Proving in combinations ... is it true ? prove that : 
 $$\sum_{k=0}^n \left ( \begin{array}{c}n\\k\end{array}\right )^2 = {2n \choose n}$$
I tried to prove it by supposing i have a class has (M) girls and n boys , and we need to choose  a team (K of students) ---- So the ways are 
$$\displaystyle {m+n \choose k}$$----------
So I have (0) boy  and (k) girl OR 1 of n and (k-1) of m OR 2 of n and (k-2) of m OR ... (k) of n and (0) of m 
SO: 
$(^{m}_{0})(^n_k)+(^m_1)(^{n}_{k-1})+(^{m}_{2})(^{n}_{k-2})+...(^{m}_{k})(^{n}_{0})$
THEN
$$ \
\sum_{i=0}^k{m \choose i}{n \choose k-i}\ $$
$$ \
\sum_{i=0}^k{m \choose i}{n \choose k-i} ={n+m \choose k}\ $$
suppose m=n=k (is it true to suppose this ? )
$$ \
\sum_{i=0}^n{n \choose i}{n \choose n-i} ={2n \choose n}\ $$
i+(n-i)=2n 
the below = the top 
what should i do ?? 
Can i prove it by Mathematical induction ?
 A: Imagine we want to solve this question. "We have $2n$ objects that we want to chose $n$ of them. How many states exist?"
The first thing we can think of is this:  $\binom{2n}{n}$
But we can solve it in another way. We can divide these objects into two parts each consisted of $n$ objects. Now we pick $k$ objects from the first group and the rest $n - k$ from the second group. So we will have $\binom{n}{k} * \binom{n}{n-k}$ possibilities. This equation is the equivalent to:
$\binom{n}{k} * \binom{n}{n - k} = \binom{n}{k} * \binom{n}{k} = \binom{n}{k}^2$
But we have to calculate these possibilities for all $k$. It is obvious that $k$ is between $0$ and $n$. So the answer will be:
$\sum_{k=0}^{n}{\binom{n}{k}}^2$
So we have two solutions for the same problem. Since this question has only one answer, these solutions must be equal. So:
$\sum_{k=0}^{n}{\binom{n}{k}}^2 = \binom{2n}{n}$
A: Another approach : 
The coefficient of $x^n$ in the product : $$(1+x)^n(x+1)^n$$
Is equal to the sum : $\displaystyle \left ( \begin{array}{c}n\\0\end{array}\right )\left ( \begin{array}{c}n\\0\end{array}\right )+\left ( \begin{array}{c}n\\1\end{array}\right )\left ( \begin{array}{c}n\\1\end{array}\right )+ \dots=\sum_{k=0}^n \left ( \begin{array}{c}n\\k\end{array}\right )^2 $
Since this product is equal to : $(1+x)^{2n}$, coefficient of $x^n$ must be same.
From here we get coefficient of $x^n$ as $\displaystyle{2n \choose n}$.
Hence : $$\sum_{k=0}^n \left ( \begin{array}{c}n\\k\end{array}\right )^2 = {2n \choose n}$$
