THE NATURE OF COUNTING Copyright © Cengage Learning. All rights reserved. 12

Copyright © Cengage Learning. All rights reserved Permutations

3 Election Problem

4 Consider a club with five members: A = {Alfie, Bogie, Calvin, Doug, Ernie} In how many ways could they elect a president and secretary? We call this the election problem. There are several ways to solve this problem.

5 Election Problem The first, and perhaps the easiest, is to make a picture, known as a tree diagram, listing all possibilities (see Figure 12.1). Tree diagram for five choices Figure 12.1

6 Election Problem We see there are 20 possibilities for choosing a president and a secretary from five club members. This method is effective for “small” tree diagrams, but the technique quickly gets out of hand. For example, if we wished to see how many ways the club could elect a president, secretary, and treasurer, this technique would be very lengthy. A second method of solution is to use boxes or “pigeonholes” representing each choice separately.

7 Election Problem Here we determine the number of ways of choosing a president, and then the number of ways of choosing a secretary. We multiply the numbers in the pigeonholes, 5  4 = 20 and we see that the result is the same as from the tree diagram.

8 Election Problem A third method is to use the fundamental counting principle.

9 Example 1 – Determine the number of ways of choosing three In how many ways could the given club choose a president, secretary, and treasurer? Solution: The answer is 5  4  3 = 60.

10 Election Problem When set symbols { } are used, the order in which the elements are listed is not important. Suppose now that you wish to select elements from A = {a, b, c, d, e} by picking them in a certain order. The selected elements are enclosed in parentheses to signify order and are called an arrangement of elements of A. For example, if the elements a and b are selected from A, then there are two pairs, or arrangements: (a, b) and (b, a)

11 Election Problem These arrangements are called ordered pairs. Remember, when parentheses are used, the order in which the elements are listed is important. This example shows ordered pairs, but you could also select an ordered triple such as (d, c, a). These arrangements are said to be selected without repetition, since a symbol cannot be used twice in the same arrangement.

12 Election Problem If we are given some set of elements and we are to choose without repetition a certain number of elements from the set, we can choose them so that the order in which the choices are made is important, or so that it is not. If the order is important, we call our result a permutation, and if the order is not important, it is called a combination. We now consider permutations. Suppose we consider the election problem for a larger set than A; we wish to elect a president, secretary, and treasurer from among {Frank, George, Hans, Iris, Jane, Karl}.

13 Election Problem We could use one of the previously mentioned three methods, but we wish to generalize, so we ask, “Is it possible to count the number of arrangements without actually counting them?” In answering this question, we note that we are selecting 3 persons in order from a group of 6 people. If an arbitrary finite set S has n elements and r elements are selected without repetition from S (where r  n ), then an arrangement of the r selected elements is called a permutation.

14 Election Problem

15 Example 3 – Number of permutations of two chosen from six How many permutations of two elements can be selected from a set of six elements? Solution: Let B = {a, b, c, d, e, f} and select two elements: (a, b), (a, c), (a, d), (a, e), (a, f), (b, a), (b, c), (b, d), (b, e), (b, f), (c, a), (c, b), (c, d), (c, e), (c, f), (d, a), (d, b), (d, c), (d, e), (d, f), (e, a), (e, b), (e, c), (e, d), (e, f), (f, a), (f, b), (f, c), (f, d), (f, e) There are 30 permutations of two elements selected from a set of six elements.

16 Election Problem

17 Example 4 – Evaluate permutations Evaluate: a. 52 P 2 b. 7 P 3 c. 10 P 4 d. 10 P 1 e. n P r

18 Example 4 – Solution We use the fundamental counting principle: a. 52 P 2 = 52  51 = 2,652 b. 7 P 3 = 7  6  5 = 210 c. 10 P 4 = 10  9  8  7 = 5,040 d. 10 P 1 = 10

19 e. n P r = n(n – 1)(n – 2)………(n – r + 1) cont’d Example 4 – Solution

20 Factorial

21 Factorial In our work with permutations, we will frequently encounter products such as 6  5  4  3  2  1 or 10  9  8  7  6  5  4  3  2  1 or 52  51  50  49 ...  4  3  2  1 Since these are rather lengthy, we use factorial notation.

22 Example 6 – Evaluate factorials Evaluate factorials for the numbers 0, 1, …….,10. Solution: 0! = 1 1! = 1 2! = 2  1 = 2 3! = 3  2  1 = 6 4! = 4  3  2  1 = 24 5! = 5  4  3  2  1 = 120 6! = 6  5  4  3  2  1 = 720

23 Example 6 – Solution 7! = 7  6  5  4  3  2  1 = 5,040 8! = 8  7  6  5  4  3  2  1 = 40,320 9! = 9  8  7  6  5  4  3  2  1 = 362,880 10! = 10  9  8  7  6  5  4  3  2  1 = 3,628,800 cont’d

24 Factorial Notice (from Example 6) that these numbers get big pretty fast. We need not wonder why the notation “!” was chosen! For example, 52!   If you actually carry out the calculations in Example 6, you will discover a useful property of factorials: 3! = 3  2!, 4! = 4  3!, 5! = 5  4!; that is, to calculate 11! you would not need to “start over” but simply multiply 11 by the answer found for 10!.

25 Factorial This property is called the multiplication property of factorials or the count-down property.

26 Example 7 – Evaluate factorial expressions Evaluate: a. 6! – 4! b. (6 – 4)! Solution: a. 6! – 4! = 720 – 24 = 696 b. (6 – 4)! = 2! = 2 c. Compare order of operations in parts a and b. Count-down property

27 Example 7 –Solution d. = 12  11  10 = 1,320 e. = 9,900 cont’d Repeated use of the count-down property This example shows that you need the count-down property even if you are using a calculator.

28 Factorial Using factorials, we can find a formula for n P r. n P r = n(n – 1)(n – 2)………(n – r + 1) = n(n – 1)(n – 2)………(n – r + 1)  = This is the general formula for n P r.

29 Factorial

30 Example 8 – Evaluate permutations by formula Evaluate: a. 10 P 2 b. n P 0 c. 8 P 8 Solution: a. 10 P 2 = = 90 b. n P 0 = = 1 c. 8 P 8 = = 8! = 40,320

31 Distinguishable Permutations

32 Distinguishable Permutations We now consider a generalization of permutations in which one or more of the selected items is indistinguishable from the others.

33 Example 12 – Find the number of distinguishable permutations Find the number of arrangements of letters in the given word: a. MATH b. HATH Solution: a. This is a permutation of four objects taken four at a time: 4 P 4 = 4! = 24 b. This is different from part a because not all the letters in the word HATH are distinguishable; that is, the first and last letters are both H’s, so they are indistinguishable.

34 Example 12 – Solution If we make the letters distinguishable by labeling them as HATH, we see that now the list is the same as in part a. Let us list the possibilities as shown below. cont’d

35 We see that there are 24 possibilities, but notice how we have arranged this listing. If we consider H = H (that is, the H’s are indistinguishable), we see that the first and the second columns are the same. Since there are two indistinguishable letters, we divide the total by 2: cont’d Example 12 – Solution

36 Distinguishable Permutations