Prove identity $\binom{2n}{n} = \sum_{k=0}^{n} \binom{n}{k}\binom{n}{n-k}$ I found hard (for me at least) example, in fact it just would be nice to prove. 
$$\binom{2n}{n} = \sum_{k=0}^{n} \binom{n}{k}\binom{n}{n-k}$$
 A: Suppose there are $n$ different red balls and $n$ different 
blue balls. 
In how many ways can we choose n balls from these $2n$ balls? 
The answer can be directly written to be $\binom{2n}{n}$.
Now, think differently. 


*

*We can choose no red balls and all blue balls. We can do this in ${\binom{n}{0}}{\binom{n}{n}}$ ways.

*We can choose $1$ red ball and $n-1$ blue balls. So total ${\binom{n}{1}}{\binom{n}{n-1}}$
.... This way we can choose $k$ red balls and $n-k$ blue balls. So, total $\binom{n}{k}\binom{n}{n-k}$ ways.
So, combining these all cases we get that there are total $\sum_{k=0}^{n} {\binom{n}{k}}{\binom{n}{n-k}}$ ways. And definitely this must be equal to $\binom{2n}{n}$. As both says the number of values of number of ways of this 'picking up $n$' game. 
So, $\binom{2n}{n}=\sum_{k=0}^{n} \binom{n}{k}\binom{n}{n-k}$.
A: Hint
$$(1+x)^{2n}=(1+x)^n (1+x)^n$$
$$\sum_{k=0}^{2n} \binom{2n}{k}x^k=\left(\sum_{k=0}^{n} \binom{n}{k}x^k \right)\left(\sum_{k=0}^{n} \binom{n}{k}x^k \right)$$
Now identify the coefficients of $x^n$.
A: Using Binomial theorem :
$$(1+x)^n=\sum_{k=0}^n{\binom{n}{k}x^k}=\sum_{k=0}^n{\binom{n}{n-k}x^{n-k}}$$
Then :
$$\sum_{k=0}^n\sum_{j=0}^n{\binom{n}{k}\binom{n}{n-j}x^{n-j+k}}=(1+x)^{2n}=\sum_{k=0}^{2n}{\binom{2n}{k}x^k}$$
From another hand the coefficient of $x^n$ in $\sum_{k=0}^n\sum_{j=0}^n{\binom{n}{k}\binom{n}{n-j}x^{n-j+k}}$ is $\sum_{k=0}^n{\binom{n}{k}\binom{n}{n-k}}$ which is equal to $\binom{2n}{n}$ usinig the other expression.  
