0

The problem of counting the number of ways $n$ people randomly take $n$ hats, assuming that they can take their own hat, is quite intuitive: the first person has $n$ hats to choose from, the second person, once the first person has chosen a hat, has $n-1$ to choose from... So there are $n!$ ways.

However, the reasoning for the case where no one takes his hat is not so simple, since to know the number of hats from among which a person can choose, it must be known whether or not his own hat is among those remaining to be chosen.

What would be the intuitive way to reason this problem? Let's say, for example, for $n=4$ people, for which the solution is $!4=9$.

Henry
  • 157,058
Asd
  • 75
  • Personally, before I made any attempt to attack this problem on my own, I would study this derangements article. – user2661923 Jul 05 '21 at 11:04
  • Principle of inclusion-exclusion can be used to solve this. It is quite a popular problem, there would be loads of articles on this. Also, I think most books cover this as theory. – Ritam_Dasgupta Jul 05 '21 at 11:07
  • 1
    As well as https://math.stackexchange.com/questions/83380/i-have-a-problem-understanding-the-proof-of-rencontres-numbers-derangements - and note there are plenty of other inclusion-exclusion questions (such as the secretary problem or couples not sitting next to each other) which lead to a limiting proportion of $\frac1e$ – Henry Jul 05 '21 at 12:01

0 Answers0