Binomial Coefficients Combinatorics For a positive integers n, prove that
$$\displaystyle\sum\limits_{v=0}^n \frac{(2n)!}{(v!)^2 ((n-v)!)^2} = \binom{2n}{n}^2.$$
If somebody could please help me with this question, I would greatly appreciated it. 
 A: A combinatorial proof. The right-hand side corresponds to choosing two (possibly overlapping) sets of size $n$ from $2n$ elements.
Now we observe that:
$$\dfrac{2n!}{v!^2(n-v)!^2} = \binom{2n}{n} \binom n v \binom n{n-v}$$
If we divide the set of $2n$ elements in two fixed sets of $n$ elements, then picking $n$ from the original set amounts to picking $v$ from the first set, and $n-v$ from the second, for $v =0 \ldots n$.
Thus we find that
$$\sum_{v=0}^n \binom nv \binom n{n-v} = \binom{2n}n$$
which finally leads to the desired
$$\sum_{v=0}^n \frac{2n!}{v!^2(n-v)!^2} = \sum_{v=0}^n\binom{2n}{n} \binom n v \binom n{n-v} = \binom{2n}n \sum_{v=0}^n \binom nv\binom n{n-v} = \binom{2n}n^2$$
A: $$ \frac{(2n)!}{(v!)^2 ((n-v)!)^2} =\frac{(2n)!}{n!n!}\cdot \left(\frac{n!}{v!(n-v)!}\right)^2=\binom{2n}n \cdot \left(\frac{n!}{v!(n-v)!}\right)^2$$
$$\sum\limits_{v=0}^n= \frac{(2n)!}{(v!)^2 ((n-v)!)^2} =\binom{2n}n \cdot\sum\limits_{v=0}^n\left(\frac{n!}{v!(n-v)!}\right)^2$$
Now, equate the coefficients of $x^n$ in the following identity $$(1+x)^n(x+1)^n=(1+x)^{2n}$$ 
 to find  $$\sum\limits_{v=0}^n\left(\frac{n!}{v!(n-v)!}\right)\cdot \left(\frac{n!}{v!(n-v)!}\right)=\binom{2n}n$$
