Expand $\binom{xy}{n}$ in terms of $\binom{x}{k}$'s and $\binom{y}{k}$'s Motivated by this question, I want to find a complete set of relations for the ring of integer-valued polynomials, where the generators are the polynomials $\binom{x}{n}$ for $n\in \mathbb{N}$. The best way to do this is would be to describe how to decompose $\binom{x+y}{n}$ and $\binom{xy}{n}$ as a sum of products of $\binom{x}{0},\dots, \binom{x}{n}$ and $\binom{y}{0},\dots,\binom{y}{n}$. This can be done in principle by peeling off the binomials starting with the highest degree and working one's way down. Playing around with Sage, one soon guesses that
$\binom{x+y}{n} = \sum_{k=0}^n \binom{x}{k}\binom{y}{n-k}$
and in fact, I think this has a combinatorial proof which straightforwardly generalizes that of the identity $\binom{m}{n} = \binom{m}{n-1} + \binom{m-1}{n-1}$.
But $\binom{xy}{n}$ seems to be not so straightforward. The first few expansions are:
$\binom{xy}{2} = 2\binom{x}{2}\binom{y}{2} + x\binom{y}{2} + y \binom{x}{2}$
$\binom{xy}{3} = 6\binom{x}{3} \binom{y}{3} + $
$\qquad ~ ~ 6 \binom{x}{3}\binom{y}{2} + 6\binom{x}{2}\binom{y}{3} + $
$\qquad ~ ~ x \binom{y}{3} + 4 \binom{x}{2} \binom{y}{2} + y \binom{x}{3}  $
$\binom{xy}{4} = 24\binom{x}{4}\binom{y}{4} + $
$\qquad ~ ~ 36\binom{x}{3}\binom{y}{4} + 36 \binom{x}{4}\binom{y}{3} + $
$ \qquad ~ ~ 14 \binom{x}{2}\binom{y}{4} + 45\binom{x}{3}\binom{y}{3} + 14 \binom{x}{4}\binom{y}{2} + $
$\qquad ~ ~ 12 \binom{x}{2}\binom{y}{3} + 12 \binom{x}{3}\binom{y}{2} + $
$\qquad ~ ~ \binom{x}{2}\binom{y}{2}$
and it is not so easy to discern a pattern.
This must be well-known: what is a closed-form expression for the expansion of $\binom{xy}{n}$?

There's another formula for the expression in equation (3) on the bottom of page 183 of my paper here: http://www.sciencedirect.com/science/article/pii/S0022404905002161 
The formula is ${{xy} \choose n} = \underset{l_1 + 2 l_2 + \cdots + nl_n = n}{\sum}  {{\sum_{i=1}^n l_i} \choose {l_1, l_2, \ldots, l_n}} {y \choose {\sum_{i=1}^n l_i}} \prod_{i=1}^n {x \choose l_i}$, proved by expanding $(1+T)^{xy} = ((1+T)^x)^y$ using the binomial and multinomial theorems.
