combinatorics - Algebraic proof of combinatorial identity

I would like to obtain the algebraic proof for the following identity. I already know the combinatorial proof but the algebraic proof is evading me.

$$\sum_{r=0}^n\binom{n}{r}\binom{2n}{n-r}=\binom{3n}{n}$$

Thanks.

Answer

We make use of the Binomial Theorem. Observe that:
\begin{align*}
\sum_{k=0}^{3n} \binom{3n}{k}x^k &= (1+x)^{3n} \\

&= (1+x)^n(1+x)^{2n} \\
&= \left[ \sum_{i=0}^{n} \binom{n}{i}x^i \right] \left[ \sum_{j=0}^{2n} \binom{2n}{j}x^j \right] \\
&= \sum_{k=0}^{3n} \left[\sum_{r=0}^n\binom{n}{r}\binom{2n}{k-r}\right]x^k \\
\end{align*}

Hence, by setting $k=n$, we compare the coefficients of $x^n$ of both sides to obtain:
$$
\binom{3n}{n} = \sum_{r=0}^n\binom{n}{r}\binom{2n}{n-r}
$$
as desired.