Vandermonde's Identity: How to find a closed formula for the given summation [duplicate]

I've stumbled upon the following challenging question. Find a closed formula for the following summation $$ S(a,b,n) = \sum_{k=0}^n {a \choose k} {b \choose n-k}$$ for all possible parameters $a,b$.

Anyone happens to see how to solve this one?

Since ${a\choose k}$ is the coefficient of $x^k$ in the polynomial $(1+x)^a$ and ${b\choose n-k}$ is the coefficient of $x^{n-k}$ in the polynomial $(1+x)^b$, the sum $S(a,b,n)$ of their products collects all the contributions to the coefficient of $x^n$ in the polynomial $(1+x)^a(1+x)^b=(1+x)^{a+b}$.

This proves that $S(a,b,n)={a+b\choose n}$.

A counting argument:

If there are $a$ items of type A and $b$ items of type B, then

$$\sum_{k=0}^{n} {a \choose k} {b \choose n-k}$$

is the number of ways to choosing $n$ items from them: choose $k$ of type A and $n-k$ of type B, vary $k$ from $0$ to $n$ and add up.

Thus the sum you seek is $${a+b \choose n}$$