This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: 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 nelement 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 kpartition 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 kpartitions 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 kpartition of { 1 , 2 ,...,n } from a kpartition 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; its 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 nelement 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 ....
View
Full
Document
 Fall '09
 WILDSTROM
 Math

Click to edit the document details