TITLE: When are products of binomial coefficients equal?
QUESTION [8 upvotes]: It's known that $\binom{n}{r} = \binom{n}{s}$ if and only if $r = s$ or $r = n - s$. 
If $n \neq m$, is it true that $\binom{n}{s} \binom{m}{r} = \binom{n}{k} \binom{m}{\ell}$ if and only if ($s = k$ or $s = n - k$) and ($r = \ell$ or $r = m - \ell$)?

REPLY [2 votes]: This is a comment, not a solution, but recall the identity $(*)$ $\binom{n}{m} {m \choose k}={n \choose k}{n-k \choose m-k}$ and letting $n=8, m=5$ we can see that  Brian's counterexample to your original claim is of this form as $k=3$ and since ${5 \choose 2}={5 \choose 3}$, $k=2$ is equivalent, thus we may interchange the products on either resulting $rhs$ of $(*)$ justified in this case by the fact that ${8 \choose 3}=2{8 \choose 2}$ as well as ${8-3 \choose 5-3}=\frac{1}{2}{8-2 \choose 5-2}$. So this would seem to suggest that $m=n-k_i$ for some $k_i$ such that for all $i,j$, ${m \choose k_i}={m \choose k_j}$. It would appear possible that all the solutions to your problem must be equivalent to an expression of the form $(*)$ by some series of relations gleaned from identities as previously mentioned, which seem to extend the conditions of your original claim, that is any solution to your problem that doesn't satisfy the conditions of your original claim must be equivalent to one that in some sense does. Just a thought though, and it could be a useless one, my apologies if thats the case.