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Unformatted text preview: Illinois Institute of Technology Department of Computer Science Lectures 7–8: February 13–18, 2008 CS 330 Discrete Mathematics Spring Semester, 2008 1 Another combinatorial identity Let us examine the identity µ n k ¶µ k r ¶ = µ n r ¶µ n r k r ¶ (1) An algebraic proof is simple, but the combinatorial proof is more interesting. As before, we demonstrate that the combinatorial problem solved by the expression on the lefthand side of the equal sign and the combinatorial problem solved by the expression on the righthand side of the equal sign are actually the same problem approached in two different fashions. The lefthand side of (1) counts the number of possible outcomes of a two stage selection process of a set R of r elements from n . First, a subset K ⊆ R of k elements is chosen and then r of these k are selected to form R . The rule of product says that this choice of R can happen in ( n k )( k r ) ways. We can count the number of ways the same event can happen by first directly choosing r of the n elements as R (this can be done in ( n r ) ways) and then choosing from the other n r elements the remaining k r elements which when added to R form K (this can be done in ( n r k r ) ways). By the rule of product the compound event can occur in ( n r )( n r k r ) ways. Since the compound event is the same in both of these applications of the rule of product, our proof is complete. 2 Vandermonde’s identity In the identities established so far, the algebraic proof has been very easy, almost eliminating the need for a combinatorial proof. We now present two identities for which the combinatorial proof is relatively simple and direct, but algebraic verification is not. Vandermonde’s identity states that µ n + m k ¶ = k X i =0 µ n i ¶µ m k i ¶ = µ n ¶µ m k ¶ + µ n 1 ¶µ m k 1 ¶ + ··· + µ n k ¶µ m ¶ (2) The lefthand side of (2) is the number of ways to select a subcommittee of k people from a committee of n men and m women. On the other hand, the righthand side counts the number of outcomes for the same problem: If the subcommittee is to have i men and k i women, the rule of product says that it can be chosen in ( n i )( m k i ) ways. By the rule of sum we must add this value for i = 0 , 1 , . . ., k to count the number of ways the subcommittee can be chosen. This proves Vandermonde’s identity. A similar identity states µ n + m n ¶ = n X i =0 µ n i ¶µ m i ¶ CS 330—Spring, 2008 2 Lectures 7–8: February 13–18, 2008 = µ n ¶µ m ¶ + µ n 1 ¶µ m 1 ¶ + ··· + µ n n ¶µ m n ¶ (3) Note that, by our convention that ( i k ) = 0 for k > i , if n > m , the last n m terms of the sum will be zero. Again, an algebraic proof is not as simple as a combinatorial one here. The lefthand side of (3) is the number of ways to select a subcommittee of n people from a committee of n men and m women. The selection of such a committee can also be done, however, by first choosing i , 0 ≤ i ≤ n , to be the number...
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 Spring '08
 Reingold,EdwardM.
 Computer Science, Binomial Theorem, n1 n1 n1, Vandermonde

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