Is there any closed form for this summation : Is there any closed form for this summation : 
$$
\sum_{i_{1}\ +\ i_{2}\ +\ i_{3}\ =\ n
      \atop
      {\vphantom{\LARGE A}i_{k}\ \geq\ 0;\quad k\ =\ 1, 2, 3}}
\frac{2n!}{i_{1}!\, i_{1}!\, i_{2}!\, i_{2}!\, i_{3}!\, i_{3}!}$$
 A: HINT :
$$\sum_{i_{1}+i_{2}+i_{3}=n} \dfrac{(2n)!}{i_{1}!\,i_{1}!\,i_{2}!\,i_{2}!\,i_{3}!\,i_{3}!} = \binom{2n}{n}\sum_{i_{1}+i_{2}+i_{3}=n}\binom{n}{i_{1},i_{2},i_{3}}^{2}$$
where $\displaystyle \binom{n}{i_{1},i_{2},i_{3}}$ is the multinomial coefficient.
As far as I know there is no simple closed form for the RHS. However one can prove that

$$\max_{i_{1}+i_{2}+\ldots +i_{d} = n} \binom{n}{i_{1},i_{2},\ldots ,i_{d}} = \mathcal{O}\left(\dfrac{d^{n}}{n^{\frac{d-1}{2}}}\right)$$

Indeed if (to simplify), $\displaystyle i_{1} + i_{2} + \ldots + i_{d} = n = md$, then
$$ \binom{n}{i_{1},i_{2},\ldots ,i_{d}} \leq \binom{md}{m,m,\ldots,m}$$
and, using Stirling formula
$$\binom{md}{m,m,\ldots,m} \sim \dfrac{d^{\frac{d}{2}}}{(2\pi)^{\frac{d-1}{2}}}\dfrac{d^{md}}{(md)^{\frac{d-1}{2}}}$$
A: $\newcommand{\+}{^{\dagger}}%
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$\ds{{\cal I} \equiv \sum_{i_{1}\ +\ i_{2}\ +\ i_{3}\ =\ n
      \atop
      {\vphantom{\LARGE A}i_{k}\ \geq\ 0;\quad k\ =\ 1, 2, 3}}
\frac{2n!}{i_{1}!\, i_{1}!\, i_{2}!\, i_{2}!\, i_{3}!\, i_{3}!}}$

\begin{align}
{\cal I} &= \sum_{k_{1} = 0}^{n}\sum_{k_{2} = 0}^{n}\sum_{k_{3} = 0}^{n}
\frac{2n!}{k_{1}!\, k_{1}!\, k_{2}!\, k_{2}!\, k_{3}!\, k_{3}!}\,
\int_{\verts{z} = 1}{\dd z \over 2\pi\ic}\,
{1 \over z^{1 - k_{1} - k_{2} - k_{3} + n}}
\\[3mm]&=
2n!\int_{\verts{z} = 1}{\dd z \over 2\pi\ic}\,{1 \over z^{n + 1}}\,
\bracks{\sum_{k = 0}^{n}{z^{k} \over \pars{k!}^{2}}}^{3}
\end{align}

The sum over $k$ is evaluated here in terms of a Bessel function and the hypergeometric function. That's what makes difficult to perform the integration.
