Ménage Problem

Actually, no. The number depends on the position at the table—rotations are distinct.

For 3 couples:

Mrs A must sit between Mr B and Mr C. Mr B can be on her left or her right, 2 possibilities. After those two seats are chosen, everyone else’s seat is determined.

But there are 6 places Mrs A might sit. Hence, for 3 couples there are 12=2x6 possible seatings, including rotations.

This [1] site addresses your intuition. The number of seatings up to rotations is the total number of seatings, divided by the number of seats (2x number of couples).

[1] http://oeis.org/wiki/M%C3%A9nage_problem

I don't believe so. I worked out the permutations for n = 3 and, accounting for rotations, you only get 2: [0, 3, 4, 1, 2, 5] [0, 5, 2, 1, 4, 3]. Of course, you get the expected answer of 12 if you multiply by 6.

It's doesn't really matter, the difference is trivial.

you can just divide out the rotations if your dinner party plans are rotation invariant.

I used to like the old slashdot conversations where someone would call shenanigans and a user with that name would show up and post. That and the lowest uid posturing where inevitably someone with a single digit uid would post. There's not so much of that kind of thing on HN, I understand it's a different kind of discussion but it's fun sometimes

/r/beetlejuicing meets HN!

Three body problem?

If you read between the lines, it's there.