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PS03-solutions

# PS03-solutions - MATH 681 Problem Set#3 1(10 points Prove...

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MATH 681 Problem Set #3 1. (10 points) Prove the following combinatorial identities: (a) (5 points) Recall that S ( n, k ) is equal to the number of ways to subdivide an n -element set into k nonempty parts. Produce a combinatorial argument to show that S ( n, k ) = kS ( n - 1 , k ) + S ( n - 1 , k - 1) . Let us consider the ways to divide { 1 , 2 , . . . , n } into k nonempty sets. There are two possibilities to be addressed: either n is in a set by itself or with other numbers. If n is in a set by itself, then removal of this set { n } leaves us with a partition of the numbers { 1 , 2 , . . . , n - 1 } into k - 1 sets. There are S ( n - 1 , k - 1) such partitions, and each of them can be identified uniquely with a k -partition of { 1 , 2 , . . . , n - 1 } by adding in the extra set { n } . On the other hand, if n is in a set with other numbers, removal of n from this set leaves k sets behind partitioning the numbers { 1 , 2 , . . . , n - 1 } . There are S ( n - 1 , k ) such possible partitions, but they are not uniquely identified with k -partitions of { 1 , 2 , . . . , n - 1 } , since the term n could have been removed from (and can be reinserted into) any of the k sets. Thus, there are k different ways to build a k -partition of { 1 , 2 , . . . , n } from a k -partition of { 1 , 2 , . . . , n - 1 } , so this possibility accounts for k · S ( n - 1 , k ) partitions. Assembling these two cases, we see that S ( n, k ) = S ( n - 1 , k - 1) + kS ( n - 1 , k ). (b) (5 points) Prove that for any m < n , m k =0 ( n k,m - k,n - m ) = 2 m ( n m ) . This is actually very much like problem #1 from the previous problem set; it’s actually a generalization, since that problem is equivalent to the m = n - 1 case of this one. The left side is fairly easy to interpret: it represents the number of ways to divide an n -element set into three distinct parts of size k , m - k , and n - m , for any value of k . So, more properly, it represents the number of ways to divide n elements into classes A, B, and C so that classes A and B together contain m elements, and class C has n - m . The right side can be interpreted as enumerating the same thing, in a different way. We start by selecting the membership of classes A and B together; we want to have m members, so there are ( n m ) ways to select these, and the remaining n - m elements are consigned to class C. Now, we want to freely distribute these m

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