Proving an equality with binomials

Question

So i have an equality: $$\binom{3n}{2n}= \sum_{k=0}^{n} \binom{2n}{2n-k}\binom{n}{k}$$

So i tried writing on longer but i couldn't find any pattern. Any help would be appreciated, or hints as well. Is there any trick in this kind of proofs that i missed?. Thank you in advance.

Answer 1

$\binom{n}{k}$ is the coefficient of $x^k$ in $(1+x)^n$ and $\binom{2n}{2n-k}$ is the coefficient of $x^{2n-k}$ in $(1+x)^{2n}$, hence:

$$\sum_{k=0}^{n}\binom{n}{k}\binom{2n}{2n-k}=[x^{2n}]\left((1+x)^n\cdot (1+x)^{2n}\right)= [x^{2n}](1+x)^{3n}=\binom{3n}{2n}$$ as wanted.

Answer 2

Say you have $3n$ numbered balls, and you want to put $2n$ of them in a box. There are $\binom{3n}{2n}$ ways of doing that.

Now imagine that the first $2n$ balls are red and the last $n$ balls are blue. Then there are $\binom{2n}{2n-k}\binom{n}{k}$ ways of putting $2n$ balls in the box, where $k$ of them are blue (and $2n-k$ are red). Sum over all possible $k$, and you get all possible ways of putting $2n$ balls into the box.

Since these two paragraphs describe the number of ways to do the same thing, they must describe the same number.