



 
Well, everybody plays 3 times with 3 others [= 9 playwiths] and there are only 7 other players, so everybody will have to play two repeats.
This has the flaw that each person plays _three_ times with one other. But we can't avoid this, if everybody is to play with everybody else! There cannot be three people who play together in both round 1 and round 2. Suppose there were:
Then none of ABC would have played with any of EFG; for round 3 to provide all the missing pairings for A, A would have to be in the same foursome as EFG in round 3. But the same is true of B, which is impossible. No player can play with all different people in round 1 and round 2. If A plays with BCD in round 1 and with none of them in round Let B be the player who plays with A in both rounds. We conclude that round 2 is (up to relabelling)
Now A and B must both play with E and F in round 3 and the solution is as given. RD  


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