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This question already has an answer here:

So I have that $\binom{n}{ c}\times \binom{n-d}{b-c} = \binom{n }{ d}\times \binom{d}{c}$.

I am trying to prove it by using number of players to prove this. What should be the best way to prove this? I can't seem to figure out what should go in n-d b-c.

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marked as duplicate by Sil, egreg, Cesareo, Javi, Lee David Chung Lin Mar 24 at 2:56

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    $\begingroup$ Please check for typos. There is something obviously wrong with the current version, with a $b$ on one side but not the other, ditto with $N$. $\endgroup$ – André Nicolas Oct 21 '13 at 23:30
  • $\begingroup$ @AndréNicolas: I think some are unknowns, but it is not clear which ones. The $N$ is most likely $n$, so I changed it. I hope that was not too presumptuous. $\endgroup$ – robjohn Oct 21 '13 at 23:39
  • $\begingroup$ There is an identity $$ \binom{n}{d}\binom{d}{c}=\binom{n}{c}\binom{n-c}{d-c} $$ that can be proven by expanding things out into factorials. However, I don't see how to make a useful substitution to get that into your form. $\endgroup$ – robjohn Oct 21 '13 at 23:43
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This might be what your are looking for, but it is hard to tell from your question.

How many ways are there from a group of $n$ people to make a committee of $d$ people with a board of $c$ people?

Pick the $d$ people for the committee, then from the committee, choose the $c$ board members: $$ \binom{n}{d}\binom{d}{c} $$ Pick the $c$ people for the board, then from the remaining $n-c$ people, choose the remaining $d-c$ committee members: $$ \binom{n}{c}\binom{n-c}{d-c} $$ Therefore, $$ \binom{n}{d}\binom{d}{c}=\binom{n}{c}\binom{n-c}{d-c} $$

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