3 ms·
Article is unclear, but I assume solutions that differ from another by just a rotation are excluded?
by mannerheim 4y ago
Article is unclear, but I assume solutions that differ from another by just a rotation are excluded?
- dernett 4y agoI 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.
- michaericalribo 4y agoActually, 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 http://oeis.org/wiki/M%C3%A9nage_problem
- lupire 4y agoIt's doesn't really matter, the difference is trivial.
- nullc 4y agoyou can just divide out the rotations if your dinner party plans are rotation invariant.