Prove that:$\binom {2n}{n}$=$\sum_{r=0}^n [\binom nr ]^2$ 
Prove that : $$\binom {2n}{n}=\sum_{r=0}^n \left[\binom nr\right]^2.$$

First of all, I tried to do in the principle of mathematical induction but I failed. Next, I expressed the binomial in algebraic form but I am not able to calculate $c$ this huge number. Somebody please help me.
 A: I'll give you a hint toward a combinatorial proof.
First, note that $\binom{2n}{n}$ is precisely the way to choose a committee of $n$ people out of a group containing $n$ men and $n$ women.
Next, note that
$$
\sum_{r=0}^{n}\binom{n}{r}^2=\sum_{r=0}^{n}\binom{n}{r}\binom{n}{n-r},
$$
since $\binom{n}{n-r}=\binom{n}{r}$.  Can you see how to interpret this last sum as the number of ways to choose such a committee?
A: There is an algebraic proof: note that
$$(x+1)^{2n}=\sum_{k=0}^{2n}\binom{2n}{k}x^{k}$$
$$(x+1)^{2n}=\left (\sum_{k=0}^{n}\binom{n}{k}x^{k}\right )^{2}=\sum_{k=0}^{2n}\left (\sum_{i=0}^{k}\binom{n}{i}\binom{n}{k-i}\right )x^{k}$$
Then,
$$ \binom{2n}{k}=\sum_{i=0}^{k}\binom{n}{i}\binom{n}{k-i}$$
for all $k$. Your equality is the case $k=n$.
A: You can use induction but first we need to rewrite as:
$\binom{m+n}{r} = \sum_{k=0}^{r} \binom{m}{r}\binom{n}{k-r}$
Assume true for n. 
$\binom{m+n+1}{r} = \sum_{k=0}^{r} \binom{m}{r}\binom{n+1}{k-r}= \sum_{k=0}^{r} \binom{m}{r}(\binom{n}{k-r} + \binom{n}{k-r-1})$
$= \sum_{k=0}^{r}\binom{m}{r}\binom{n}{k-r} + \sum_{k=0}^{r-1} \binom{m}{r}\binom{n}{k-r-1}$
$= \binom{m+n}{r} + \binom{m+n}{r-1} = \binom{m+n+1}{r}$
