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Question Number 109954 by Study last updated on 26/Aug/20
!3=????
$$!\mathrm{3}=???? \\ $$
Answered by mathdave last updated on 26/Aug/20
solution recall that !x=x!Σ_(n=0) ^x (((−1)^n )/(n!)) !3=3!Σ_(n=0) ^3 (((−1)^n )/(n!))=6((((−1)^0 )/(0!))+(((−1)^1 )/(1!))+(((−1)^2 )/(2!))+(((−1)^3 )/(3!))) !3=6(1−1+(1/2)−(1/6))=6(((3−1)/6))=2 ∵ !3=2 this mean that the how many way 3 object can be disarrange is 2ways
$${solution} \\ $$$${recall}\:{that} \\ $$$$!{x}={x}!\underset{{n}=\mathrm{0}} {\overset{{x}} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}!} \\ $$$$!\mathrm{3}=\mathrm{3}!\underset{{n}=\mathrm{0}} {\overset{\mathrm{3}} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}!}=\mathrm{6}\left(\frac{\left(−\mathrm{1}\right)^{\mathrm{0}} }{\mathrm{0}!}+\frac{\left(−\mathrm{1}\right)^{\mathrm{1}} }{\mathrm{1}!}+\frac{\left(−\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{2}!}+\frac{\left(−\mathrm{1}\right)^{\mathrm{3}} }{\mathrm{3}!}\right) \\ $$$$!\mathrm{3}=\mathrm{6}\left(\mathrm{1}−\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{6}}\right)=\mathrm{6}\left(\frac{\mathrm{3}−\mathrm{1}}{\mathrm{6}}\right)=\mathrm{2} \\ $$$$\because\:\:!\mathrm{3}=\mathrm{2}\:\:\:{this}\:{mean}\:{that}\:{the}\:{how}\:{many}\:{way}\:\mathrm{3} \\ $$$${object}\:{can}\:{be}\:{disarrange}\:{is}\:\mathrm{2}{ways} \\ $$
Commented by mathdave last updated on 26/Aug/20
i mean that 3 object can be disarrage in 2ways
$${i}\:{mean}\:{that}\:\mathrm{3} \\ $$$${object}\:{can}\:{be} \\ $$$${disarrage}\:{in} \\ $$$$\mathrm{2}{ways} \\ $$

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