+198 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!
+6250 I don't know what the committee forming argument is.
However proving this should be easy enough using the definition
\(k \dbinom{n}{k} = k \dfrac{n!}{k!(n-k)!} = \\ \dfrac{n!}{(k-1)!(n-k)!} = \\ n \dfrac{(n-1)!}{(k-1)!((n-1)-(k-1))!} = \\ n\dbinom{n-1}{k-1}\)
+198 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??
