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Unformatted text preview: Counting III Margaret M. Fleck 12 April 2010 This lecture covers the rest of section 5.4 of Rosen, including combinato- rial proof, plus some material from section 5.5. 1 Overview We’ll see a bunch of combinatorial formulas, as well as some extended cases of counting combinations and permutations. For all of them, there is a good picture or construction that can help you remember the formula, or recon- struct it if necessary. These formulas would not be so pleasant to memorize blindly. 2 Combinatorial proofs There are a large number of useful identities involving binomial coefficients. A very simple example is: Claim 1 ( n k ) = ( n n- k ) There are two ways to prove this. First, we could convert both sides to expressions involving factorials and show that they are equal. 1 Proof: ( n n ) = n ! k !( n- k )! Also, ( n n- k ) = n ! ( n- k )!( n- ( n- k ))! = n ! ( n- k )! k ! So ( n k ) = ( n n- k ) Alternatively, you can prove this by choosing two sets and discussing why the two numbers of combinations must be the same. Suppose we have a set S containing n elements. We can match up the subsets containing k elements with the subsets containing n- k elements. Specifically, each subset A of S is matched to S- A . This is a bijection, so there are the same number of subsets of size k as subsets of size n- k . So C ( n, k ) and C ( n, n- k ) must be equal. That is, ( n k ) = ( n n- k ) square The second proof technique is called a “combinatorial proof.” 3 Vandermonde’s Identity Here’s another useful identity: Claim 2 (Vandermonde’s Identity) ( n + m r ) = ∑ r k =0 ( n k )( m r- k ) We can prove this using a combinatorial argument, as follows: Proof: Suppose that we have a set A with n elements and a set B with m elements, where A and B don’t overlap. If we want to choose r elements from A ∪ B , we will have to pick some number k from A and then the remaining r- k from B ....
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- Spring '10
- ways, combinatorial proof