Ed wrote: << Alejandro Aguado solved this longstanding problem. I'll rephrase it, then give his solution. 16 couples (A-P & a-p) play golf, in foursomes. Each day, all 32 of them play. At the end of 10 days, all of them have played exactly once with everyone else, except for their spouse. Day0 Day1 Day2 Day3 Day4 Day5 Day6 Day7 Day8 Day9 1pm ABCD AEIm Alof AciL AgNK AhJp AFde APbn AjMG AkHO 2pm EFGH BgcH Bhkm BjdK BElP BFin BoGJ BNap BIeO BMfL 3pm IJKL DFKp DNGi DkbJ DIoH DcmO DEaf DglM DhPL Djen 4pm MNPO kPdf FaLO FgoP hdiO INbf hlNc koce gaJm Flbm 5pm abcd Maei MbHK fGOp beLp leGK PiKm fHiJ cfKn IPcG 6pm efgh ChbG CPeJ CIla Cjfm CMod CHLn CEKO CFkN Cgip 7pm ijkl jNoL Ejcp EhMn FMcJ EgkL gjbO FhIj Eobi ENdJ 8pm mnop lJnO gIdn NeHm kaGn jPaH IkMp dGLm ldHp hoaK Best solutions for 28, 36, or 40 golfers are all unknown. http://www.cs.brown.edu/~sello/golf.html

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Aha -- so each person has played in 10 foursomes, i.e., with potentially 30 different other people, which in this solution are in fact all the 30 people other than themself & their spouse -- so this solution must be optional. Is there a combinatorial framework into which this kind of question falls (e.g., could it be "block designes" ?) ? --Dan