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Let $n\ge k\ge m\ge 0$ be integers. Consider the following formula:

$$\binom{n}k\binom{k}m=\binom{n}m\binom{n-m}{k-m}$$

Give two different proofs. On proof should use the factorial formula for $\binom{n}k$. The other proof should be combinatorial. Try to develop a question that both sides of the equation answer.

So this question asked us to prove the formula below in two ways, and while I got it for the 1st way [using the factorial formula for $\binom{n}k$], I still don't understand how I can prove this using a combinatorial proof. I'm having trouble constructing a counting argument. Can anyone help me out?

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merged by quid Feb 28 at 23:50

This question was merged with Proving an identity with a combinatorial proof: $\binom{n}{k}\binom{k}{r}=\binom{n}{r}\binom{n-r}{k-r}$ because it is an exact duplicate of that question.