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I’m trying to formulate a combinatorial interpretation of the identity

$$\sum_{k=0}^{\min(p,q)}{p\choose k}{q\choose k}{n+k\choose p+q}={n\choose p}{n\choose q}.$$

This has a proof here and there is a combinatorial interpretation of a sort-of similar identity here.

I interpreted the RHS as the answer to: “given $n$ balls, how many ways can you paint black the top half of $p$ of them and the bottom half of $q$ of them?” For example, for $(n, p, q) = (11, 4, 6)$ one such way would be

$$⬒ ⬓ ⬜ ⬓ ⬛ ⬜ ⬓ ⬒ ⬛ ⬓ ⬜.$$

My idea, thus far, was to interpret the $k$ on the LHS as the number of fully painted balls. (Here $k=2$.) Then the challenge is to find an interpretation of the selections in ${p\choose k}{q\choose k}{n+k\choose p+q}$ as “encoding” a way to do what was asked, leaving $k$ balls fully painted.

I suppose the selection of $k$ out of $p$ could be interpreted as: “which of the top-painted balls is also bottom-painted?” And the selection of $k$ out of $q$ is the same for the bottom-painted balls. So the first two choices tell us what the above sequence looks like when only looking at the top-painted and bottom-painted balls respectively:

$$⬒ ⬛ ⬒ ⬛ \quad\text{and}\quad ⬓ ⬓ ⬛ ⬓ ⬛ ⬓.$$

From this info (by matching up the ⬛s) we know the sequence looks like this:

$$\{\substack{⬒\\⬓⬓}\} ~⬛~ \{\substack{⬒\\⬓}\} ~⬛~ \{⬓\} $$

where the expressions in braces expand in some order to the boxes inside them, possibly padded with unpainted boxes, but not with ⬛s. In our case

\begin{align*} \{\substack{⬒\\⬓⬓}\} &\to ⬒ ⬓ ⬜ ⬓ \\ \{\substack{⬒\\⬓}\} &\to ⬜ ⬓ ⬒ \\ \{⬓\} &\to ⬓⬜. \end{align*}

And now there should be some way to encode precisely this info in a selection of $(p+q)$ elements out of $(n+k)$ elements! I’m totally mystified as to how this would work.

Is there indeed a way to do this? Or an easier combinatorial interpretation of this identity?

I feel like I am on the right track, as my choice of an interpretation for $k$ is “too nice not to work” (☺) and also worked for the similar identity I linked. But I just can’t find a collection of $(n+k)$ “things” to look at so that choosing $(p+q)$ from them yields a solution.

Lynn
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  • My comment at https://math.stackexchange.com/questions/280481/proof-of-the-identity-sum-k-0-minp-qp-choose-kq-choose-knk-choose cites a paper giving (or purporting to give) combinatorial proofs (I haven't read it). – darij grinberg Feb 18 '18 at 02:43
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    Your "idea, thus far", sadly, doesn't work. If it did, then ${p\choose k}{q\choose k}{n+k\choose p+q}$ would be divisible by ${n \choose k}$ (since you can arbitrarily choose the $k$ balls to be fully painted first), which it is not (for $n = 5$, $p = 2$, $q = 2$ and $k = 2$). – darij grinberg Feb 18 '18 at 02:47

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