Among the 5 married couples,if the names of 5 | vedclass

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(C) The total number of ways to match $5$ men with $5$ wives is given by $5! = 120$. The number of ways in which no man is matched with his own wife i

Vedclass · Accepted Answer

$\frac{11}{30}$

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(C) The total number of ways to match $5$ men with $5$ wives is given by $5! = 120$. The number of ways in which no man is matched with his own wife is a derangement of $5$ objects,denoted by $D_5$. The formula for derangement is $D_n = n! \left( 1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \dots + \frac{(-1)^n}{n!} \right)$. For $n = 5$: $D_5 = 5! \left( 1 - 1 + \frac{1}{2} - \frac{1}{6} + \frac{1}{24} - \frac{1}{120} \right) = 120 \left( \frac{60 - 20 + 5 - 1}{120} \right) = 44$. Therefore,the probability that no man is matched with his wife is $\frac{D_5}{5!} = \frac{44}{120} = \frac{11}{30}$.

Among the $5$ married couples,if the names of $5$ men are matched with the names of their wives randomly,then the probability that no man is matched with the name of his wife is

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