35 Permutations, Combinations and Proba bility（35排列,组合Proba能力）.pdf

35 Permutations, Combinations and Proba- bility Thus far we have been able to list the elements of a sample space by drawing a tree diagram. For large sample spaces tree diagrams become very complex to construct. In this section we discuss counting techniques for finding the number of elements of a sample space or an event without having to list them. Permutations Consider the following problem: In how many ways can 8 horses finish in a race (assuming there are no ties)? We can look at this problem as a decision consisting of 8 steps. The first step is the possibility of a horse to finish first in the race, the second step the horse finishes second, ... , the 8th step the horse finishes 8th in the race. Thus, by the Fundamental Principle of counting there are 8 · 7 · 6 · 5 · 4 · 3 · 2 · 1 = 40, 320 ways This problem exhibits an example of an ordered arrangement, that is, the order the objects are arranged is important. Such ordered arrangement is called a permutation. Products such as 8 · 7 · 6 · 5 · 4 · 3 · 2 · 1 can be written in a shorthand notation called factoriel. That is, 8 · 7 · 6 · 5 · 4 · 3 · 2 · 1 = 8! (read ”8 factoriel”). In general, we define n factoriel by n! = n(n − 1)(n − 2) · · · 3 · 2 · 1, if n ≥ 1 1, if n = 0. where n is a whole number n. Example 35.1 Evaluate the following expressions: 10! (a) 6! (b) 7! . Solution. (a) 6! = 6 · 5 · 4 · 3 · 2 · 1 = 720 (b) 10! = 10·9·8·7·6·5·4·3·2·1 = 10 · 9 · 8 = 720 7! 7·6·5·4·3·2·1 Using factoriels we see that the number of permutations of n objects is n!. 1 Example 35.2 There are 6! permutations of the 6 letters of the word ”square.” In how many of them is r the second letter? Solution. Let