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For fixed positive integers $m$ and $n$, let $$S=\sum_{k=0}^m (-1)^k\binom{m}{k}(m-k)^n.$$Show that $S = 0$ if $n < m$ and $S = n!$ if $n = m$.

I have asked this question before: Proof Involving Sigmas, but have not received any answer that I could relate to. I know that I have to use PIE(principle of inclusion and exclusion) and a counting argument. Here's my work so far:

There are $m$ people, $\binom{m}{k}$ represents all the ways to pick $k$ people out of $m$ people. $(m-k)^n$ represents all the ways of giving the people that weren't chosen $n$ distinguishable objects. What I don't get is how to apply PIE, why the signs are alternating, and how does $S = 0$ if $n < m$ and $S = n!$ if $n = m$. Could anyone offer an answer by the counting argument I started? Thanks!

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  • You should have added this to the earlier question or left me a comment under my answer rather than repeating the question. I’ll expand my answer there a bit, but I’ll have to think a little about just how to do it most usefully. Did you look at the two earlier answers to which I linked? – Brian M. Scott Aug 15 '16 at 18:12
  • Notice:

    $$\text{S}=\sum_{k=0}^m (-1)^k\binom{m}{k}(m-k)^n=m!\mathcal{S}_{n}^{(m)}$$

    Where $\mathcal{S}_{n}^{(m)}$ is the Stirling number of the second kind.

    http://mathworld.wolfram.com/StirlingNumberoftheSecondKind.html

     – Jan Eerland Aug 15 '16 at 18:13
  • In any case, posting the same question again is inappropriate. –  Aug 15 '16 at 18:13

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