Todd Trimble
Some combinatorics on trees and functions

Throughout, we let |S|{|S|} denote the cardinality of a finite set SS.

Recall that the Stirling numbers of the second kind S(n,k)S(n, k) can be defined as the coefficients that express powers of an indeterminate xx as linear combinations of falling powers of xx:

x n= kS(n,k)x k̲x^n = \sum_k S(n, k) x^{\underline{k}}

where x 0̲1x^{\underline{0}} \coloneqq 1 and x k̲x(x1) k1̲x^{\underline{k}} \coloneqq x \cdot (x-1)^{\underline{k-1}}.

Lemma

Let Surj(n,k)Surj(n, k) be the number of surjections from an nn-element set to a kk-element set. Then S(n,k)=Surj(n,k)k!S(n, k) = \frac{Surj(n, k)}{k!}.

Proof

The assertion is the polynomial identity

x n= kSurj(n,k)(xk)x^n = \sum_k Surj(n, k) \binom{x}{k}

but this is clear since it holds whenever xx is an integer jj, by a bijective proof: j nj^n counts the number of functions from an nn-element set to a jj-element set, and each such function ff factors as a surjection onto some kk-element subset (of the jj-element set), followed by the subset inclusion. There are \binom(j}{k} such subset inclusions.

It is well-known (and can be proven by any number of methods, e.g., generating functions) that

S(n,k)=\binom(j}S(n, k) =

A MathOverflow question

Created on April 27, 2013 at 00:44:04 by Todd Trimble