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Question:

Assmue that $d$ is give postive integer numbers,and $$(x_{1}+x_{2}+\cdots+x_{d})^n=\sum_{k_{1}+\cdots+k_{d}=n} \binom{n}{k_{1},k_{2},\cdots,k_{d}}x^{k_{1}}_{1}x^{k_{2}}_{2}\cdots x^{k_{d}}_{d}$$ can see: Multinomial theorem

Find the value $$\sum_{k_{1}+k_{2}+\cdots+k_{d}=n}\binom{n}{k_{1},k_{2},\cdots,k_{d}}^2$$

for example

when $d=2$,then $$(x_{1}+x_{2})^n=\sum_{k_{1}+k_{2}=n}\binom{n}{k_{1},k_{2}}x^{k_{1}}x^{k_{2}}$$ we know $$\binom{n}{k_{1},k_{2}}=\binom{n}{k}$$ so $$\sum_{k_{1}+k_{2}=n}\binom{n}{k_{1},k_{2}}^2=\sum_{k=0}^{n}\binom{n}{k}^2=\binom{2n}{n}$$ can see:Non-inductive, not combinatorial proof of $\sum_{i \mathop = 0}^n \binom n i^2 = \binom {2 n} n$

But my problem I can't it,can you help?

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    Where did you get this problem from? It's not obvious to me that it's tractable, especially since the $n=2$ case is essentially a consequence of the Vandermonde identity and the symmetry $\binom{n}{k}=\binom{n}{n-k}$ (which doesn't help here). – Semiclassical Oct 27 '14 at 17:30

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