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I have to give a combinatorial proof and the meaning of the following identity.

$$\sum_{k = 1}^n (-1)^k k !S(n,k) = (-1)^n,$$ where $S(n,k)$ is the Stirling number of the second kind.

Could anyone give me a hint on how to go about this? I suspect it has something to do with surjective maps, since I know that $k!S(n,k)$ is the number of bijective maps from $n$ to $k$.

urpi
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1 Answers1

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Let $n$ be odd. The number of ways to partition $n$ elements into an odd number of sets is $1$ more than the number of ways to partition the elements into an even number of sets.

Proof idea: for each partitioning of $n$ into an even number of sets, find the smallest non-singleton set, and split it into two sets at the smallest element. This forms a bijection between the even number sets and the odd number sets, except for the (odd) partitioning consisting of all $n$ elements. A similar argument holds if $n$ is even.

T. Silver
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  • I know it was just a proof idea, but perhaps you can flesh out a detail nevertheless: If the smallest non-singleton set has two elements, it gets split into two singleton sets. But if the partition has singletons, any two singletons could have formed the two-element set that gets split, so many even partitions get mapped to the same odd partition? – joriki May 02 '16 at 17:29