3
$\begingroup$

This question already has an answer here:

I'm trying to prove this combinatorially. $$k\binom{n}{k} = n\binom{n-1}{k-1}$$ I know the first step is to relate a question to the equation. My question was if you have $n$ friends how many ways can you choose $k$ of them. I know this isn't correct because that would be the question if the left side wasn't being multiplied by k. Can anyone help me figure out what multiplying ${n \choose k}$ by $k$ means?

$\endgroup$

marked as duplicate by Martin Sleziak, Jean-Claude Arbaut, Claude Leibovici, Davide Giraudo, Grigory M May 15 '14 at 10:33

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

10
$\begingroup$

The standard example is that you have a group of $n$ people, and want to pick a committee of $k$, including a designated committee Chair. We count the number of ways to do this in two ways.

(a) You can pick $k$ people, and then pick one of these to be Chair. This can be done in $\binom{n}{k}\cdot k$ ways.

(b) You can pick a Chair, and pick $k-1$ people to join her. This can be done in $n\binom{n-1}{k-1}$ ways.

Remark: A less bureaucratic example would be nice. You have $n$ friends and a large van, and want to pick $k$ friends to go drinking, with one of them the designated driver.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.