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 ! 

Although a MI method may be used, a Combinatorial argument is much more simple. We assume that we want to select n distinct objects of a total of m objects. We can always select them directly and get RHS.Alternatively, we can separate the m objects into two groups, one of n and the other of m-n.
Now assume we take 0 from the group with n, we must choose n from the group with m-n. If we take 1 from the group with n, we must Take (n-1) from the group with (m-n). We continue repeating this & sum up the result, and we get RHS.
