# Reason for breakdown of a nice binomial identity

One has the nice identities $${xy\choose 1}={x\choose 1}{y\choose 1},$$ $${xy+1\choose 2}={x+1\choose 2}{y+1\choose 2}+{x\choose 2}{y\choose 2}$$ and $${xy+2\choose 3}={x+2\choose 3}{y+2\choose 3}+4{x+1\choose 3}{y+1\choose 3}+{x\choose 3}{y\choose 3}.$$

(The proof is essentially trivial by interpreting $${z\choose k}$$ as a polynomial of degree $$k$$.)

This sequence of identities stops: There seems to be no nice expression of $${xy+k-1\choose k}$$ as a linear combination of $${x+k-i\choose k}{y+k-i\choose k},i=1,\ldots,k$$ for $$k\geq 4$$.

Is there a good reason for this breakdown? (Probably a better question is: Is there a reason for these identities to hold for $$k=2$$ and, especially, for $$k=3$$?)

• Is it a coincidence that 1; 1,1; 1,4,1 is the Eulerian numbers sequence? Aug 3 at 14:59
• No idea if this is coincidental. Coefficients on the right hand side should however add up to $k!$ (by comparing coefficients of $(xy)^k$ on both sides). Aug 3 at 15:30

$$\def\des{\operatorname{des}}$$Let $$\des(\pi)$$ be the number of descents of the permutation $$\pi$$. Then for any permutation $$\pi$$ in $$S_k$$, we have $$\begin{equation*}\binom{xy+k-\des(\pi)-1}{k} =\sum_{\sigma\tau=\pi}\binom{x+k-\des(\tau)-1}{k} \binom{y+k-\des(\sigma)-1}{k}.\tag{*}\label{star} \end{equation*}$$
If $$\pi$$ is the identity then $$\des(\pi)=0$$, so the left side of \eqref{star} is $$\binom{xy+k-1}{k}$$ and on the right, $$\tau=\sigma^{-1}$$. The OP's identities correspond to the fact that if $$k\le 3$$ then the number of descents of $$\pi$$ is the same as the number of descents of $$\pi^{-1}$$, but this is not true for $$k>3$$. This also explains the occurrence of the Eulerian numbers as coefficients for $$k\le3$$.
• MathJax notes: as with TeX, you can use \label+\eqref with \tag and get nice hyperlinks to boot. Unlike TeX, MathJax is not aggressive about discarding whitespace, so a preamble line $\def\des{…}$ followed by a newline will force a space in the text. Unfortunately, as ugly as it is in the source, the only way to get rid of the space is to put the closing $ directly next to the first line of the post, with no intervening whitespace. I have edited accordingly. Aug 3 at 18:56 • is it implied that$\pi = \tau \sigma$? Aug 4 at 12:03 • @NooneAtAll If$\sigma\tau$is the identity then so is$\tau\sigma$but in general$\tau\sigma\ne\sigma\tau\$. Aug 4 at 20:44