Combinatorical identity $$\sum _{i=0}^n\left(\binom{n}{i}\binom{m}{i}\right)=\binom{m + n}{m}$$
I tried to find a problem that describes both sides. The right is the simplest, so it can be described by choosing $m$ different to a group, but the left is the problem. I have a multiply in the sigma, so it should choosing $i$ elements from $n$, and for each chosen element from the first choice there are $i$ elements to choose from another $m$ elements (so it is cross), and then from each different $i$ it must be co-prime situations. 
 A: There's a more general result known as Vandermonde's identity:
For non negative integers $m,n$, and $r$ we have,
$$\sum_{k=0}^{r}\binom{m}{k}\binom{n}{r-k}=\binom{m+n}{r}.$$
Proof. The RHS is the number of ways to in which out of $m+n$ people we select $r$ to form a committee. Let this be $m$ men in the committee and $n$ women. If there are $k$ men then there are $r-k$ women in the committee. Thus $\binom{m}{k}\binom{n}{r-k}$ is also the number of ways to form the same committee. Summing over all the values of $k$ we get the desired result. $\square$ 
A: Expands your left hand side. It is $$ \binom {n}0 \binom {m}0+  \binom {n}1 \binom {m}1+ \cdots \binom{n}{n} \binom{m}{n}$$ Now replace $\binom {n}{r}$ with $\binom {n}{n-r}$. So now, your left hand side manipulates to $$\binom {n}n \binom {m}0+  \binom {n}{n-1} \binom {m}1+ \cdots \binom{n}{0} \binom {m}{n}$$ So, now your left hand side is selecting a total of $n$ things using different ways. But wait! That is what your right side is(in a much more simplified expression). There you go.
