In 1993 Strehl showed that
$$
\sum_k\binom nk^3=\sum_k\binom nk^2\binom{2k}n.
$$
I’m interested in a combinatorial proof.
Upd (Jan '14). Maybe the original question was too restrictive — I'm now asking for any proof.
I've written a combinatorial proof (see answers), but it's somewhat convoluted. Maybe there is a nice proof using generating functions / residues (something like one for the Dixon's identity), for example?..
Comments and thoughts
- There is of course an algebraic proof (any such identity can be proved using WZ-magic of hypergeometric functions AFAIK). Feel free to post any proof, but I’m mainly interested in a combinatorial proof.
- It’s well known that $\sum_k\binom nk=2^n$ and $\sum_k\binom nk^2=\binom{2n}n$. Perhaps, one should think about Stehl identity as the next member of this family. (Is there any analogue for degree 4, I wonder, btw?)
- Quadratic case suggests that it’s, perhaps, better to write LHS as $\binom nk^2\binom n{n-k}$. And maybe instead of one $n$ one should introduce two or three parameters (s.t. Strehl identity corresponds to n=m[=l]). A naive approach is $\sum\binom nk\binom mk\binom m{m-k}$ but looks like $\sum\binom nk\binom mk\binom n{m-k}$ might be a better choice — actually, it's equal to $\sum\binom nk\binom mk\binom{2k}m$.
- arXiv:0712.3946 might be of some relevance. There LHS is interpreted as the number of ways to deal a deck of 3n cards (n denominations, 3 colors) to 3 players so that no player receives a card of her own color.
- Let's write the identity in the form $$ \sum_{i+j=n}\binom Bi\binom Wj\binom ni=\sum_k\binom Bk\binom Wk\binom{2k}n, $$ where $B$ and $W$ are two $n$-element sets. What I want to say is that both sides count ways to choose an $n$-element subset of $B\sqcup W$ with some kind of... extension (cf. two formulas for a trinomial coefficient, perhaps).