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

2. Copyright © Cengage Learning. All rights reserved. 12.2 Combinations

3. 3 Committee Problem

4. 4 Let us consider the following example: A = {Alfie, Bogie, Calvin, Doug, Ernie } In how many ways could they elect a committee of two persons? We call this the committee problem. One method of solution is easy (but tedious) and involves the enumeration of all possibilities: {a, b} {b, c} {c, d} {d, e} {a, c} {b, d} {c, e} {a, d} {b, e} {a, e}

5. 5 Committee Problem We see there are ten possible two-member committees. Do you see why we cannot use the fundamental counting principle for this committee problem in the same way we did for the election problem? We have presented two different types of counting problems, the election problem and the committee problem. For the election problem, we found an easy numerical method of counting, using the fundamental counting principle. For the committee problem, we did not; further investigation is necessary.

6. 6 Committee Problem Let us compare the committee and election problems. While considering the election problem, we can see that the following arrangements are different: That is, the order was important. However, in the committee problem, the three-member committees listed above are the same. In this case, the order is not important. When the order in which the objects are listed is not important, we call the list a combination and represent it as a subset of A.

7. 7 Example 1 – List all arrangements of elements Select two elements from the set A = {a, b, c, d, e}, and list all possible arrangements as well as all possible subsets. Solution:

8. 8 Committee Problem We now state a definition of a combination.

9. 9 Committee Problem The notation n C r is similar to the notation used for permutations, but since is more common in your later work in mathematics, we will usually use this notation for combinations. The notation is read as “n choose r.” The formula for the number of permutations leads directly to a formula for the number of combinations since each subset of r elements has r! permutations of its members.

10. 10 Committee Problem Thus, by the fundamental counting principle,

11. 11 Example 2 – Evaluate combinations Evaluate: a. b. c. Solution: a. b. c.

12. 12 Example 2 – Solution cont’d

13. 13 Pascal’s Triangle

14. 14 Pascal’s Triangle For many applications, if n and r are relatively small, you should notice the following relationship:

15. 15 Pascal’s Triangle That is, is the number in the nth row, rth diagonal of Pascal’s triangle.

16. 16 Example 8 – Evaluate combinations using Pascal’s triangle Find a. b. c. d. 12 C 10 Solution: a. = 56 b. = 6 c. = 3,003 d. 12 C 10 = 66 row 8, diagonal 3 row 6, diagonal 5 row 14, diagonal 6 row 12, diagonal 10

17. 17 Counting with the Binomial Theorem

18. 18 Counting with the Binomial Theorem

19. 19 Example 10 – Find the number of subsets How many subsets of {1, 2, 3, 4, 5} are there? Solution: There are subsets of 5 elements, subsets of 4 elements, and so forth. Thus, the total number of subsets is

20. 20 Example 10 – Find the number of subsets Instead of carrying out all of the arithmetic in evaluating these combinations, we note that so that if we let a = 1 and b = 1, we have

21. 21 Example 10 – Solution We see that there are 2 5 = 32 subsets of a set containing five elements. cont’d

22. 22 Counting with the Binomial Theorem

23. 23 Example 11 – Find the number of subsets of a set of 10 elements How many subsets are there for a set containing 10 elements? Solution: There are 2 10 = 1,024 subsets.