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Question Number 202257 by cortano12 last updated on 23/Dec/23 | ||
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Commented by Frix last updated on 23/Dec/23 | ||
Each wife is specific | ||
Commented by mr W last updated on 22/Apr/24 | ||
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$${p}=\frac{\left(\mathrm{2}{n}\right)!}{\mathrm{2}^{{n}} \left(\mathrm{2}{n}\right)!}=\frac{\mathrm{1}}{\mathrm{2}^{{n}} }=\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{6}} }=\frac{\mathrm{1}}{\mathrm{64}} \\ $$ | ||