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three men drop their hats off at the cloakroom when entering a restaurant and choose at random when taking out 1 of the 3 hats.

Calculating the probability that none of the 3 men take their own hats.

The total outcomes are 3! = 6

There are favorable outcomes: those that have neither 1 in the first position, nor 2 in second, nor 3 in third.

With manual counting, there are then 2 favorable outcomes.

The probability is, therefore 2/6

All is good, but how can I find the number of favorable outcomes when 50 drop their hats off at the cloakroom?

Edit: After research, the precise answer is the nearest integer to n!/e, 50!/e in my case.

Lioneds
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    The correct search term here is "Derangement." – JMoravitz Oct 04 '19 at 21:46
  • The question was confusing when first posted. If you are one of the early birds, the question is corrected now. – Lioneds Oct 04 '19 at 21:46
  • The end result is that with $n$ people, the probability that none of them get their correct hat back will be $\dfrac{!n}{n!}$ where $!n$ is the number of derangements on $n$. Check the linked question and the wikipedia article on derangements for more information. – JMoravitz Oct 04 '19 at 21:48
  • The search term Derangement provides a satisfactory answer. Thanks, JMoravitz. question should be closed. – Lioneds Oct 04 '19 at 21:51
  • I do encourage you to try to come up with one or more of the formulas for derangements yourself for practice as well. The most accessible one usually being arrived at through direct application of inclusion-exclusion. – JMoravitz Oct 04 '19 at 21:53

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