# hw4sol - Math 146 Algebraic Combinatorics Homework 4...

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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 ) α = k ( α 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 α = - 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 ( 0 ) = 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 0 - xD 0 - D = - e - x ( 1 - x ) D 0 = D - e - x 1

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which as power series becomes n 0 D ( n + 1 ) x n n ! - n 0 ( n + 1 ) D ( n + 1 ) x n + 1 ( n + 1 ) ! = n 0 D ( n ) x n n ! + n 0 ( - 1 ) n + 1 x n n ! . Examining the coefficient of x n yields the result. (c) Prove, by any method, that D ( n + 1 ) = n ( D ( n )+ D ( n - 1 )) for n 1 , D ( 0 ) = 1 , D ( 1 ) = 0. Solution. The entry 1 can map to n different things. There are two cases: If 1 is sent to i and i is sent to 1, then the remaining elements form a derangement in D ( n - 1 ) ways.
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