Proving $\binom{n + m}{r} = \sum_{i = 0}^{r} \binom{n}{i}\binom{m}{r - i}$

Note that$$(x+1)^{n+m}=(x+1)^n(x+1)^m$$ Then by binomial theorem and collecting terms \begin{align} \sum_{r=0}^{n+m}\binom{n+m}{r}x^r &= \sum_{i=0}^{n}\binom{n}{i}x^i\sum_{j=0}^{m}\binom{m}{j}x^j \\&= \sum_{r=0}^{n+m}\sum_{i+j=r}\binom{n}{i}\binom{m}{j}x^r \\ &= \sum_{r=0}^{n+m}\sum_{i=0}^r\binom{n}{i}\binom{m}{r-i}x^r \end{align} Then compare the coefficient

You can think of a combinatoric proof as follows.

Suppose we have $n$ men and $m$ women, and we wish to form a committee of $r$ people. We can count in two different ways.

Case 1 $\binom{n+m}{r}$ is the number of r-subsets of a set with n+m elements.

Case 2 First we pick $i$ males. That leaves $r-i$ female to choose. So given $0\le i \le r$ We have $\binom{n}{i}\binom{m}{r-i}$ such committees. If we let $i$ range, we add up all these committees, so we get $\displaystyle\sum_{i=0}^r \binom{n}{i}\binom{m}{r-i}$ .

Since both cases count the same number, they must be equal.