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n people can be seated on n chairs in n! different ways. These are permutations.

If the people get up and sit down again in such a way that nobody sits on the same chair they sat on before, this can be done in !n different ways. These are derangements.

Now, if the people get up and sit down again in such a way that nobody sits on a chair they have sat on before, the number of possibilities depends on the choice in the previous step.

For example, with n=4:

First seating: ABCD
Second seating: BADC
Possible third seatings: CDAB CDBA DCAB DCBA, a total of 4.

First seating: ABCD
Second seating: BCDA
Possible third seatings: CDAB DABC, a total of 2.

As far as I can determine, with four objects, the number of possible third seatings is either 2 or 4, and with five objects, the number of possible third seatings is either 12 og 13.

Is there a formula for (or can anything intelligent be said about) the number of possible third (and fourth, and fifth ...) seatings?

oz1cz
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  • Related question: What is the probability that a randomly-selected derangement of a randomly-selected derangement is still a derangement of the original set? In other words, if everybody gets up and sits down in a different chair and then does that again, what is the chance that no one is sitting in their original chair? – John Barber May 03 '23 at 14:54
  • Related: https://math.stackexchange.com/questions/3674756/double-derangement-and-the-other-unfamous-kind-of-derangement?rq=1 – JMoravitz May 03 '23 at 15:18
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    Bruteforce gives that for 6 it's 80 or 82, for 7 it's 578, 579 or 580, for 8 - 4738, 4740, 4744 or 4752. – mihaild May 03 '23 at 18:02

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