Counting two ways, $\sum \binom{n}{k} \binom{m}{n-k} = \binom{n+m}{n}$ prove by counting two ways:

I though to prove the right hand side I would say:
Let n represent a number of boys and m a number of girls. We want to choose a group of n from boys and girls. But for the left hand side I want to keep the variables m and n girls and boys. But I don't know what k will be. Suggestions?

If you are choosing $k$ girls, then exactly $k$ of the boys will not be chosen (since you are choosing as many children as there are boys).
