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# What is committee forming argument, and how do i use it on this?

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Prove that \(k\binom{n}{k} = n\binom{n-1}{k-1}\).

I have no idea how to start whatsoever, so help, hints, or the proof would be appreciated!

Sep 15, 2018

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I don't know what the committee forming argument is.

However proving this should be easy enough using the definition

\(k \dbinom{n}{k} = k \dfrac{n!}{k!(n-k)!} = \\ \dfrac{n!}{(k-1)!(n-k)!} = \\ n \dfrac{(n-1)!}{(k-1)!((n-1)-(k-1))!} = \\ n\dbinom{n-1}{k-1}\)

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Sep 15, 2018
edited by Rom  Sep 15, 2018
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Thanks for your reply. However, i have already done it with algebra, so i already know how to do it the way you are doing. I need help on doing it specifically with a committee forming argument. I am stuck on how to start. Hints please? Also, can any mod take the check mark off??

Sep 15, 2018
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Rom  Sep 15, 2018
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Thank you very much! :)

Sep 15, 2018