Question Number 134905 by bramlexs22 last updated on 08/Mar/21
$$\mathrm{Combinatorics} \\ $$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
$$\:\mathrm{5}^{\mathrm{9}} \:−\:^{\mathrm{5}} \mathrm{C}_{\mathrm{1}} \mathrm{4}^{\mathrm{9}} \:+\:^{\mathrm{5}} \mathrm{C}_{\mathrm{2}} \mathrm{3}^{\mathrm{9}} \:−\:^{\mathrm{5}} \mathrm{C}_{\mathrm{3}} \mathrm{2}^{\mathrm{9}} \:+\:^{\mathrm{5}} \mathrm{C}_{\mathrm{4}} \mathrm{1}^{\mathrm{9}} \:=\:\mathrm{843120} \\ $$
Answered by math55 last updated on 08/Mar/21
$$\underline{{Solution}} \\ $$$${Given}, \\ $$$$\:{Total}\:{objects}\:'{n}'=\mathrm{9} \\ $$$$\:{Total}\:{person}\:'{r}'=\mathrm{5} \\ $$$${then}, \\ $$$${Total}\:{way}\:{for}\:{the}=^{{n}} {P}_{{r}} \\ $$$${distribution} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=^{\mathrm{9}} {P}_{\mathrm{5}} =\frac{\mathrm{9}!}{\left(\mathrm{9}−\mathrm{5}\right)!} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{9}!}{\mathrm{4}!} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left[\:\:\:=\mathrm{15120}\:{ways}\right. \\ $$
Commented by bramlexs22 last updated on 08/Mar/21
$$\mathrm{wrong} \\ $$
Commented by mr W last updated on 08/Mar/21
$${correct}\:{is}\:\mathrm{834120},\:{i}\:{think}. \\ $$
Commented by math55 last updated on 08/Mar/21
$${how}\:{sir}\:{please}\:{explain} \\ $$
Answered by mr W last updated on 08/Mar/21
$${to}\:{distribute}\:{n}\:{identical}\:{objects} \\ $$$${among}\:{r}\:{persions},\:{the}\:{number}\:{of} \\ $$$${ways}\:{is}\:{the}\:{coefficient}\:{of}\:{x}^{{n}} \:{term}\:{in} \\ $$$$\left({x}+{x}^{\mathrm{2}} +{x}^{\mathrm{3}} +…\right)^{{r}} \\ $$$$\left({each}\:{person}\:{gets}\:{at}\:{least}\:{one}\:{object}\right) \\ $$$$ \\ $$$${if}\:{the}\:{n}\:{objects}\:{are}\:{different}, \\ $$$${similarly}\:{the}\:{number}\:{of}\:{ways}\:{is}\:{the} \\ $$$${coefficient}\:{of}\:{x}^{{n}} \:{term}\:{in} \\ $$$${n}!\left({x}+\frac{\mathrm{1}}{\mathrm{2}!}{x}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{3}!}{x}^{\mathrm{3}} +…\right)^{{r}} ={n}!\left({e}^{{x}} −\mathrm{1}\right)^{{r}} \\ $$$$ \\ $$$${with}\:{n}=\mathrm{9},\:{r}=\mathrm{5}\:{we}\:{get}\:{the}\:{coef}.\:{of} \\ $$$${x}^{\mathrm{9}} \:{in}\:\mathrm{9}!\left({e}^{{x}} −\mathrm{1}\right)^{\mathrm{5}} \:{is}\:\mathrm{834120}. \\ $$
Commented by mr W last updated on 08/Mar/21
