proof that { (k),(n) :}}=((k!)/(n!×(k−n)!))


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Question Number 134270 by abdurehime last updated on 01/Mar/21

$proof that$ ${\begin{cases} k \\ n \end{cases}} = \frac{k!}{n! \times (k - n)!}$
Commented by mr W last updated on 02/Mar/21
$you mean (_n ^k )=((k!)/(n!×(k−n)!)). {_n ^k } has different meaning. it is the stirling number of the second kind. {_n ^k } is the number of ways to divide k distinct objects into n non− empty groups. for example {_2 ^4 }=7.$
$y o u m e a n (_{n}^{k}) = \frac{k!}{n! \times (k - n)!} .$ ${_{n}^{k}} h a s d i f f e r e n t m e a n i n g . i t i s t h e$ $s t i r l i n g n u m b e r o f t h e s e c o n d k i n d .$ ${_{n}^{k}} i s t h e n u m b e r o f w a y s t o d i v i d e$ $k d i s t i n c t o b j e c t s i n t o n n o n -$ $e m p t y g r o u p s . f o r e x a m p l e {_{2}^{4}} = 7 .$
Commented by mr W last updated on 02/Mar/21
$per definition (_n ^k ) is the number of ways to choose n objects from k objects. to choose the first object there are k ways, to select the second object there are (k−1) ways. etc. to select the n−th object, there are (k−n+1) ways. totally there are k(k−1)(k−2)...(k−n+1) ways. since the order of the n objects is not of concern, {_n ^k }=((k(k−1)(k−2)...(k−n+1))/(n!)) this can be simplified to {_n ^k }=((k(k−1)(k−2)...(k−n+1)(k−n)(k−n−1)...1)/(n!(k−n)(k−n−1)...1)) =((k!)/(n!×(k−n)!))$
$p e r d e f i n i t i o n (_{n}^{k}) i s t h e n u m b e r o f$ $w a y s t o c h o o s e n o b j e c t s f r o m k$ $o b j e c t s .$ $t o c h o o s e t h e f i r s t o b j e c t t h e r e a r e$ $k w a y s, t o s e l e c t t h e s e c o n d$ $o b j e c t t h e r e a r e (k - 1) w a y s . e t c . t o$ $s e l e c t t h e n - t h o b j e c t, t h e r e a r e$ $(k - n + 1) w a y s . t o t a l l y t h e r e a r e$ $k (k - 1) (k - 2) . . . (k - n + 1) w a y s . s i n c e$ $t h e o r d e r o f t h e n o b j e c t s i s n o t o f$ $c o n c e r n,$ ${_{n}^{k}} = \frac{k (k - 1) (k - 2) . . . (k - n + 1)}{n!}$ $t h i s c a n b e s i m p l i f i e d t o$ ${_{n}^{k}} = \frac{k (k - 1) (k - 2) . . . (k - n + 1) (k - n) (k - n - 1) . . . 1}{n! (k - n) (k - n - 1) . . . 1}$ $= \frac{k!}{n! \times (k - n)!}$
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