What is the value of $\sum\limits_{i,j,k=0}^n{n \choose i+j}{n \choose i+k}{n \choose k+j}$ The question was asked by a friend. I tried using that ${n \choose k}={n-1 \choose k}+{n-1 \choose k-1}$ an to obtain some inductive expression but that doesn't seem to work because you get some nasty terms when expanding everything.
Another hope is that it has an intuitive answer, that is, perhaps you can explain it in terms of picking elements from some set(s). 
I indeed assume that ${n \choose k}=0$ for $k>n$.
 A: For the sake of an answer: calculating the first four terms with Alpha (because I make far too many errors doing such things by hand) yields the sequence $(1,4,29,229,\ldots)$ and an OEIS lookup on this yields http://oeis.org/A087809 , "The number of triangulations having $3+3n$ vertices of a triangle with each side subdivided by $n$ additional points" which includes a link to this relatively recent paper on the arXiv, as well as this older one. Given that there's no more explicit formula listed there, and that even the GF expressions are pretty messy, it seems likely that no cleaner formulation exists.
ETA: Some years later, now, I suspect that this can be formalized: it might be possible to use some version of Zeilberger's algorithm for definite sums of hypergeometric terms to find a polynomial recurrence relation in $n$ for this sum and then show that there's no term hypergeometric in $n$ that corresponds to it, similar to how one can prove that e.g. $\sum_k{n\choose k}^3$ has no 'closed-form' representation. In fact, one alternate formula in the paper, that the sum is equal to $2^{3n-1}-3\sum_{j=0}^{n-2}{3n-1\choose j}$, is so close to established 'first third of binomial coefficients' formulas that are known to have no closed form that I suspect it would be an easy proof: see theorem 8.8.1 of A=B.
