922 CHAPTER 13 Counting and Probability 3 Solve Counting Problems Using Permutations Involving n Nondistinct Objects Forming Different Words How many different words (meaningful or not) can be formed using all the letters in the word REARRANGE? Solution EXAMPLE 10 Each word formed will have 9 letters: 3 R’s, 2 A’s, 2 E’s, 1 N, and 1 G.To construct each word, we need to fill in 9 positions with the 9 letters: 1 2 3 4 5 6 7 8 9 The process of forming a word consists of five tasks. Task 1: Choose the positions for the 3 R’s. Task 2: Choose the positions for the 2 A’s. Task 3: Choose the positions for the 2 E’s. Task 4: Choose the position for the 1 N. Task 5: Choose the position for the 1 G. Task 1 can be done in ( ) C 9, 3 ways. There then remain 6 positions to be filled, so Task 2 can be done in ( ) C 6, 2 ways. There remain 4 positions to be filled, so Task 3 can be done in ( ) C 4, 2 ways. There remain 2 positions to be filled, so Task 4 can be done in ( ) C 2, 1 ways. The last position can be filled in ( ) C 1, 1 way. Using the Multiplication Principle, the number of possible words that can be formed is C C C C C 9, 3 6, 2 4, 2 2, 1 1, 1 9! 3! 6! 6! 2! 4! 4! 2! 2! 2! 1! 1! 1! 0! 1! 9! 3! 2! 2! 1! 1! 15,120 ( )( )( )()() ⋅ ⋅ ⋅ ⋅ = ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ = ⋅ ⋅ ⋅ ⋅ = 15,120 possible words can be formed. The form of the expression before the answer to Example 10 is suggestive of a general result. Had all the letters in REARRANGE been different, there would have been ( ) = P 9, 9 9! possible words formed. This is the numerator of the answer. The presence of 3 R’s, 2 A’s, and 2 E’s reduces the number of different words, as the entries in the denominator illustrate. This leads to the following result: THEOREM Permutations Involving n Objects That Are Not Distinct The number of permutations of n objects of which n1 are of one kind, n2 are of a second kind, ,… and nk are of a kth kind is given by … n n n n ! ! ! !k 1 2 ⋅ ⋅ ⋅ (3) where = + + + n n n n .k 1 2 Arranging Flags How many different vertical arrangements are there of 8 flags if 4 are white, 3 are blue, and 1 is red? Solution EXAMPLE 11 We seek the number of permutations of 8 objects, of which 4 are of one kind, 3 are of a second kind, and 1 is of a third kind. Using formula (3), we find that there are 8! 4! 3! 1! 8 7 6 5 4! 4! 3! 1! 280 different arrangements ⋅ ⋅ = ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ = Now Work PROBLEM 51
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