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Why does (in words) $${n \choose k }{k \choose l} = {n \choose l}{n - l \choose k - l}$$

Using the binomial theorem, it's easy to show equality. I just don't understand how to equate the two in words. The LHS appears to be "The number of subsets of size $k$ from a set of size $n$ times the number of subsets of size $l$ from a set of size $k$. It is not obvious to me how the right side is counting the same thing.

Any hints are helpful, thanks

Cotton Headed Ninnymuggins
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  • Out of $n$ people, form a team of $k$ and from the team choose $l$ as starters. Or you can choose $l$ starters from $n$ people, then choose the remainder of the team from the remaining people. this kind of solution is called combinatorial proofs or double counting – acat3 Feb 04 '23 at 06:14

1 Answers1

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A club has $n$ members. They have to choose a committee of $k$ members, and a sub-committee of $l$ members.

  • Method 1: choose the committee, $\binom nk$ ways, then the subcommittee, $\binom kl$ ways.
  • Method 2: choose the subcommittee first, $\binom nl$ ways, then another $k-l$ committee members from the remaining $n-l$ club members, $\binom{n-l}{k-l}$ ways.

Therefore $$\binom nk \binom kl = \binom nl \binom{n-l}{k-l}\ .$$

David
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