counting

# 0 1 2 n since subsets of s need to have between 0

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Unformatted text preview: with n elements. The number of subsets of S is 2n. The number of subsets with 0, 1, 2,..., n elements is given by ￿ ￿￿ ￿￿ ￿ ￿￿ n n n n , , ,..., 0 1 2 n Since subsets of S need to have between 0 and n elements, the claim of the theorem follows. 31 Binomial Theorem Theorem: Let x and y be variables. Let n be a nonnegative integer. Then n (x + y ) = n￿ ￿ k=0 ￿ n k n− k xy k Proof: Let us expand the left hand side. The terms of the product in expanded form are xkyn-k for 0 &lt;= k &lt;= n. To obtain the term xkyn-k one￿ ￿ choose k x’s from must the n (x+y) terms. There are n ways to do that. k 32 Another Binomial Coefficient Identity Corollary: Let n be a positive integer. Then ￿￿ n ￿ kn =0 (−1) k k=0 Proof: We have 0 = 0n = (-1+1)n. Expanding the right hand side with the help of the binomial theorem, we obtain the claim. [This also says that the number of subsets with an even number of elements is equal to the number of subsets with an odd number of elements.] 33 Pascal’s Identity (1) Theorem: Let n and k be positive integers with n&gt;=k. Then ￿ n+1 k ￿ = ￿ ￿ ￿￿ n n + k−1 k Proof: We are going to prove this by counting the number of subsets with k elements of a set T with n+1 elements in two different ways. First way of counting: The set T clearly contains ￿ n+1 k 34 ￿ subsets of size k. Pascal’s Identity (2) Second way of counting: Recall that T is a set with n+1 elements. Let us consider an element t of T. We will count the subsets of T of size k that (a) contain the element t, and (b) do not contain the element t. (a) There are ￿ ￿ n k−1 subsets of T that contain t, since t is already chosen, but the remaining k-1 elements need to be chosen from T-{t}, a set of size n. (b) There are ￿￿ n k subset of T not containing t, since one can choose any k elements from the set T-{t} with n elements. Since the two cases are exhaustive, C(n+1,k) = C(n,k-1) + C(n,k). Pascal’s Identity (3) The first way of counting yields t...
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