Show that $\sum_{k=0}^n \frac{(2n)!}{{k!^2(n-k)!}^2}= \binom{2n}{n}^2$ 
Show that
  $$\sum_{k=0}^n \frac{(2n)!}{k!^2(n-k)!^2}  = \binom{2n}{n}^2.$$

I tried canceling $2n!$ from both sides then moving $k!$ to right but still not sure how to proceed.
 A: $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
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\begin{align}
\color{#f00}{\sum_{k=0}^{n}{(2n)! \over k!^{2}\pars{n - k}!^{2}}}  = 
\sum_{k=0}^{n}{n! \over k!\pars{n - k}!}\,{n! \over k!\pars{n - k}!}\,{\pars{2n}! \over n!\,n!} =
{2n \choose n}\
\underbrace{\ \sum_{k = 0}^{n}{n \choose k}^{2}}_{\ds{2n \choose n}\ }\ =\
\color{#f00}{{2n \choose n}^{2}}
\end{align}

Note that
  \begin{align}
{2n \choose n} &= \bracks{x^{n}}\pars{1 + x}^{2n} =
\bracks{x^{n}}\braces{\pars{1 + x}^{n}\pars{1 + x}^{n}}
\\[3mm] & =
\bracks{x^{n}}\sum_{k = 0}^{n}\,\sum_{k' = 0}^{n}{n \choose k}
{n \choose n - k'}x^{k + k'} =
\sum_{k = 0}^{n}{n \choose k}{n \choose n - k}
\end{align}

A: For a combinatorial approach, the right-hand side counts ways of splitting $2n$ people into two groups of size $n$, twice; i.e., we assign half of the group $0$ and half of them $1$, and we assign half of them $a$ and half of them $b$.
All together, each person now is labeled either '$0a$', '$0b$', '$1a$', or '$1b$'.
On the left-hand side, the $k$th term is $\binom{2n}{k,k,n-k,n-k}$: it counts the number of ways to split $2n$ people into $k$ people with '$0a$', $k$ people with '$1b$', $n-k$ people with '$0b$', and $n-k$ people with '$1a$'.
This is summed up over all possible $k$.
Note that no matter how the $2n$ people are assigned into halves of $0/1$ and $a/b$ on the right-hand side, there will be the same number of '$0a$'s as '$1b$'s.
Thus this will correspond to some $k$ term on the left-hand side, so the two counts are the same, and so the equality is proved.
