Combinatorics


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Question Number 134905 by bramlexs22 last updated on 08/Mar/21

$Combinatorics$ What is the number of ways of distributing 9 different objects among 5 persons such that each gets at least 1?\n
Commented bysoumyasaha last updated on 08/Mar/21

$5^{9} -^{5} C_{1} 4^{9} +^{5} C_{2} 3^{9} -^{5} C_{3} 2^{9} +^{5} C_{4} 1^{9} = 843120$
	Answered by math55 last updated on 08/Mar/21

	$\underset{―}{S o l u t i o n}$ $G i v e n,$ $T o t a l o b j e c t s^{'} n^{'} = 9$ $T o t a l p e r s o n^{'} r^{'} = 5$ $t h e n,$ $T o t a l w a y f o r t h e =^{n} P_{r}$ $d i s t r i b u t i o n$ $=^{9} P_{5} = \frac{9!}{(9 - 5)!}$ $= \frac{9!}{4!}$ $[= 15120 w a y s$
	Commented bybramlexs22 last updated on 08/Mar/21

	$wrong$
	Commented bymr W last updated on 08/Mar/21

	$c o r r e c t i s 834120, i t h i n k .$
	Commented bymath55 last updated on 08/Mar/21

	$h o w s i r p l e a s e e x p l a i n$
	Answered by mr W last updated on 08/Mar/21

	$t o d i s t r i b u t e n i d e n t i c a l o b j e c t s$ $a m o n g r p e r s i o n s, t h e n u m b e r o f$ $w a y s i s t h e c o e f f i c i e n t o f x^{n} t e r m i n$ ${(x + x^{2} + x^{3} + . . .)}^{r}$ $(e a c h p e r s o n g e t s a t l e a s t o n e o b j e c t)$ $i f t h e n o b j e c t s a r e d i f f e r e n t,$ $s i m i l a r l y t h e n u m b e r o f w a y s i s t h e$ $c o e f f i c i e n t o f x^{n} t e r m i n$ $n! {(x + \frac{1}{2!} x^{2} + \frac{1}{3!} x^{3} + . . .)}^{r} = n! {(e^{x} - 1)}^{r}$ $w i t h n = 9, r = 5 w e g e t t h e c o e f . o f$ $x^{9} i n 9! {(e^{x} - 1)}^{5} i s 834120 .$
	Commented bymr W last updated on 08/Mar/21

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