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Unformatted text preview: Math 146: Algebraic Combinatorics Homework 4 Solutions 2.7 Let f ( n , k , h ) be the number of ordered representations of n as a sum of exactly k integers, each of which is h . Find n f ( n , k , h ) x n . Solution. First, consider ( x h + x h + 1 + x h + 2 + . . . ) k = r h x r ! k . An ordered representation of n can be viewed as the coefficient of x n in the expansion of the product above. This is because any way of obtaining an x n above as a product x n = x m 1 x m 2 x m k yeilds a sum n = m 1 + m 2 + + m k . Thus f ( n , k , h ) = [ x n ] ( r h x r ) k . 2.26 Derive ( 2 . 39 ) from ( 2 . 38 ) . That is, show that n k = ( 1 ) k n + k 1 k . Solution. To prove the identity above: n k = ( n )( n 1 ) ( n k + 1 ) k ! = ( 1 ) k n ( n + 1 )( n + 2 ) ( n + k 1 ) k ! = ( 1 ) k n + k 1 k . Equation ( 2 . 38 ) is ( 1 + x ) = k ( k ) x k and ( 2 . 39 ) is 1 ( 1 x ) n + 1 = k ( n + k k ) x k . Derive ( 2 . 39 ) from ( 2 . 38 ) by letting = n and replacing x with x . 2.27 Let D ( n ) be the number of derangements of n letters, discussed in Example 2.18. (b) Prove, by any method, that D ( n + 1 ) = ( n + 1 ) D ( n )+( 1 ) n + 1 for n 0; D ( ) = 1. Solution. From Example 2.18 and from what we covered in discussion section, the generating function for D ( n ) is D ( x ) = e x 1 x . Taking a derivative we get D xD D = e x ( 1 x ) D = D e x 1 which as power series becomes n D ( n + 1 ) x n n ! n ( n + 1 ) D ( n + 1 ) x n + 1 ( n + 1 ) ! = n D ( n ) x n n ! + n ( 1 ) n + 1 x n n !...
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This note was uploaded on 05/19/2010 for the course MATH mat 146 taught by Professor Gregkuperberg during the Spring '10 term at UC Davis.
 Spring '10
 GregKuperberg

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