Question Number 207787 by mr W last updated on 26/May/24
$$\boldsymbol{{n}}\:{married}\:{couples}\:{are}\:{invited}\:{to} \\ $$$${a}\:{dance}\:{party}.\:{for}\:{the}\:{first}\:{dance} \\ $$$$\boldsymbol{{n}}\:{paires}\:{are}\:{radomly}\:{selected}.\: \\ $$$${what}'{s}\:{the}\:{probability}\:{that}\:{no}\:{woman} \\ $$$${dances}\:{with}\:{her}\:{own}\:{husband}? \\ $$$$\left.\mathrm{1}\right)\:{if}\:{a}\:{pair}\:{must}\:{be}\:{of}\:{different} \\ $$$$\:\:\:\:\:{genders}. \\ $$$$\left.\mathrm{2}\right)\:{if}\:{a}\:{pair}\:{can}\:{also}\:{be}\:{of}\:{the}\:{same}\: \\ $$$$\:\:\:\:\:{gender}. \\ $$
Answered by A5T last updated on 26/May/24
$$\left.{For}\:{case}\:\mathrm{1}\right) \\ $$$${The}\:{number}\:{of}\:{ways}\:{such}\:{that}\:{no}\:{woman} \\ $$$${dances}\:{with}\:{her}\:{own}\:{husband}\:{is}\: \\ $$$${D}_{{n}} ={n}!\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}!}+\frac{\mathrm{1}}{\mathrm{2}!}−\frac{\mathrm{1}}{\mathrm{3}!}+\frac{\mathrm{1}}{\mathrm{4}!}+…\underset{−} {+}\frac{\mathrm{1}}{{n}!}\right) \\ $$$$\left({ends}\:{with}\:+\frac{\mathrm{1}}{{n}!}\:{when}\:{n}\:{is}\:{even},\:−\:{otherwise}\right) \\ $$$$\Rightarrow{Probability}=\frac{{D}_{{n}} }{{n}!}=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}!}+\frac{\mathrm{1}}{\mathrm{2}!}−\frac{\mathrm{1}}{\mathrm{3}!}+…\underset{−} {+}\frac{\mathrm{1}}{{n}!} \\ $$
Commented by mr W last updated on 26/May/24
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Commented by A5T last updated on 27/May/24
![D_n =n![Σ_(n=0) ^n (−1)^n (1/(n!))]](https://www.tinkutara.com/question/Q207800.png)
$${D}_{{n}} ={n}!\left[\underset{{n}=\mathrm{0}} {\overset{{n}} {\sum}}\left(−\mathrm{1}\right)^{{n}} \frac{\mathrm{1}}{{n}!}\right] \\ $$