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Let $n \in \mathbb{N}_{\geq 1}$ be a number of pairs of people, so in total there are $2n$ people and every person has exactly $1$ partner. Furthermore, if person $A$ is the partner of person $B$, then person $B$ is the partner of person $A$.

The people want to play the game "Secret Santa", which involves giving gifts to people. Every person must give away exactly $1$ gift and must receive exactly $1$ gift. It is not allowed to give a gift to oneself or to ones respective partner.

How many possibilities are there to play this game?

The first few values of the sequence we are looking for are $a_1 = 0, a_2 = 4, a_3 = 80,...$

If we drop the restraint that the people are organised in pairs, then the wanted number is just the number of fixed-point free permutations of $2n$ elements, but the pair requirement seems to substantially complicate things.

Joker123
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    Attempted proofs? Claims? Wildly incorrect reasoning? – Aweygan Dec 20 '19 at 09:49
  • Just curious: how did you calculate $a_3 = 80$? Exhaustive listing (maybe by writing code)? Tedious application of inclusion-exclusion? – antkam Dec 20 '19 at 15:17
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    Also, I don't understand the downvote... The problem is quite interesting, and if OP can explain how $a_3 = 80$ was computed (which I think is non-obvious), that should count as making an attempt. – antkam Dec 20 '19 at 15:21
  • I computed it via exhaustive listing using Python :) – Joker123 Dec 21 '19 at 16:04

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