Wikipedia:Reference desk/Archives/Mathematics/2015 December 23

= December 23 =

Combinatorics
Can someone explain why in the permutation formula $$n(n-1)(n-2)\cdots(n-k+1)=\frac{n!}{(n-k)!}$$ we have $$n-(k-1)$$ choices of picking the $$k$$ element? I don't see the logic.

This proof will help me get the binomial coefficient. יהודה שמחה ולדמן (talk) 08:02, 23 December 2015 (UTC)
 * First you need to choose the 1st element. You have n choices.
 * Then you choose the 2nd element. 1 of the n elements was already chosen and you can't choose it again. So there are $$n-1$$ elements remaining, and this is the number of choices.
 * Then you choose the 3rd element. 2 of the n elements were already chosen and you can't choose them again. So there are $$n-2$$ elements remaining, and this is the number of choices.
 * Then you choose the 4th element. 3 of the n elements were already chosen and you can't choose them again. So there are $$n-3$$ elements remaining, and this is the number of choices.
 * Then you choose the kth element. $$k-1$$ of the n elements were already chosen and you can't choose them again. So there are $$n-(k-1)$$ elements remaining, and this is the number of choices.
 * -- Meni Rosenfeld (talk) 11:01, 23 December 2015 (UTC)
 * At the $$k$$th row here, do you mean $$k-1$$ elements were already chosen? יהודה שמחה ולדמן (talk) 12:09, 23 December 2015 (UTC)
 * Correct, sorry. Fixed now. -- Meni Rosenfeld (talk) 12:45, 23 December 2015 (UTC)
 * Correct, sorry. Fixed now. -- Meni Rosenfeld (talk) 12:45, 23 December 2015 (UTC)

Generalization of &Gamma; function
Can anyone find an expression for $$\Gamma_a(x+1)=x^a~\Gamma_a(x)$$ ? (Definite integral, infinite series or product, continued fraction, etc). If not in the general case, then at least for one or more special cases, such as $$a=\pm2,~\pm3,~\pm\frac12,$$ etc. Thank you. — 79.118.191.221 (talk) 22:45, 23 December 2015 (UTC)
 * Simply $$(\Gamma(x))^a$$ will do the trick. -- Meni Rosenfeld (talk) 00:03, 24 December 2015 (UTC)
 * Man, I feel silly sometimes... :-) — 79.118.170.51 (talk) 14:19, 24 December 2015 (UTC)