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So far I have that $\binom{n}{m} \binom{m}{k}$ is equivalent to the number of ways you can take $m$ elements from $n$ elements and then $k$ elements from those $m$ elements.

Counting in the second way for

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

I find that we are still counting the number of ways that we can chose $m$ elements from $n$ elements but partitioning it into two sets on of size $n$ and the other of size $n-k$ with set difference $k$

would it be possible to use something like $S_{m,2}$ the sterling numbers of the second kind in this proof? How do I tie it all together?

Any help would be greatly appreciated.

Martin Sleziak
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Pablo
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1 Answers1

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Try thinking like this:

Suppose you are selecting a squad of $m$ players from a pool of $n$ candidates, of whom $k$ will form the team and the remainder the reserves.

Alternatively you are selecting a team of $k$ players from the original $n$, and $m-k$ reserves from the remaining $n-k$ candidates.

Mark Bennet
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