Then r balls distributed in the n bins determine an r

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Unformatted text preview: Proof. Let S be a set with n elements. Order the elements of this set. We can specify an r-multisubset T of S by specifying the multiplicity of the first element, the multiplicity of the second element, and so on. 43 Proof (continued) Let us use n bins and mi balls in the ith bin to specify that the ith element of S occurs with multiplicity mi in T . Then r balls distributed in the n bins determine an r-multisubset of S . Therefore, it suffices to count the number of ways we can distribute r balls into n bins. This number corresponds to the number of strings consisting of r bullets • (representing the balls), and n − 1 bars | (representing the walls between the bins. Since the positions of the r balls among the n − 1 + r balls and bars suffice to ￿ specify￿such a string, we can n−1+r conclude that there are strings consisting of r n − 1 bars and r balls. By construction, this coincides with the number multisubsets with r elements of a set with n elements, so the theorem is proved. Example Let S = {a, b, c, d} be a set with four elements. Then the number of 3-combinations with repetitions from S is given by ￿￿ ￿￿ ￿ ￿ 4 4+3−1 = = 20 3 3 45 Permutations with Repetition Let S be a set with n elements. An r-permutation with repetition of S is an arrangement of r elements of S with repetitions allowed. Theorem: The number of r-permutations with repetition of a set with n elements is given by nr. Proof. This follows directly from the product rule. 46 Permutations under Indistinguishability Constraints 47 Motivation So far, we assumed that (a) the elements are clearly distinguishable and (b) each element is chosen at most once for a permutation and combinations or (b’) that each element can be chosen repeatedly in permutations and combinations. Now we will allow that (a’) the elements from which we are allowed to choose are not necessarily distinguishable. [For example, we could choose from a multiset of m red balls and n blue balls.] 48 Permutations with Indistinguishable Objects In how many ways can we reorder SUCCESS? We can choose 3 locations for the Ss in C(7,3) ways. Then we can choose 1 location for the U in C(4,1) way. Then we can choose 2 locations for the Cs in C(3,2) ways. Finally, we can choose 1 location for the E in C(1,1) way. C(7,3)C(4,1)C(3,2)C(1,1)= 7! 4! 3! 1! 3!4! 1!3! 2!1! 1!0! 49 = 7! 3!2!1!1! = 420 Permuting Indistinguishable Objects Theorem. Suppose we have n different objects that are of k different types, where the objects of the same type are indistinguishable. If there are ni indistinguishable objects of type i for 1 ≤ i ≤ k , then the total number of permutations is n! . n 1 ! n 2 ! · · · nk ! 50...
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