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.
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?
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 $k=n$.
You can use induction but first we need to rewrite as:
$\binom{m+n}{k} = \sum_{r=0}^{k} \binom{m}{r}\binom{n}{k-r}$
Assume true for $n\ge 1.$ Consider the case for $n+1$:
$\binom{m+n+1}{k} = \sum_{r=0}^{k} \binom{m}{r}\binom{n+1}{k-r}= \sum_{r=0}^{k} \binom{m}{r}(\binom{n}{k-r} + \binom{n}{k-r-1})$
$= \sum_{r=0}^{k}\binom{m}{r}\binom{n}{k-r} + \sum_{r=0}^{k} \binom{m}{r}\binom{n}{k-1-r}$
$= \binom{m+n}{k} + \binom{m+n}{k-1} = \binom{m+n+1}{k}$
