# How find this binomial-coefficients sum $\sum_{k_{1}+k_{2}+\cdots+k_{d}=n}\binom{n}{k_{1},k_{2},\cdots,k_{d}}^2$ [duplicate]

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 closed form $$\sum_{k_{1}+k_{2}+\cdots+k_{d}=n}\binom{n}{k_{1},k_{2},\cdots,k_{d}}^2$$

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 $$\sum_{k_{1}+k_{2}=n}\binom{n}{k_{1},k_{2}}^2=\sum_{k=0}^{n}\binom{n}{k}^2=\binom{2n}{n}$$ this proof can see:http://en.wikipedia.org/wiki/Binomial_coefficient

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

• Have a look at: mathoverflow.net/questions/128249/… – Suvrit Oct 27 '14 at 17:09
• In the OEIS you can consult A245397 and A033935. – juan Oct 27 '14 at 19:13
• When d=3, the sequence is A002893 – juan Oct 27 '14 at 19:28