from his collection? 1) Draw a box with a label of "jackets" on top: 2) Ask yourself - "how many Jackets can I choose from?" In this simple case, we have 5 jackets, so the box contains 5 possible jackets to select from. The SeBox will look like this: 5 is indeed the answer to the question.

We continue our journey into the land of Permutations and Combinations. In this lesson we'll discuss an OR relationship between SeBoxes. Take a look at this problem: A certain class has 5 boys and 3 girls. If the teacher wishes to choose a boy OR girl to be hall monitor, how many ways of choosing a hall monitor does she have? This is a Combinations & Permutations question, as it asks us to count the number of ways of choosing something out of something. Remember our prime directive : Every Item deserves it own SeBox. In this simple question there are two "items": Boys and Girls. Draw two SeBoxes, with the appropriate labels:

Now ask yourself "How many "items" can I choose from in this box?" We have 5 boys and 3 girls to choose from - write those numbers in the appropriate box: The teacher needs to choose one hall monitor from either the boys or the girls. An OR relation means we're basically combining the two SeBoxes into one big "Box" containing 5 boys and 3 girls, and choosing one from the (5+3) choices available. Since the two Seboxes have a relationship of OR between them, simply add the two numbers: Remember this: "OR" means "ADD" - if a question presents multiple SeBoxes with an OR relation, add the number of options in the SeBoxes.

We continue our journey into the land of Permutations and Combinations. In this lesson we'll discuss an AND relationship between SeBoxes. Take a look at this problem: A certain class has 5 boys and 3 girls. If the teacher wishes to choose one boy AND one girl to be hall monitors, how many ways of choosing a hall monitor team does she have? This is a Combinations & Permutations question, as it asks us to count the number of ways of choosing something out of something. Remember our prime directive for : Every Item deserves it own SeBox. In this simple question there are two "items": Boys and Girls. Draw two SeBoxes, with the appropriate labels:

Now ask yourself for each SeBox "How many "items" can I choose from in this box?" We have 5 boys and 3 girls to choose from - write those numbers in the appropriate boxes: The teacher needs to choose one boy AND one girl to be hall monitors. For every one of 5 boys, there will be a choice of 3 girls to partner with (e.g. Danny and one of three girls, Oliver and one of three, etc.). Therefore, the teacher will have [5 boys × 3 girls] choices for hall monitor team. Since the two Seboxes have a relationship of AND between them, simply multiply the two numbers: Remember this: "AND" means "Multiply" - if a question presents multiple SeBoxes with an AND relation, multiply the number of options in the SeBoxes.

Remember our business exec? A business executive is packing for a conference. He has 5 jackets, 4 pairs of shoes, 3 pairs of pants, 2 suitcases and a carry bag. He needs to choose 1 jacket, 1 pair of shoes, and 1 pair of pants to wear on the flight, and one piece of luggage (suitcase or carry bag) to carry the rest of his clothes. How many combinations of clothing and luggage can he choose from? First question: How many SeBoxes does the problem present? There are 5 "items" in the above question: jackets, shoes, pants, suitcases and carry bag. Each item deserves its own SeBox.

Now, work the problem Step-by-Step, or SeBox after SeBox: Look at the first SeBox (Jackets): Ask yourself, "how many "jackets" can I choose from?" That's right, 5 Jackets. Draw the box, label it, and write 5 inside the Box. You can use shorthand - J for jacket, Sh for shoes, etc. Now, take the next step - move on to the next Box (Shoes): "how many options of shoes?". There are 4 pairs of shoes to choose from, so write 4 inside the box. Continue in this way, Step-by-Step, until you've filled all the SeBoxes. Your notebook should look like this: ...J....Sh....P....Su....C ..5.....4.....3....2.....1 Now, add the relationships between the SeBoxes. According to the info in the question, our executive needs a jacket AND Shoes AND Pants, AND a piece of luggage, which can be either a suitcase OR a Carry bag. Place brackets around the Su and C SeBoxes to visually show that they're lumped under the same "luggage" category. Therefore: J(5) AND Sh(4) AND P(3) AND {Su(2) OR C(1)} Finally, multiply/add the number of choices in each SeBox according to its relationship (AND/OR). J(5) ₓ Sh(4) ₓ P(3) ₓ {Su(2) + C(1)} Now for the actual calculation. There are 5 ways of choosing 1 jacket out of 5, so we can simply replace the "J" SeBox with the number inside it. The same goes for Shoes - there are only 4 ways to choose 1 pair of shoes out of 4 pairs, so replace the Sh SeBox with the number inside. Repeat the process until you reach this calculation, then work it out: 5ₓ4ₓ3ₓ3 = 180 So our executive has a total of 180 possible choices of flight wear and gear.

How do I tell if the order matters or doesn't matter?

OK, pay attention: In questions where the order of choice MATTERS, the chosen group must have "titles": President, Vice President, and Chairman; gold, silver, and bronze medals (in a race); etc. Look for these words in the question: how many Arrangements / Orderings. In questions where the order of choice DOESN'T MATTER, the chosen group has no titles and all members are equal: members of a chosen committee; winners of certificates (without a definition of first, second third places); etc. Look for these words in the question: how many Combinations / Committees / Groups. Important tip: Order DOESN'T MATTER is by far the more common case in GMAT C & P questions, so when in doubt, divide by k!

Combinations problems where the order of choice matters are actually known as Permutations. You may remember the Permutations formula for calculating the number of ways of ordering k items out of n possible items:

P(n,k) = n! / (n-k)! In our case, we want to order k=3 items out of n=8 possible items. Therefore: P(8,3) = 8! / (8-3)! = 8! / 5! = 8×7×6 = 336. If you don't remember the formula, don't sweat it - we reach the same solution, either by using the formula, or by breaking the problem into k steps and multiplying the number of options for each step - 8 for the first 1, 7 for the 2nd, 6 for the 3rd. Working out Permutations problems one step at a time works just as well as using the formula. In fact, getting used to analyzing the problem and working it step-by-step is BETTER than just blindly remembering formulas - not every question will exactly fit the formula, and some flexibility is required.

Back to our Hall monitor problems, with a significant twist: A certain class has 5 boys and 4 girls. If the teacher wishes to choose a hall monitor team of four students, containing at least one girl, how many ways of choosing a hall monitor team does she have?

Pay extra attention to the phrase "at least", as it is the main cause of difficulty in this version of the question. Let's break it down and see what it means to have at least one girl on a team of four. Having one girl on the team is a possible scenario, but "at least" one means we can't stop there. Our possible combinations are: One girl, three boys OR Two girls, two boys OR Three girls, one boy OR Four girls, no boys. Each of these is a different scenario. The obvious way to deal with this question is the same method we've seen for dealing with multiple scenarios: Work out the number of combinations for each one, then add the results, as these scenarios have an OR relationship.

What if we had to choose a greater number of friends? Is there an easy way to calculate the number of ways to arrange a large number of items, without drawing numerous SeBoxes?

Sure thing. Think of the general pattern: If Alice had to arrange 5 friends, we'd be drawing 5 SeBoxes with reducing number from 5 to 1; 10 friends - 10 SeBoxes, numbers beginning with 10 and coming down to 1. In fact, there's a general formula for this case you'd do well to memorize (we'll use it later): If we're ordering k items, the number of ways of arranging k items is simply k!

Now that we've discussed the SeBox and the possible (AND / OR) relationships between SeBoxes, it's time to get to the crux of our general method for dealing with Combinations & Permutations Questions. As we mentioned in the C & P overview, the main problem with C & P questions is finding the right model for the question - forming the right calculation. We advocate a step-by-step approach to analyzing C & P questions: 1) break down the question (any combinations question) into a series of SeBoxes, one SeBox at a time for each "item". 2) For each SeBox, find the number of choices it represents. 3) Multiply / add the SeBoxes using AND/OR relationships.

That last bit of arithmetic is the answer to the question correctly. Adhering to these simple steps will solve over 90% of the C&P questions in the test. We'll now focus on how to implement them on increasingly more complex problems. As for the other 10%, don't worry, we'll deal with them too. Later. The remainder of our Combinations and Permutations lessons will present a series of questions you need to ask yourself in order to correctly identify the SeBoxes and the number of choices in each SeBox.

A company has installed a combination lock on its front door. The lock requires a 4-digit passkey. The company's paranoid security officer has decided to change the passkey every day. In how many days will he run out of new passkeys? This is an example of a fairly simple combinations question of choosing from a single source. Unlike our business executive problem, where we chose from multiple sources (1 each from the sources of jacket, shoes, pants..etc), this question presents choices from the SAME SOURCE - the source of "digits". We're bringing this question up in order to discuss our basic method for solving Combination questions choosing several items from a single source.

The first question we ask ourselves is the same question: How many SeBoxes does the problem present? Although we have only one source, we're choosing 4 times from that source - 1st digit, 2nd digit, 3rd digit, and 4th digit. This bit is crucial for solving more complicated questions - Every digit merits its own SeBox! Four digits = four SeBoxes. Your notebook should look like this: 1st Digit( ) 2nd Digit( ) 3rd Digit( ) 4th Digit( ) Next, calculate the number of options for each place in the code. How many options are there to choose a digit for the 1st place? Since a single digit can be anything between 0-9, the answer is "10 options". Write that down under 1st digit. Good. Suppose we chose "8" for the first digit. How many options are there to place a digit in the second SeBox? Is the answer still 10, any digit between 0-9? No Yes Indeed! Since the problem does not specify that each digit can only be used once, we still have 10 options for the second digit - 0-9. In this case, we say that the problem allows repetition - each digit can repeat itself. For example, our code could be "8888". Therefore, the 2nd digit will have the same SeBox as the first: 1st Digit(10) 2nd Digit(10) 3rd Digit( ) 4th Digit( ) Since this is a problem with repetition, the same process applies to the 3rd and 4th digits of our code - we have 10 possible options each, and therefore will all have the same SeBox: 1st Digit(10) 2nd Digit(10) 3rd Digit(10) 4th Digit(10) Finally, let's calculate the number of options for our 4 digit code: Since we have 10 options for the first digit, AND 10 options for the 2nd digit, AND so on and so forth, multiply the numbers together. Remember, AND means '×'. Therefore, the number of possible combinations is: 1st Digit(10)×2nd Digit(10)× 3rd Digit(10)×4thDigit(10)=10^4

OK. So how do we get rid of these redundancies? By how much do we need to reduce 336? Well, first we need to find out how many redundancies are there. ABC and BAC are the same - we got that. But they're not alone. Think of all the different ways to order A, B, and C: ABC ACB BAC BCA CAB CBA All of these orderings of A, B, and C are counted as "different ways of choosing" in our 336 options, but if the order of choosing doesn't matter, they all mean the same thing - A, B, C are chosen to be Alice's BFFs. They should be counted as only one choice (One group): {A, B, C} The same could be done for a different group of friends. For example, the 6 different ways if ordering D, E and F are also considered just 1 option: {D, E, F} are now the group of Alice's BFFs. In other words, if the order of choosing doesn't matter, every six "internal" combinations of three friends become just one "group".

The general rule: If the order of choosing doesn't matter, calculate the number of orderings (permutations), then divide by the number of internal arrangements of each group of choices. In this case, the number of internal arrangements of 3 friends is six, so the final calculation is: (8×7×6) / 6 = 336 / 6 = 56. Got it. In this case, we could just list and count the number of combinations of 3 friends. What if we had to choose a greater number of friends? Is there an easy way to calculate the number of internal arrangements? Why, of course there is. We've already seen that the number of internal arrangements of k items is simply k!. In our case, k=3, so divide by k!=3!=3·2·1=6. The final general rule: If the order of choosing doesn't matter, calculate the number of orderings (permutations), then divide by the k! = the number of internal arrangements of each group of choices. You may remember the Combinations formula for calculating the number of ways of choosing k items out of n possible items: C(n,k) = n! / (n-k)! k! In our case, we want to choose k=3 items out of n=8 possible items. Therefore: C(8,3) = 8! / [(8-3)!·3!] = 8! / [5!·3!] expand 8! in the numerator and 5! in the denominator and reduce: (8×7×6×5×4×3×2×1) / 5×4×3×2×1·3! = (8×7×6) / 3! = (8×7×6) / 3×2×1 = 56. Take a closer look at the Combinations formula. Hmm, this looks like our Permutations formula: P(n,k) = n! / (n-k)! Only, did you notice that we added k! in the bottom? We must divide by the number of internal arrangements of k items since the order doesn't matter. Yet again, we reach the same solution, either by using the formula, or by breaking the problem into k steps, multiplying the number of options for each step - 8 for the first 1, 7 for the 2nd, 6 for the 3rd, then dividing by k! because the order doesn't matter. Working out Combinations problems one step at a time works just as well as using the formula. In fact, getting used to analyzing the problem and setting it up step-by-step is better than just blindly remembering formulas - not every question will exactly fit the formula, and some flexibility is required.

This lesson wraps up our basic work order for dealing with GMAT Combinations (order doesn't matter) and Permutations (order matters) questions. We will also use the opportunity to discuss one of the key issues within the C & P topic - discriminating between these two question types. The two subcategories are similar in the sense that in both of them the question requires choosing a small group out of a larger group - pulling red marbles out of a bag of mixed marbles, choosing boys and girls in a class for a hall monitor team, etc.

The initial stages of dealing with those questions are therefore quite similar: 1) Identify the number of sources of to choose from: Are we choosing from a single source (such as children), or multiple distinct sources (such as boys and girls)? For each source: 2) Find the size of the small group that must be chosen. In other words, how many SeBoxes? Or a similar question: what is the value of k? 3) Find the size of the larger group that is chosen from: How many options can I choose from for the first SeBox? Or a similar question: what is the value of n? Fill in the rest of the SeBoxes, Step-by-Step, paying attention to: 4) Repetition: does the problem present a case With repetition (such as digits in a code, where each digit can be used more than once) - SeBox size remains the same: or n × n × n × n ×... Without repetition (such as choosing children for a hall monitor team, where each child can only be chosen once) - SeBox size changes: n × (n-1) × (n-2) ×... 5) Finally, after all SeBoxes have been filled in, comes the crucial question: Order of choice MATTERS or DOESN'T MATTER? If the order of choice matters, leave the calculation as is. If the order of choice doesn't matter, divide by k! - the number of internal arrangements of k items.

A business executive is packing for a conference. He has 5 jackets, 4 pairs of shoes, 3 pairs of pants, 2 suitcases and a carry bag. He needs to choose 1 jacket, 1 pair of shoes, and 1 pair of pants to wear on the flight, and one piece of luggage (suitcase or carry bag) to carry the rest of his clothes. How many combinations of clothing and luggage can he choose from? Welcome to the wonderful world of GMAT Combinations and Permutations questions. This subject is rightly considered one of the toughest and least communicative topics introduced by the test. The questions here will ask you to count the number of ways of choosing combinations from one or more sources. Some of them may also ask you to calculate the number of ways of arranging or ordering these choices (number of ways of arranging books on a shelf, for example).

The reason why Combinations and Permutations questions are difficult is that they usually require a straightforward calculation. Our host of alternative solving methods - Ballparking, POE, Plugging In in all its forms - will usually not be applicable to C & P questions. The final calculations themselves are not that hard - most of the difficulty comes from the initial process of analyzing a question and finding WHICH one of several calculation models fits that particular question.

The above is an example of choosing from a single source with repetition. We've seen that With repetition, the SeBoxes remain the same. The question can be set up as: n × n × n × n..., where n is the number of items to choose from. The number of n's in the above setup is determined by the number of SeBoxes. In the case of our 4 digit code, for example, we had 4 SeBoxes - one for each digit. Since n=10 digits, we ended up with 10×10× repeating itself 4 times, once for each SeBox. In general terms, we can say the question is modeled as n*n*... repeating itself k times, 1 for each item selected.

This can be described algebraically as the following rule: Number of combinations for choosing multiple items from a single source, with repetition: n^k Where k is the number of choices to be made (number of SeBoxes), and n is the number of options to choose from. In our above example, we need to choose 4 digits, so k=4. Each selection has 10 possible digits to choose from, with repetition, so n=10. Therefore, the number of combinations for a 4 digit code is n^k = 10^4 = 10,000.

Working the problem SeBox after SeBox is our basic approach for any combinations problem. In this problem, each SeBox represented a different item, so the method involved simply determining their AND/OR relationship and multiplying / adding accordingly. Other problems will present further difficulties: bad choices (forbidden choices), or choosing more than 1 item from each source (which will require further SeBOxes within that source). We'll deal with those soon. For now, remember this: The first question we ask ourselves when approaching a Combinations question: How many SeBoxes does the problem present?

Work the problem Step-by step - 1) break down the question (any combinations question) into a series of SeBoxes, one SeBox at a time for each "item". 2) For each SeBox, find the number of choices in the box. 3) Multiply / add the SeBoxes using AND/OR relationships.