Proof by Combinatorial Argument

Question

By a combinatorial argument, prove for $r \le n$ and $r \le m$,

$${n+m \choose r} = { m\choose 0}{n \choose r} + {m \choose 1} {n \choose r-1}+\cdots+{m \choose r}{n \choose 0}$$

Besides knowing that ${m \choose 0}=1$ and ${m \choose 1}=m$, I am completely at a loss. Can someone please advise?

The answer by Pedro Tamaroff uses a combinatorial argument at the duplicate. — Dietrich Burde, Mar 15 '19 at 19:52

score 0 · Answer 1 · answered Mar 15 '19 at 19:53

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The congress has $n$ rebublicans and $m$ democrats. The possibilities to create a team of $r$ persons consists of $r$ democrats and $0$ replublicans, $r-1$ democrats and $1$ republican, ..., $0$ democrats and $r$ republicans.

answered Mar 15 '19 at 19:53

New2Math

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Proof by Combinatorial Argument

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