# Proof by Combinatorial Argument [duplicate]

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?

• Is this homwwork ? – user645636 Mar 15 '19 at 19:48
• The answer by Pedro Tamaroff uses a combinatorial argument at the duplicate. – Dietrich Burde Mar 15 '19 at 19:52

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.