Proving $\sum_{k=0}^n {n \choose k}{m-n \choose n-k} = {m \choose n}$ I have tried using the principle of mathematical induction. The base case is simple, but I am having trouble with the induction step. Any help would be great, thanks ! 

Another way of proving this is by comparing the coefficients of $x$ in
$$(1+x)^m = (1+x)^n(1+x)^{m-n}$$
i.e. $$\sum_{k=0}^m \binom{m}{k} x^k = \sum_{l=0}^n \binom{n}{a}x^l\sum_{l=0}^{m-n} \binom{m-n}{l}x^l$$
Where the product of two series $∑a_nx^n, ∑b_nx^n$ has coefficients $c_n = ∑_{k+l=n}a_kb_{n-l}$.
