Combinatorics-What-is-the-number-of-ways-of-distributing-9-different-objects-among-5-persons-such-that-each-gets-at-least-1-

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?

Commented by soumyasaha 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 by bramlexs22 last updated on 08/Mar/21

wrong

Commented by mr W last updated on 08/Mar/21

c o r r e c t i s 834120, i t h i n k .

Commented by math55 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} + \dots)}^{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} + \dots)}^{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 by mr W last updated on 08/Mar/21

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