$5$ persons play with their car keys such that after the game none leaves with his own key and everybody has exactly one key. It is not possible for all five of them to meet at one place again. If the probability that they can still have their right key back assuming each one of them meets one or more friends at most once is $P$, then $P$ equals
We know that derangement of $n$ distinct object in $n$ places as
$\displaystyle n!\bigg(1-\frac{1}{2!}+\frac{1}{3!}-\frac{1}{4!}+\frac{1}{5!}-\cdots \bigg)$
So derangement of $5$ distinct object
in $5$ places as
(All $5$ persons not getting its eaxct key)
$\displaystyle n!\bigg(1-\frac{1}{2!}+\frac{1}{3!}-\frac{1}{4!}+\frac{1}{5!} \bigg)=44$
And arrangement of $5$ distinct
object in $5$ places is $5!=120$
So probability in which all persons not getting its exact key
$\displaystyle \frac{44}{120}=\frac{11}{30}$
And probability of at least one person getting its exact key
$\displaystyle =1-\frac{11}{30}=\frac{19}{30}$
But here I did not know how I find probability $P$
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