# hw7sol - Math 146: Algebraic Combinatorics Homework 7...

This preview shows pages 1–2. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Math 146: Algebraic Combinatorics Homework 7 Solutions 3.11 Let S , T be fixed sets of positive integers. Let f ( n ; S , T ) be the number of partitions of [ n ] whose class sizes all lie in S and whose number of classes lies in T . Show that { f ( n ; S , T ) } n has egf e T ( e S ( x )) , where e S ( x ) = s S x s s ! . Solution. This is a direct application of the exponential formula (Theorem 3.11). More pre- cisely we use Corollary 3.13. Define the the deck D i in the following way: D i will contain zero cards if i / S and one card if i S . In the latter case the card will correspond to the partition of [ i ] that places everything in one class: { 1 , 2 ,..., i } . Thus the deck enumerator is D ( x ) = n d n x n n ! = s S x s s ! = e S ( x ) . By making hands that draw from the decks D i we ensure the partition of [ n ] has class sizes all in S . To ensure the number of classes (that is, the number of cards in the hand) is in T we apply the exponential formula with the condition that h ( n , k ) = 0 if k / T and h ( n , k ) = x n n ! D ( x ) k k ! if k T , where h ( n , k ) is the number of partitions of [ n ] with class sizes in S with k classes where k T . This is exactly the coefficient of x n n ! in e T ( e S ( x )) . 3.12 Fix k > 0. Let f ( n , k ) be the number of permutations of n letters whose longest cycle has length k . Find the egf of { f ( n , k ) } n for k fixed. Solution. Let the deck D i have a card for each cyclic permutation of i letters. Thus | D i | = ( i- 1 ) !. If we restrict to cyclic permutations of length at most k , then the deck enumerator is x + x 2 2 + + x k k . Let F ( n , k ) be the number of permutations of n letters whose cycles have lengths at most k . By the exponential formula F has the egf exp ( x + + x k / k ) . But f ( n , k ) counts those whose longest cycle has length exactly equal to k . So if k 1 then f ( n , k ) = F ( n , k )- F ( n , k- 1 ) , and the required egf is e x + + x k- 1 k- 1 e x k k- 1 . GK7.1. The aim of this exercise is yet another proof of the exponential formula. Let A ( x ) = k = 1 a k x k be the exponential generating function for some types of animals of positive sizes. Youre suppose to prove everything in this exercise directly, rather than by using theorems in chapter...
View Full Document

## 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.

### Page1 / 5

hw7sol - Math 146: Algebraic Combinatorics Homework 7...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online