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

Question

prove by counting two ways:

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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?

See also: Combinatorial interpretation for the identity $\sum\limits_i\binom{m}{i}\binom{n}{j-i}=\binom{m+n}{j}$? — Martin Sleziak, Oct 10 '19 at 06:37

score 3 · Accepted Answer · answered Feb 15 '14 at 13:06

Suppose $n \leq m $.

Then if you want to choose $n$ people, you can choose $k$ boys in $\binom{n}{k}$ ways and $n-k$ girls in $\binom{m}{n-k}$ ways , for $k \in \{0,1, \dots n\}$.

So we have $$\sum_{k = 0}^{n} \binom{n}{k} \binom{m}{n-k} = \binom{n+m}{n}$$

score 0 · Answer 2 · answered Feb 15 '14 at 12:59

0

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).

answered Feb 15 '14 at 12:59

Marc van Leeuwen

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

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