<s>[INST] 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 $\binom{xy}{n}$?
 [/INST] A beautiful combinatorial answer was found by Gjergji Zajmi. For an expanded version, see the solution to Exercise 3.9 in my Notes on the combinatorial fundamentals of algebra. (At least, it is Exercise 3.9 in the version of 10 January 2019. In future versions, the numbering can shift.)
Unlike Martin's answer, this one directly gives a formula for $\dbinom{xy}{n}$ as a nonnegative linear combination of products $\dbinom{x}{k}\dbinom{y}{\ell}$, as opposed to a polynomial in the $\dbinom{x}{k}$ and the $\dbinom{y}{\ell}$. This, of course, does not work for arbitrary $\lambda$-rings.
</s>