Ch. 4_Combinatorics

Fundamental Counting Principle

*If you must make a number of separate decisions, then MULTIPLY the numbers of ways to make each individual decision to find the number of ways to make all the decisions.

*For problems in which certain choices are restricted and/or affect other choices,*

*choose the most restricted options first.*

*If a GMAT problem requires you to choose two or more sets of items from separate pools,*

*count the arrangements separately - using a different anagram grid each time. Then multiply the numbers of possibilities for each step.*

Factorial MEMORIZE 1!

1! = 1

5. A women's health store offers its own blend of trail mix. If the store uses 4 different ingredients, how many bins will it need to hold every possible blend, assuming that each blend must have at least two ingredients?

11. I said 48. Trail mix blends contain either 2, 3, or 4 ingredients. Consider each case separately. First, figure out the number of 2-ingredients blends as anagrams of the word YYNN: Thus, 4!/2!(2!) = 4 Then consider the unique blend that includes all 4 ingredients. Thus, there are 6 + 4 + 1 = 11 blends. You can also solve by using the formula for combination 3 separate times: once for all combinations of 2 different flavors, once for all combinations of 3 different flavors, and once for all combinations of 4 different flavors. Thus, 4!/(2!) x 2 + 4!/3! x 1 + 4!/4! x 0 = 6 + 4 + 1 = 11

Factorial MEMORIZE 2!

2! = 2 x 1 = 2

3. A men's basketball league assigns every player a two-digit number for the back of his jersey. If the league uses only digits, 1-5, what is the max number of players that can join the league such that no player has a number with a repeated digit, and no two players have the same number?

20. I said 60. In this problem, order matters. Each number can be either the tens digit, the units digit, or not a digit in the number. Therefore, this problem is a question about the number of anagrams that can be made from the word TUNNN. Thus, 5!/(5 - 2)! = 5!/3! = 20. You can also use the slot method. I said: 5!/2! = 60

Factorial MEMORIZE 3!

3! = 3 x 2 x 1 = 6

9. There are 10 soloists featured on an album, but the liner notes are only 5 pages long, and therefore only have room for 5 of the soloists. The soloists are fighting over which of them will appear in the liner notes, as well as who will be featured on which page. How many different liner note arrangements are possible?

30,240. I said 252. In this problem, the order in which the soloists appear is important. Order matters. Therefore, the problem can be modeled with anagrams of the word 12345NNNNN, in which each number represents the page on which a soloist might appear: 10!/(10 - 5)! = 30,240 The 5! in the denominator corresponds to the 5 soloists not chosen.

11. A student's teacher has given them a list of 6 birds they can choose to write about. If Lizzie wants to write a report that includes two or three of the birds, how many different reports can she write?

35. I had no idea. I got stuck. First, solve the 2-bird report as anagram of the word YYNNNN: 6!/4!(2!) = 15 Then, consider the number of 3-bird reports of the word YYYNNN: 6!/3!(3!) = 20 Thus, there are 15 + 20 = 35 possible bird combinations. I said: 6!/3! = 120 6!/2! = 360

7. If the mayor of Metropolis chooses 3 members of the 7-person delegation to meet with Superman, how many different 3-person combinations can he choose?

35. I said 6. Model this problem with anagrams for the word YYYNNNN, in which three people are in the delegation and 4 are not: Thus, 7!/3!(4!) = 35 Order does not matter. I said: 3! = 6

Factorial MEMORIZE 4!

4! = 4 x 3 x 2 x 1 = 24

15. Gordon buys 5 dolls for his 5 nieces. The gifts include two identical beach dolls, one dress up doll, one GI doll, and one troll doll.

48. I guessed. I said 4. First, solve the problem without considering that the youngest does not want the GI doll. Gordon's niece could get either one of the beach dolls. This problem can be modeled with anagrams for the word SSETG. Thus, 5!/2! = 60 Divide by 2! because of the two identical beach dolls. There are 60 ways in which Gordon can give the gifts to his nieces. However, we know the youngest does not want the GI doll. So we calculate the number of arrangements in which the youngest girl *does* get the GI doll. If niece E gets doll G, then we still have 2 S dolls, 1 E doll, and 1 T doll to give out to nieces A, B, C, and D. Model this situation with the anagrams of the word SSET: 4!/2! = 12 There are 12 ways in which the youngest niece *will* get the GI doll. Therefore, there are 60 - 12 = 48 ways in which Gordon can give the dolls to his nieces. I said: 4!/(4 - 1)! = 4

12. A mother bird has 6 babies. Every time she returns to the nest, she feeds half. If it takes her 5 minutes to feed each baby, how long will it take her to feed all the possible combinations of three babies?

5 hours. I didn't finish solving. I said 20. The mother bird can feed 3 babies out of six. Model the situation with the word YYYNNN. Find the number of anagrams for this word: 6!/3!3! = 20. If it takes 5 mins to feed each baby, then it will take 15 minutes to feed each combination. Thus, it will take 20 x 15 = 300 mins or 5 hours. I said: 6!/3!3! = 20 Didn't finish.

Factorial MEMORIZE 5!

5! = 5 x 4 x 3 x 2 x 1 = 120

Factorial MEMORIZE 6!

6! = 6 x 5 x 4 x 3 x 2 x 1 = 720

8. A spy is trying to escape from his prison cell. The lock requires him to enter one number, from 1-9, and then push a pair of colored buttons simultaneously. He can make one attempt every 3 seconds. If there are 6 colored buttons, what is the longest possible time it could take the spy to escape from the prison cell?

6.75 minutes. I had no idea. I said 720. First, consider how many different pairs of colored buttons there are with the anagram YYNNNN: Thus, 6!/4!(2!) = 15 For each spy tries, he must try all 15 button combinations. Therefore, there are 15 tries per number. With 9 numbers, there are 15 x 9 = 135 tries. If each try takes 3 seconds, it will take the spy a maximum of 135 x 3 = 405 seconds, or 6.75 minutes. I said: 9! x 6! = 720

14. 3 gnomes and 3 elves sit down in a row of 6 chairs. If no gnome will sit next to another gnome, and no elf will sit with another elf, in how many different ways can the elves and gnomes sit?

72. I had no idea. I said 8. You can have either: GEGEGE or EGEGEG. Use the slot method to assign seats to gnomes or elves. If the first gnome has 6 choices and sits in an odd numbered chair, the second gnome can sit in either of the two remaining odd numbered chairs. The last gnome has only 1 option. Then sit the elves. The first elf can sit in any of the three empty chairs, the second in any of the other two, and the third in the final chair. Therefore, the first elf has 3 choices, second has 2 choices, and third has 1 choice. Thus, find the product of the number of choices for each: 6 x 2 x 1 x 3 x 2 x 1 = 72 OR Each of the subsequent choices has 3! = 6 options. Therefore, 2 x 3! x 3! = 2 x 6 x 6 = 72. I said: 3! = 6 3! = 6 Thus, 12 - 4 = 8

10. The principal of a school needs to observe 6 teachers. She plans to visit one teacher each day from Mon-Fri. She will only have time to see 5 of her teachers. How many different observation schedules can she create?

720. I said 6. Model this problem with anagrams of the word 12345N where N means not at all. Thus, 6! = 720 Do not divide since no letters are repeated. Or, since order matters. 6!/(6 - 5)! = 6!/1! = 720 I said: 6!/5!(1!) = 6

Arrangements with constraints Example: Greg, Marcia, Peter, Jan, Bobby, and Cindy go to a movie and sit next to each other in 6 adjacent seats in front row of theater. If Marcia and Jan will not sit next to each other, in how many different arrangements can the six people sit?

Constraint: Marcia and Jan will not sit next to each other. Step 1: Solve to problem, ignoring the constraint for now. Just find the number of ways in which six people can sit in 6 chairs. 6! = 720 Step 2: Count the arrangements in which Jan is sitting next to Marcia (constraint), and *subtract* them from that total of 720. Imagine Jan and Marcia and one person. Thus, 5! = 120. Each of those 120 ways represents two different possibilities, because they are stuck together - either J-M or M-J. Therefore, the total # of seating arrangements with Jan next to Marcia is 2 x 120 = 240. Thus, 720 - 240 = 480.

Example: A restaurant menu features five appetizers, six entrees, and three desserts. If a dinner special consists of one appetizer, one entree, and one dessert, how many different dinner specials are possible?

Since the choices are separate, the total is the product of dinner specials. 5 x 6 x 3 = 90

Example: An office manager must choose a five-digit lock code for the office door. The first and last digits of the code must be odd, and no repetition of digits is allowed. How many different lock codes are possible?

Step 1: Pick the most restricted. The first and last digits must be odd. There are 5 different odd digits (1, 3, 5, 7, and 9). Thus, there are 5 ways to pick the first number. Since there is no repetition, the last digit has only 4 ways. Step 2: Choose the less restricted. You can pick the other three digits in any order, but make sure you account for lack of repetition. For those choices, you only have 8, 7, and 6 digits available. Step 3: Fundamental Counting Principle. The total number of lock codes is 4 x 5 x 8 x 7 x 6 = 6,720

Example: An office manager must choose a five-digit lock code for the office door. The first and last digits of the code must be odd, and no repetition of digits is allowed. How many different lock codes are possible? Slot Method

Step 1: Set up slots. We have five. _ _ _ _ _ Step 2: Fill in the restricted slots. 5 _ _ _ 4 Step 3: Fill in the remaining slots. 5 8 7 6 4 Therefore, 5 x 8 x 7 x 6 x 4 = 6,720

Anagram example: If three of the seven standby passengers are selected for a flight, how many different combinations of standby passengers can be selected?

Step 1: Use an anagram grid. Person: A B C D E F G Seat: F F F N N N N *The four nonflying passengers are designated as N. Here, we divide by two factorials, 3! for the F's and 4! for the N's, yielding a much smaller number. 7!/(3!)(4!) = (7 x 6 x 5 x 4 x 3 x 2 x 1)/[(3 x 2 x 1) x (4 x 3 x 2 x 1)] = 7 x 5 = 35

Multiple arrangements

The GMAT will often *combine* the Fundamental Counting Principle and the anagram approach on more difficult combinatorics problems, requiring you to choose successive or multiple arrangements.

Anagram example: How many different anagrams are possible for the word GMAT?

There are 4 distinct letters; therefore, 4! = 24 anagrams of the word.

Anagram example: How many different anagrams are possible for the word ATLANTA?

This features *multiple* repetitions: three A's and two T's. Each of these sets creates redundancy. There are 3! ways to arrange the three A's and there are 2! to arrange the two T's. Thus, you must divide by each factorial. Therefore, 7!/(3!)(2!) = (7 x 6 x 5 x 4 x 3 x 2 x 1)/[(3 x 2 x 1) x (2 x 1)] = 7 x 5 x 4 x 3 = 420 When we have repeated items in the set, we reduce the number of arrangements.

Multiple arrangements Example: ETA fraternity must choose a delegation of three senior members and two junior members for a conference. If ETA has 12 senior members and 11 junior members, how many different delegations are possible?

This problem involves two genuinely different arrangements: three seniors chosen from a pool of 12 seniors, and two juniors chosen from a separate pool of 11 juniors. These arrangements should be calculated separately. The three spots in the delegation are not distinguishable, choosing the seniors is equivalent to choosing an anagram of three Y's and nine N's, which equals: 12!/(9!)(3!) = 220 The juniors equals to choosing an anagram of two Y's and nine N's, which equals: 11!/(9!)(2!) = 55 Since the choices are successive and independent, multiply the numbers: 220 x 55 = 12,100 different delegations are possible.

Factorials "!"

Use factorials to count arrangements. *The number of ways of putting n distinct objects in order, if there are no restrictions, is n! (n factorial).*

Arrangements with constraints

Use the Glue Method Pretend that items are stuck together and are actually one larger item.

Slot Method

Use when there are restrictions. Apply the restrictions first then make the less restricted choices. Step 1: Draw empty slots corresponding to each of the choices you have to make. Step 2: Fill in each slot with the number of options for that slot. Fill in whatever order make the most sense, picking the *most restrictive choices first.* Step 3: Follow to Fundamental Counting Principle, multiply the numbers in the slots to find the total number of nominations.

Fundamental Counting Principle E.g., Making a sandwich with: -One type of bread out of two types (rye or whole wheat) -One type of filling out of three types (chicken, peanut butter, or tuna).

We see that there are 6 possible sandwiches. You can simply multiply the number of bread choices by the number of fillings, as dictated by the fundamental counting principle. 2 breads x 3 fillings = 6 possible sandwiches

Anagram is...

a rearrangement of the letters in a word or phrase. *rarely the subject of a GMAT question.*

Combinatorics is basically

advanced counting we are often counting the number of possibilities how many different ways you can arrange things.

Many combinatorics can be solved using

an anagram grid. Remember to divide by the factorial of factorials of repeated letters in the bottom row.

When you have repeated items,

divide the total factorial by each repeat factorial to count the different arrangements.

The number of anagrams of a word is the

factorial of the total number of letters, divided by the factorial(s) corresponding to each set of repeated letters.

Anagram example: How many different anagrams are possible for the word PIZZAZZ?

≠ 7! due to repetition. The four Z's are indistinguishable from each other. If the four Z's were all different letters, then we would have 7!. In this case, Total!/Repeats! Because there are 4! = 24 ways to rearrange the four Z's in the word PIZZAZZ without changing anything. We should divide by 4! (= 24) to account for the 4 repeated Z's. Therefore, 7!/4! = (7 x 6 x 5 x 4 x 3 x 2 x 1)/(4 x 3 x 2 x 1) = 7 x 6 x 5 = 210