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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)[reply]

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)[reply]

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)[reply]

Correct, sorry. Fixed now. -- Meni Rosenfeld (talk) 12:45, 23 December 2015 (UTC)[reply]

Generalization of Γ 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 $a=\pm 2,~\pm 3,~\pm {\frac {1}{2}},$ etc. Thank you. — 79.118.191.221 (talk) 22:45, 23 December 2015 (UTC)[reply]

Simply

(\Gamma (x))^{a}

will do the trick. -- Meni Rosenfeld (talk) 00:03, 24 December 2015 (UTC)[reply]

Man, I feel silly sometimes... :-) — 79.118.170.51 (talk) 14:19, 24 December 2015 (UTC)[reply]

Resolved

– Thanks! :-)