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# hw6sol - Math 146 Algebraic Combinatorics Homework 6...

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Unformatted text preview: Math 146: Algebraic Combinatorics Homework 6 Solutions 3.2 Fix integers n , k . Let f ( n , k ) be the number of permutations of n letters whose cycle lengths are all divisible by k . Find a simple, explicit egf for { f ( n , k ) } n ≥ . Find a simple, explicit formula for f ( n , k ) . Solution. Let n and k be fixed. The deck D mk has a card corresponding to each cyclic per- mutation of length mk . All other decks are empty. The number of cyclic permutations on mk letters is ( mk- 1 ) !. Thus the deck enumerator is D ( x ) = ∑ m ≥ 1 ( mk- 1 ) ! ( mk ) ! x mk = ∑ m ≥ 1 x mk mk = 1 k log 1 1- x k . Now apply Corollary 3.12 to get the hand enumerator, without regard to the number of cards in the hand: H ( x ) = exp 1 k log 1 1- x k = 1 ( 1- x k ) 1 / k = ∑ m ≥- 1 k m (- 1 ) m x mk . The required number is the coefficient of x n n ! from H ( x ) , which is 0 if k does not divide n , and is (- 1 ) r- 1 k r n ! = n ! r ! k r ( k + 1 )( 2 k + 1 ) ··· (( r- 1 ) k + 1 ) if n / k = r is an integer. 3.5 Let T n be the number of involutions of n letters. (a) Find a recurrence formula that is satisfied by these numbers. Solution. From Section 3.8, and more specifically the comment immediately after The- orem 3.16, we see the exponential generating function for T n is T ( x ) = ∑ n ≥ T n x n n ! = e x + 1 2 x 2 . Take the natural log on both sides, then take a derivative on both sides to obtain ∑ n ≥ T n + 1 x n n ! = ∑ n ≥ T n x n n ! + ∑ n ≥ 1 nT n- 1 x n n ! from which we see T n + 1 = T n + nT n- 1 for n ≥ 1 and T = T 1 = 1. (b) Compute T 1 , T 2 ,..., T 6 . 1 Solution. Using the recurrence above, T 1 = 1 , T 2 = 2 , T 3 = 4 , T 4 = 10 , T 5 = 26 , T 6 = 76. (c) Give a combinatorial and constructive interpretation of the recurrence. That is, af- ter having derived it from the generating function, re-derive it without the generating function. Solution. Let σ be an involution on n letters. There are two cases: either n is a fixed point ( σ ( n ) = n ) or it is not ( σ ( n ) 6 = n ) . In the first case there are T n- 1 ways to make the remaining n- 1 elements into an involution. In the second case suppose σ ( n ) = i ....
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hw6sol - Math 146 Algebraic Combinatorics Homework 6...

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