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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?

$$\mathrm{Combinatorics} \\ $$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

$$\:\mathrm{5}^{\mathrm{9}} \:−\:^{\mathrm{5}} \mathrm{C}_{\mathrm{1}} \mathrm{4}^{\mathrm{9}} \:+\:^{\mathrm{5}} \mathrm{C}_{\mathrm{2}} \mathrm{3}^{\mathrm{9}} \:−\:^{\mathrm{5}} \mathrm{C}_{\mathrm{3}} \mathrm{2}^{\mathrm{9}} \:+\:^{\mathrm{5}} \mathrm{C}_{\mathrm{4}} \mathrm{1}^{\mathrm{9}} \:=\:\mathrm{843120} \\ $$

Answered by math55 last updated on 08/Mar/21

Solution Given, Total objects ′n′=9 Total person ′r′=5 then, Total way for the=^n P_r distribution =^9 P_5 =((9!)/((9−5)!)) =((9!)/(4!)) [ =15120 ways

$$\underline{{Solution}} \\ $$$${Given}, \\ $$$$\:{Total}\:{objects}\:'{n}'=\mathrm{9} \\ $$$$\:{Total}\:{person}\:'{r}'=\mathrm{5} \\ $$$${then}, \\ $$$${Total}\:{way}\:{for}\:{the}=^{{n}} {P}_{{r}} \\ $$$${distribution} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=^{\mathrm{9}} {P}_{\mathrm{5}} =\frac{\mathrm{9}!}{\left(\mathrm{9}−\mathrm{5}\right)!} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{9}!}{\mathrm{4}!} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left[\:\:\:=\mathrm{15120}\:{ways}\right. \\ $$

Commented by bramlexs22 last updated on 08/Mar/21

wrong

$$\mathrm{wrong} \\ $$

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

correct is 834120, i think.

$${correct}\:{is}\:\mathrm{834120},\:{i}\:{think}. \\ $$

Commented by math55 last updated on 08/Mar/21

how sir please explain

$${how}\:{sir}\:{please}\:{explain} \\ $$

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

to distribute n identical objects among r persions, the number of ways is the coefficient of x^n term in (x+x^2 +x^3 +...)^r (each person gets at least one object) if the n objects are different, similarly the number of ways is the coefficient of x^n term in n!(x+(1/(2!))x^2 +(1/(3!))x^3 +...)^r =n!(e^x −1)^r with n=9, r=5 we get the coef. of x^9 in 9!(e^x −1)^5 is 834120.

$${to}\:{distribute}\:{n}\:{identical}\:{objects} \\ $$$${among}\:{r}\:{persions},\:{the}\:{number}\:{of} \\ $$$${ways}\:{is}\:{the}\:{coefficient}\:{of}\:{x}^{{n}} \:{term}\:{in} \\ $$$$\left({x}+{x}^{\mathrm{2}} +{x}^{\mathrm{3}} +...\right)^{{r}} \\ $$$$\left({each}\:{person}\:{gets}\:{at}\:{least}\:{one}\:{object}\right) \\ $$$$ \\ $$$${if}\:{the}\:{n}\:{objects}\:{are}\:{different}, \\ $$$${similarly}\:{the}\:{number}\:{of}\:{ways}\:{is}\:{the} \\ $$$${coefficient}\:{of}\:{x}^{{n}} \:{term}\:{in} \\ $$$${n}!\left({x}+\frac{\mathrm{1}}{\mathrm{2}!}{x}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{3}!}{x}^{\mathrm{3}} +...\right)^{{r}} ={n}!\left({e}^{{x}} −\mathrm{1}\right)^{{r}} \\ $$$$ \\ $$$${with}\:{n}=\mathrm{9},\:{r}=\mathrm{5}\:{we}\:{get}\:{the}\:{coef}.\:{of} \\ $$$${x}^{\mathrm{9}} \:{in}\:\mathrm{9}!\left({e}^{{x}} −\mathrm{1}\right)^{\mathrm{5}} \:{is}\:\mathrm{834120}. \\ $$

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

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