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December 23

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Can someone explain why in the permutation formula we have choices of picking the 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 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 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 elements remaining, and this is the number of choices.
...
Then you choose the kth element. of the n elements were already chosen and you can't choose them again. So there are elements remaining, and this is the number of choices.
-- Meni Rosenfeld (talk) 11:01, 23 December 2015 (UTC)[reply]
At the th row here, do you mean 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

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Can anyone find an expression for  ? (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 etc. Thank you. — 79.118.191.221 (talk) 22:45, 23 December 2015 (UTC)[reply]

Simply 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! :-)