Proving $ \binom n 0 ^2 + \binom n 1 ^2 + \dots + \binom n n ^2 = \binom { 2n} n $ without induction [duplicate]

I have to prove that: $$ \binom n 0 ^2 + \binom n 1 ^2 + \dots + \binom n n ^2 = \binom { 2n} n $$

I don't want a complete solution, but only a hint.

The secret is to expand $(1+x)^{2n}$ in two ways: $$ \sum_{k=0}^{2n}\binom{2n}{k}x^k=(1+x)^{2n}=\left(\sum_{k=0}^{n}\binom{n}{k}x^k\right)\left( \sum_{k=0}^{n}\binom{n}{k}x^{n-k}\right). $$ In the left sum the coefficient of $x^n$ is $\binom{2n}{n}$ while the the right one in $$ \sum_{k=0}^n\binom{n}{k}\binom{n}{k}=\sum_{k=0}^n\binom{n}{k}^2. $$ Hence $$ \binom{2n}{n}=\sum_{k=0}^n\binom{n}{k}^2 $$

Generalization. Similiarly one can obtain that $\,\,\,\displaystyle \binom{mn}{n}=\sum_{k_1+\cdots+k_m=n}\binom{n}{k_1}\cdots\binom{n}{k_m}. $

A combinatorial proof:

Suppose you have $2n$ elements, and you want to choose $n$ among them.

Of course you can do that in $\binom {2n}{n}$ ways.

Let's count them in another way: consider the first half of the elements and the second half separately.

To choose $n$ elements overall, you can choose $k$ from the first half and $n-k$ from the second half, and then sum for all $k = 0 .... n$

That is $$\sum_{k=0}^n \binom nk \binom n{n-k} = \binom{2n}n$$

But of course $$ \binom nk \binom n{n-k} = \binom nk ^ 2$$ hence you are done.