how prove $\sum_{m=0}^{n}\left(\frac{n!}{m!(n-m)!}\right)^2=\frac{(2n)!}{(n!)^2}$? How to prove $\forall n \in \mathbb N$
$$\sum_{m = 0}^{n} \left(\frac{n!}{m!(n-m)!}\right)^2=\frac{(2n)!}{(n!)^2}$$
 A: Note that 
$$\left(\frac{n!}{m!(n-m)!}\right)^2=\binom{n}{m}\binom{n}{n-m}.$$
Now imagine that we have a group of $n$ boys and $n$ girls, and want to choose a committee of $n$ people. 
It is clear that  the number of ways to choose a committee of $n$ from our $2n$ people  is $\binom{2n}{n}=\frac{(2n)!}{n!n!}$.
Now let us count this another way. For any $m$ with $0\le m\le n$, there are $\binom{n}{m}\binom{n}{n-m}$ ways to choose a committee with $m$ boys and $n-m$ grls. Sum over all $m$. 
Remark: Counting something in two different ways can be a powerful method for proving combinatorial identities.
A: In terms of binomial coefficients the proposed identity is
$$\sum_{m=0}^n\binom{n}m^2=\binom{2n}n\;.$$
Once you realize that $\dbinom{n}m^2=\dbinom{n}m\dbinom{n}{n-m}$, this becomes a special case of Vandermonde’s identity, and you’ll find a combinatorial proof in the linked article. (I’m pretty sure that you’ll also find combinatorial proofs on this site.)
A: First of all
$$
\frac{n!}{m!(n-m)!} = \binom{n}{m}
$$
is a binomial coefficient, and so is
$$
\frac{(2n)!}{(n!)^2} = \binom{2n}{n}
$$
Then consider the coefficient of $x^{n}$ in
$$
(1 + x)^{2n},
$$
which is $\dbinom{2n}{n}$, but can also be computed via
$$
(1 + x)^{2n} = 
((1 + x)^{n})^{2}
=
(\sum_{m=0}^{n} \binom{n}{m} x^{m})^{2}
$$
as
$$
\sum_{m = 0}^{n} \binom{n}{m} \binom{n}{n-m}
=
\sum_{m = 0}^{n} \binom{n}{m}^{2}
$$
