Verify the following combinatorial identity: $\sum_{k=0}^{r} \binom{m}{k}\binom{n}{r-k} = \binom{m+n}{r}$ $$\sum_{k=0}^{r} \binom{m}{k}\binom{n}{r-k} = \binom{m+n}{r}$$
Nice, so I've proven some combinatorial identities before via induction, other more simple ones by committee selection models.... But this one is weird, induction doesn't even seem feasible here without things getting nasty, and the summation on the left is not making things easier. Can anyone help?

Note that there exist different $m+n$ balls and two bags. 
One bag contains $m$ balls and the other has $n$. 
What is the number of choosing $r$ balls ? Clearly $$ _{m+n}C_r$$
But we divided into two bag. : The number of choosing $k$ balls in the first bag is $ _mC_k $ And we must choose $r-k$ balls in second. The number of possibility is $ _nC_{r-k}$ So we have $$ _mC_k\ _nC_{r-k}$$ 
