Lecture on Permutation Combination

# Theobjectsmaybeeitherdifferentfromeachother

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Unformatted text preview: ix cookies be chosen? Solution: The number of ways to choose six cookies is the number of 6‐combinations of a set with four elements. By Theorem 2 is the number of ways to choose six cookies from the four kinds. 29 Summarizing the Formulas for Counting Permutations and Combinations with and without Repetition 30 Permutations with Indistinguishable Objects Example: How many different strings can be made by reordering the letters of the word SUCCESS. Solution: There are seven possible positions for the three Ss, two Cs, one U, and one E. The three Ss can be placed in C(7,3) different ways, leaving four positions free. The two Cs can be placed in C(4,2) different ways, leaving two positions free. The U can be placed in C(2,1) different ways, leaving one position free. The E can be placed in C(1,1) way. By the product rule, the number of different strings is: The reasoning can be generalized to the following theorem. 31 Permutations with Indistinguishable Objects Theorem 3: The number of different permutations of n objects, where there are n1 indistinguishable objects of type 1, n2 indistinguishable objects of type 2, …., and nk indistinguishable objects of type k, is: Proof: By the product rule the total number of permutations is: C(n, n1 ) C(n n1, n2 ) ··· C(n n1 n2 ··· nk, nk) since: The n1 objects of type one can be placed in the n positions in C(n, n1 ) ways, leaving n n1 positions. Then the n2 objects of type two can be place...
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## This document was uploaded on 03/29/2014 for the course COT 3100h at University of Central Florida.

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