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3 Certain classes of counting problems come up frequently, and it is useful to develop formulas to deal with them.

4 Example 1 – Casting Ms. Birkitt, the English teacher at Brakpan Girls High School, wanted to stage a production of R. B. Sheridans play, The School for Scandal. The casting was going well until she was left with five unfilled characters and five seniors who were yet to be assigned roles. The characters were Lady Sneerwell, Lady Teazle, Mrs. Candour, Maria, and Snake; while the unassigned seniors were April, May, June, Julia and Augusta. How many possible assignments are there?

5 Example 1 – Solution To decide on a specific assignment, we use the following algorithm: Step 1: Choose a senior to play Lady Sneerwell; 5 choices. Step 2: Choose one of the remaining seniors to play Lady Teazle; 4 choices. Step 3: Choose one of the now remaining seniors to play Mrs. Candour; 3 choices.

6 Example 1 – Solution Step 4: Choose one of the now remaining seniors to play Maria; 2 choices. Step 5: Choose the remaining senior to play Snake; 1 choice. Thus, there are 5 4 3 2 1 = 120 possible assignments of seniors to roles. contd

7 Permutations and Combinations What the situation in Example 1 has in common with many others is that we start with a sethere the set of seniors and we want to know how many ways we can put the elements of that set in order in a list. In this example, an ordered list of the five seniorssay, 1. May 2. Augusta 3. June

8 Permutations and Combinations 4. Julia 5. April corresponds to a particular casting: Cast Lady Sneerwell May Lady Teazle Augusta Mrs. Candour June Maria Julia Snake April

9 Permutations and Combinations We call an ordered list of items a permutation of those items. If we have n items, how many permutations of those items are possible? We can use a decision algorithm similar to the one we used in the Example 1 to select a permutation.

10 Permutations and Combinations Step 1: Select the first item; n choices. Step 2: Select the second item; n – 1 choices. Step 3: Select the third item; n – 2 choices.... Step n – 1: Select the next-to-last item; 2 choices. Step n: Select the last item; 1 choice. Thus, there are n (n – 1) (n – 2)... 2 1 possible permutations. We call this number n factorial, which we write as n!

11 Permutations and Combinations Permutations A permutation of n items is an ordered list of those items. The number of possible permutations of n items is given by n factorial, which is n! = n (n – 1) (n – 2)... 2 1 for n a positive integer, and 0! = 1.

12 Permutations and Combinations Visualizing Permutations Permutations of 3 colors in a flag:

13 Permutations and Combinations Quick Example The number of permutations of five items is 5! = 5 4 3 2 1 = 120. Sometimes, instead of constructing an ordered list of all the items of a set, we might want to construct a list of only some of the items. So, we can generalize our definition of permutation to allow for the case in which we use only some of the items, not all.

14 Permutations and Combinations Check that, if r = n below, this is the same definition we give for permutation of n items. Permutations of n items taken r at a time A permutation of n items taken r at a time is an ordered list of r items chosen from a set of n items. The number of permutations of n items taken r at a time is given by P(n, r ) = n (n – 1) (n – 2)... (n – r + 1) (there are r terms multiplied together). We can also write

15 Permutations and Combinations Quick Example The number of permutations of six items taken two at a time is P(6, 2) = 6 5 = 30 which we could also calculate as

16 Permutations and Combinations For ordered lists we used the word permutation; for unordered sets we use the word combination. Permutations and Combinations A permutation of n items taken r at a time is an ordered list of r items chosen from n. A combination of n items taken r at a time is an unordered set of r items chosen from n. Visualizing

17 Permutations and Combinations Note Because lists are usually understood to be ordered, when we refer to a list of items, we will always mean an ordered list. Similarly, because sets are understood to be unordered, when we refer to a set of items we will always mean an unordered set. In short: Lists are ordered. Sets are unordered.

18 Permutations and Combinations Quick Example There are six permutations of the three letters a, b, c taken two at a time: 1. a, b; 2. b, a; 3. a, c; 4. c, a; 5. b, c; 6. c, b. There are three combinations of the three letters a, b, c taken two at a time: 1. {a, b}; 2. {a, c}; 3. {b, c}. There are six lists containing two of the letters a, b, c. There are three sets containing two of the letters a, b, c.

19 Permutations and Combinations How do we count the number of possible combinations of n items taken r at a time? The number of permutations is P(n, r ), but each set of r items occurs r ! times because this is the number of ways in which those r items can be ordered. So, the number of combinations is P(n, r )/r !.

20 Permutations and Combinations Combinations of n items taken r at a time The number of combinations of n items taken r at a time is given by We can also write

21 Permutations and Combinations Quick Example The number of combinations of six items taken two at a time is which we can also calculate as

22 Permutations and Combinations Note There are other common notations for C(n, r ). Calculators often have n C r. In mathematics we often write which is also known as a binomial coefficient. Because C(n, r ) is the number of ways of choosing a set of r items from n, it is often read n choose r.