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We have 12 golfers and 3 foursomes everyday for 4 days. What is the formula so each golfer plays with everyone else at least once, but no more than twice. Obviously each golfer will only play twice with one other golfer one time. Help!

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It is not possible.

Let's label the players so that on the first day we have the groups:

ABCD  EFGH  IJKL

Let's color the first group (A,B,C,D) red.

On every subsequent day, at least two red players will play together. Let's consider the 3 groups, one for each subsequent day, where two red players play together. In these groups there are 6 appearances of red players. One or two red players must appear twice.

Let's say, WLOG, that A appears twice, on day 2 and 3. If we look at A's schedule, it must look like this:

1: ARRR  2: AR??  3: AR??  4: A???

where R is a red player and ? any player, red or not.

From there we can see that A can play with at most 7 different non-red players. But there are 8 of them. Therefore A doesn't play with everybody.

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  • $\begingroup$ Very nice proof! $\endgroup$ – Trenin Jan 5 '15 at 15:54
  • $\begingroup$ Thanks so much for helping. I greatly appreciate it. Here's how I started: A plays with BCD, A plays with EFG, A plays with HIJ and A plays with KL and ? So this scenario cannot be repeated by any other player(s)? With no regard for repeats, is there then a formula so everybody at least plays with everyone else at least once? $\endgroup$ – Gary Jan 6 '15 at 4:06
  • $\begingroup$ @Gary It might work for A, but not for all of A,B,C,D. And I never use the fact that nobody plays more than twice with each other. $\endgroup$ – Florian F Jan 6 '15 at 9:28

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