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

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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)

Generalization of Γ function

Can anyone find an expression for ${\mathrm{\Gamma }}_a(x+1)=x^a{\mathrm{\Gamma }}_a(x)$ ? (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 $a=\pm 2,\pm 3,\pm \frac{1}{2},$ etc. Thank you. — 79.118.191.221 (talk) 22:45, 23 December 2015 (UTC)

Simply $(\mathrm{\Gamma }(x){)}^a$ 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)