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Show that $\begin{align}{n \choose k}{k \choose m} = {n \choose m}{n-m \choose k-m}.\end{align}$

Not sure how to approach this exactly. I've tried to use the property $\begin{align}{n \choose k} = {n \choose n-k} \end{align}$, which seems like it could be useful, but doesn't gett me anywhere.

Martin Sleziak
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St Vincent
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    http://math.stackexchange.com/questions/912126/help-finding-a-combinatorial-proof-of-k-n-choose-k-n-n-1-choose-k-1/912151#912151 – Shivang jindal Oct 24 '14 at 19:09

3 Answers3

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From $X$ with $|X| =n$, choose two subsets $A$ and $B$ such that $A \subset B$ and $|A| = m, |B| = k$

You can firstly choose $B$ from $X$(${n \choose k}$ ways), then choose $A$ from $B$(${k\choose m}$ ways).

Alternatively, you can firstly choose $A$ from $X$(${n \choose m}$ ways), then add $k-m$ anther elements into $A$ to form $B$( ${n-m\choose k-m}$ ways).

Petite Etincelle
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Hint: $$\binom{a}{b} = \frac{a!}{b!(a-b)!}.$$

user133281
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HINT: You can do it by a straightforward calculation with factorials, using the fact that $\binom{n}k=\frac{n!}{k!(n-k)!}$, or you can do it combinatorially. For the combinatorial reasoning in the same problem in a slightly different guise see this answer.

Community
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Brian M. Scott
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