Ch. 15 Combinations and Permutations
15.2.8 Combinations with Restrictions
15.2.8 Some Items Must Be Chosen In combination problems where it is already decided that items MUST be chosen, simply remove those items from the overall set and remove them from the number of spots considered as well. Then perform combination formula as usual. 15.2.9 Some Items Must NOT Be Chosen Similar to the strategy above, except that you do not remove the number of available spots to be chosen.
15.3 Permutations
A permutation is a selection of objects in which the order of the objects matters. For example, if we wanted to determine the number of ways in which the letters A, B, C could be arranged in a line, we would have a permutation problem. Question: In a competition involving ten competitors, one Gold, one Silver, and one Bronze medal will be awarded. How many different medal arrangements are possible? Question: To open a digital combination lock, a student must correctly enter a four-digit number sequence on a ten-digit keypad. How many different four-digit number sequences are possible? Question: A librarian must arrange four books on a bookshelf. How many sequences of books are possible? Question: In how many different ways can the letters of the word "love" be arranged?
15.3.9 Some items can't be next to each other
Advice: if you cannot remember how to do this, calculate the total number of all permutations and remember that the answer must be some number smaller than that. Most likely the next lowest in the group. Remember, total number with restriction = total possible - total of just the restricted elements.
15.3.8 When some items must be together, link those items together
Another helpful technique is to link together any items that can be considered one unit. Notice, in the next example, how John and Sara can form a single unit. Also pay close attention to the way in which this alters how we visualize the number of items being arranged. In how many ways can Abby, John, Sara, Mark and Todd be arranged to form a line if John and Sara must always stand next to each other? Simply consider John and abby as one unit, and solve like a normal permutation. Then, since Abby/John can also be arranged as John/Abby, you need to multiply the result by the ! number of potential combinations of that unit (2!).
15.2.5 Choosing multiple items from multiple groups using the word "and"
Choosing multiple items from multiple groups using the word "and" Apply combination method to each individual choice scenario. Multiply each of the outcomes together. Mutually Exclusive Events "or" Two or more events are mutually exclusive if they cannot occur together. If there are x ways to accomplish event A and y ways to accomplish event B and if A and B are mutually exclusive, then there are x + y ways to accomplish A or B. Essentially, look for the word "or" to clue you into mutually exclusive events. If mutually exclusive, then add the resulting combinations together rather than multiply.
15.1 Introduction
Combinations Combinations are used when the order in which a task is completed does not matter. Order is not important! Combination problems ask one to count the number of possible ways in which a task can be performed when the order in which the task is accomplished does not matter. Permutations Permutations, in contrast, are used when the order in which a task is completed does matter. Order is important!
15.2.14 Dependent combinations
Example #17, 15.2.14: A manager must assign nine out of ten workers to three different teams. If each team is composed of three workers and no worker is allowed to be on more than one team, in how many different ways can the three teams be formed? Essentially, run combinations on team 1 only --> 10C3, then use the unchosen workers for group 2 --> 7C3, then finally the last group --> 4C3. Multiply each of the combinations together and divide by the number of groups !
15.4 Creating codes
For example, let's assume we have to create a three-digit code to label some products to be sold. Let's say that these codes must be created from only the 26 letters of the alphabet and that any of the letters may be repeated. For example, a code could be ABC, or XXZ, or RST. If we had 5,000 different products, would we be able to give each product a unique letter code? Thus, 26 × 26 × 26 = 263 codes are possible. Notice that since 263 > 203 = 20 × 20 × 20 = 8,000, the 263 codes will be more than enough for 5,000 products.
15.2.4 The fundamental counting principle
If there are m ways to perform task 1 and n ways to perform task 2 and the tasks are independent, then there are m × n ways to perform both of the tasks together. For more than 2 tasks, just multiply them all together. # of task 1 options x # of task 2 options x # of task 3 options...
15.3.4 The permutation formula for indistinguishable items
In permutation problems, count only the number of distinguishable permutations. P=N!/(r1!)×(r2!)×(r3!)×.... Where r! is the number of times a specific repeated item is in the set. Complete this for all items which repeat.
15.2.12 Some items can never be together in the same subgroup
In problems where items cannot be together in a group, remember that: The Number of 3-Person Club Configurations in which Both People Are Not Together in the Club = The Total Number of 3-Person Club Configurations That Can be Made - The Number of 3-Person Club Configurations in which Both People Are Together in the Club Total considering exclusions = Total considerations - total of exclusions specifically
15.2.2 Solving combination problems
Method 1: The Basic Combination Formula Combination formula = n!/k!(n-k)! n = number of objects from which we will choose from. k = the actual number we choose.
15.3.2 Solving permutation problems
Method 1: The Basic Permutation Formula Permutation = nPk = n!/(n-k)! n = number of objects from which a choice can be made k = number of objects that are to be chosen.
15.3.5 Circular arrangements
The formula to compute the number of ways in which items can be arranged in a circle is known as the circular permutation formula: # of ways to arrange a set of items in a circle = (K - 1)! Where K represents the number of items to be arranged in that circle.
15.3.6 Arrangements with restrictions examples
The most basic permutation problems ask students to order items without asking any additional restrictions, but more challenging permutation problems tend to add layers of restrictions or stipulations to the order. Question: In how many ways can three men and three women be ordered to form a line if the men must be placed only in the odd positions of the line? Question: In how many ways can four consonants and one vowel be arranged if the vowel must always be first? Question: In how many ways can Adam, Bob, Candy, Doug, Elli, and Fritz be arranged to form a line if Adam and Bob refuse to stand next to each other? The Anchor Method When using the anchor method, first place the restricted items into their specific spots in the arrangement, then handle the additional items that can be arranged in the remaining spots. For instance, if you know Jack is always in the front of the line out of 5 people. Simply ignore Jack (as he is a constant for all permutations) and perform the permutation on the remaining 4 only.
15.2.7 Choosing "at least" some number of items
The phrase "at least" means equal to or greater than. "At least" problems typically involve the addition of outcomes. Essentially, run combinations analysis for each possible event from the lowest value to the highest value and add the results.
15.2.11 Collectively exhaustive events
Two events are collectively exhaustive if together, those events represent all of the possible outcomes of a scenario. Let's assume that on a certain day, there are only two states of weather: rain and snow. Let's also assume that the two can never coexist at the same time. The two states of rain and snow are thus mutually exclusive because they cannot occur together and are collectively exhaustive since together the two states represent all of the possible weather outcomes. Original total = total including + total excluding
