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Suppose that each of N men at a party throws his hat into the center of the room.
The hats are first mixed up, and then each man randomly selects a hat. What is the
probability that none of the men selects his own hat?

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+1 vote

There is a concept called Derangement. Using that,

Number of ways in which N men can be deranged with n hats(say $nD$) =  $n! \sum_{k = 0}^{n} \frac{(-1)^{k}}{k!}$

Total number of ways in which n men can pick hats(say $T$) =  $n!$

Therefore, $P = \frac{nD}{T} = \sum_{k = 0}^{n} \frac{(-1)^{k}}{k!}$

by Active (1.8k points)
+1

from sheldon ross.