feb13-18

# feb13-18 - Illinois Institute of Technology Department of...

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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 left-hand side of the equal sign and the combinatorial problem solved by the expression on the right-hand side of the equal sign are actually the same problem approached in two different fashions. The left-hand 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 left-hand 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 right-hand 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 left-hand 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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feb13-18 - Illinois Institute of Technology Department of...

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