Prove combinatorial identity

Question

Prove the following identity:

$$ {{i+j}\choose{i}}\left\{{n}\atop{i+j}\right\} = \sum_{k=0}^n{{n}\choose{k}}\left\{{k}\atop{i}\right\}\left\{{n-k}\atop{j}\right\} $$

Answer 1

Suppose we are trying to prove that $${n\brace p+q} {p+q\choose p} = \sum_{k=0}^n {n\choose k} {k\brace p} {n-k\brace q}.$$

We will use exponential generating functions in $n$. The RHS is a convolution of two generating functions call them $A(z)$ and $B(z)$. The first is $$\sum_{m\ge 0} {m\brace p}\frac{z^m}{m!}.$$

Recall the bivariate generating function of the Stirling numbers of the second kind: $$\exp(u(\exp(z)-1))$$ so that $${n\brace k} = n! [z^n] [u^k] \exp(u(\exp(z)-1)). $$

This yields $$A(z) = \sum_{m\ge 0} \frac{z^m}{m!} m! [z^m] \frac{1}{p!} (\exp(z)-1)^p.$$ which is $$\sum_{m\ge 0} z^m [z^m] \frac{1}{p!} (\exp(z)-1)^p.$$ The sum cancels the coefficient extractions and we get $$A(z) = \frac{1}{p!} (\exp(z)-1)^p.$$ Similarly, $$B(z) = \frac{1}{q!} (\exp(z)-1)^q.$$ Therefore $$A(z) B(z) = \frac{1}{(p+q)!} {p+q\choose p} (\exp(z)-1)^{p+q}.$$

The generating function of the LHS with respect to $n$ is $${p+q\choose p} \frac{1}{(p+q)!} (\exp(z)-1)^{p+q}.$$

The two generating functions are the same, QED.

Observation. A combinatorial proof goes like this. The left counts the number of ways of partitioning $n$ labelled elements into $p+q$ sets and marking $p$ of these. This is the same as the right, where we choose $k$ elements from the $n$ elements for the $p$ marked sets and $n-k$ for the umarked ones.

This MSE link I illustrates the convolution technique, as does this MSE link II.

Observation II. Instead of substituting the coefficients from the bivariate generating function to obtain $A(z)$ and $B(z)$ we could have argued from first principles that they represent the species $$\mathfrak{P}_p(\mathfrak{P}_{\ge 1}(\mathcal{Z}))$$ and $$\mathfrak{P}_q(\mathfrak{P}_{\ge 1}(\mathcal{Z}))$$ and obtained them by direct translation.