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 Question from Mike: Regarding arranging golf players so no person plays with anyone more than once. You have given examples for 16 and 24 players. If it can be done, i need a solution for 20 players, 4 players per team one round per day for 5 days Thank you Mike

Mike,

For 20 players, there are 20 choose 2 = 190 pairings. Each foursome accounts for 4 choose 2 = 6 of them. Since 190 is not evenly divisible by 6, there is no collection of foursomes where each person plays with every other person exactly once.

If you have 5 foursomes per day, then you account for 30 pairings per day, and 150 pairings in all, so some people
will not play together.

You can construct a schedule as follows. First make a schedule for 25 golfers with 5 fivesomes per day for 6 days. This exists (it is an affine plane of order 5). You can make one in the same way as the 16 player example. It works because 5 is prime.

After making the schedule for 25 players, choose one of the fivesomes that play on day 6 -- say 21,22,23,24,25. Then, delete these players from every fivesome where they appear on days 1 through 5. This is your schedule. (That is, forget about day 6 -- these are the players that are not in your tournament, or the pairs who will will never play together.)