1

I am trying to show that $\sum_{j=0}^k {k \choose j}j!S(n,j) = k^n$ by showing $\sum_{j=0}^k {k \choose j}j!S(n,j)$ counts functions from $[n]$ to $[k]$. I am not very adept at this type of proof and am hoping to get help on which direction to take. ($S(n,j)$ is stirling number counting partitions of $[n]$ into $j$ blocks.)

Mroog
  • 360
  • Choose $j$ values from the available $k$ values that will form the range. Partition the $n$ available slots into $j$ non-empty subsets. Match these to a permutation of the $j$ values. – Marko Riedel Nov 02 '15 at 17:10

1 Answers1

2

To expand Marko’s comment a bit, for $j\in\{0,\ldots,k\}$ let $\mathscr{F}_j$ be the set of functions from $[n]$ to $[k]$ whose range has cardinality $j$.

  • If $S\subseteq[k]$ with $|S|=j$, how many surjections from $[n]$ to $S$ are there? HINT: Each of these surjections defines a partition of $[n]$ into how many parts?
  • How many $S\subseteq[k]$ are there such that $|S|=j$?
  • Combine the first two results to get an expression for $|\mathscr{F}_j|$, and sum over $j$.
Brian M. Scott
  • 631,399