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Show by combinatorial arguments that:

$$\sum_{k = 0}^m \binom{m}{k}\binom{n}{r + k} = \binom{m + n}{m + r}$$

Hello , I want to prove this argument by double counting method, i kinda have an idea of proving by comb args. but not by double counting... Can someone give any clues or ideas? thanks :)

Martin Sleziak
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Johnathan
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    It might help to first look at $\binom{m}{k}$ instead as $\binom{m}{m-k}$. Now... Suppose that you have $m$ men and $n$ women. In how many ways can you form a committee of size _____? By picking simultaneously? By picking genders separately? – JMoravitz Mar 24 '17 at 22:34
  • You can write directly your question by using $\LaTeX$ or/and $\texttt{MathJax}$. For instance, $$\texttt{\sum_{k = 0}^{m}{m \choose k}{n \choose r + k} = {m + n \choose m + r}}$$ yields $\sum_{k = 0}^{m}{m \choose k}{n \choose r + k} = {m + n \choose m + r}$. – Felix Marin Mar 24 '17 at 22:36
  • @FelixMarin That's interesting, I've never seen the syntax for \choose done correctly and have been in the habit of using \binom{ }{ } instead. (I've seen too many people use it incorrectly though, I guess its more noticeable when its wrong than when its right) Regardless, @ Johnathan, a link for resources on typing with MathJax can be found here. – JMoravitz Mar 24 '17 at 22:39
  • @JMoravitz $\texttt{binom}$ is fine too !!!. – Felix Marin Mar 24 '17 at 22:41

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This is essentially Vandermonde's identity. Consider $m$ men and $n$ women. In how many ways may we pick a committee of $m+r$ members?

On the one hand, this is just $\binom{m+n}{m+r}$.

On the other hand, if we pick $m-k$ men, then we need to pick $r+k$ women. Hence we have $$\sum_{k=0}^m\binom{m}{k}\binom{n}{r+k}=\sum_{k=0}^m\binom{m}{m-k}\binom{n}{r+k}=\binom{m+n}{m+r}.$$

jlammy
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