Consider the identity
$$\sum_{k=0}^n (-1)^kk!{n \brace k} = (-1)^n$$
where ${n\brace k}$ is a Stirling number of the second kind. This is slightly reminiscent of the binomial identity
$$\sum_{k=0}^n(-1)^k\binom{n}{k} = 0$$
which essentially states that the number of even subsets of a set is equal to the number of odd subsets.
Now there is an easy proof of the binomial identity using symmetric differences to biject between even and odd subsets. I am wondering if there is an analogous combinatorial interpretation for the Stirling numbers. The term $k!{n\brace k}$ counts the number of set partitions of an $n$ element set into $k$ ordered parts. Perhaps there is something relating odd ordered partitions with even ordered partitions?
As an added note, there is a similar identity
$$\sum_{k=1}^n(-1)^k(k-1)!{n\brace k}=0$$
A combinatorial interpretation of this one would also be appreciated.
Best Answer
There is indeed. Let's try to construct an involution $T_n$, mapping odd ordered partitions of $n$-element set to even and vice versa: if partition has part $\{n\}$, move $n$ into previous part; otherwise move $n$ into new separate part.
Example: $(\{1,2\},\{\mathbf{5}\},\{3,4\})\leftrightarrow(\{1,2,\mathbf{5}\},\{3,4\})$.
This involution is not defined on partitions of the form $(\{n\},\ldots)$, but for these partitions one can use previous involution $T_{n-1}$ and so on.
Example: $(\{5\},\{4\},\{1,2\},\{\mathbf{3}\})\leftrightarrow(\{5\},\{4\},\{1,2,\mathbf{3}\})$.
In the end only partition without pair will be $(\{n\},\{n-1\},\ldots,\{1\})$. So our (recursively defined) involution gives a bijective proof of $\sum_{\text{k is even}}k!{n \brace k}=\sum_{\text{k is odd}}k!{n \brace k}\pm1$ (cf. 1, 2).
Upd. As for the second identity, the involution $T_n$ is already defined on all cyclically ordered partitions, so $\sum_{\text{k is even}}(k-1)!{n \brace k}=\sum_{\text{k is odd}}(k-1)!{n \brace k}$.
P.S. I can't resist adding that $k!{n \brace k}$ is the number of $(n-k)$-dimensional faces of an $n$-dimensional convex polytope, permutohedron (the convex hull of all vectors formed by permuting the coordinates of the vector $(0,1,2,\ldots,n)$). So $\sum(-1)^{n-k}k!{n \brace k}=1$ since it's the Euler characteristic of a convex polytope.