724 CHAPTER 11 Probability principle. For the first position, there are three choices. There are then two choices for the second position, and only one choice is left for the third position. Number of permutations 3 2 1 6 = ⋅ ⋅ = The product 3 2 1 ⋅ ⋅ is referred to as 3 factorial, and is written 3!. Thus, 3! 3 2 1 6 = ⋅ ⋅ = Number of Permutations The number of permutations of n distinct items is n factorial, symbolized n!, where n n n n ! ( 1)( 2) (3) (2) (1) = − − It is important to note that 0! is defined to be 1. Many calculators have the ability to determine factorials. Often to determine factorials you need to press the 2nd or INV key. If necessary, do an internet search using your model of calculator to determine how to calculate factorials on your calculator. Example 4 Dogs on a Poster Sumiko works at an animal shelter that has seven dogs available for adoption. Sumiko is preparing a poster that will feature a photograph of each of the seven dogs. a) How many different arrangements of the seven photographs are possible? b) If one of these arrangements of photographs is randomly selected, what is the probability that the photographs are in alphabetical order by the dogs’ names? Assume the dogs all have different names. Solution a) Since there are seven different photographs, the number of permutations is 7! 7! 7654321 5040 = ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ = The seven photographs can be arranged in 5040 different ways. b) Since the dogs all have different names, there is only one of 5040 arrangements of photographs in which the dogs’ names are in alphabetical order. Therefore, P(alphabetical order) 1 5040 . = 7 Now try Exercise 33 Example 5 illustrates how to use the fundamental counting principle to determine the number of permutations possible when only a part of the total number of items is to be selected and arranged. Example 5 Permutations of Three of Five Letters Consider the five letters a b c d e , , , , . In how many distinct ways can three letters be selected and arranged if repetition is not allowed? Solution We are asked to select and arrange only three of the five possible letters. Fig. 11.20 shows some possibilities. Using the fundamental counting principle, we determine that there are five possible letters for the first choice, four possible letters for the second choice, and three possible letters for the third choice: 5 4 3 60 ⋅ ⋅ = Thus, there are 60 different possible ordered arrangements, or permutations. On the left, we show 5 of the 60 possible permutations. 7 Now try Exercise 41 b a d d a b e c a c d e a b c Some of the many permutations of 3 of the 5 letters Figure 11.20
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