By a combinatorial argument, prove that for $\displaystyle r \leq n$ and $\displaystyle r \leq m$, $\displaystyle \binom {n+m} {r} = \binom {m} {0} \binom {n} {r} + \binom {m} {1} \binom {n} {r-1} + ... + \binom {m} {r} \binom {n} {0}$.
By a combinatorial argument, prove that for $\displaystyle r \leq n$ and $\displaystyle r \leq m$, $\displaystyle \binom {n+m} {r} = \binom {m} {0} \binom {n} {r} + \binom {m} {1} \binom {n} {r-1} + ... + \binom {m} {r} \binom {n} {0}$.