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 Question from Leon: I have a group of 11 golfers wanting to play 10 rounds of golf in grouping of 4,4,3 .What is the best solution so that everyone plays each other as many times as possible

Leon,

It is difficult to compute the best schedule as the number of possibilities to consider is about 10 to the power 30. I have a program that will give pretty good solutions (close to best) in cases like this. Here is what it found:

Day 1 : (0, 0, 1, 2, 0, 1, 1, 2, 0, 1, 2)
Day 2 : (0, 1, 0, 2, 0, 1, 2, 1, 2, 1, 0)
Day 3 : (0, 1, 1, 2, 2, 0, 1, 1, 0, 2, 0)
Day 4 : (0, 1, 2, 0, 1, 1, 0, 2, 2, 0, 1)
Day 5 : (0, 1, 2, 2, 0, 2, 1, 0, 1, 0, 1)
Day 6 : (0, 0, 1, 0, 1, 0, 2, 2, 1, 1, 2)
Day 7 : (0, 1, 1, 2, 2, 0, 2, 0, 0, 1, 1)
Day 8 : (0, 1, 2, 1, 2, 2, 0, 1, 1, 0, 0)
Day 9 : (0, 0, 0, 1, 1, 2, 0, 1, 2, 2, 1)
Day 10 : (0, 0, 0, 0, 1, 1, 1, 1, 2, 2, 2)

The 11 positions in each sequence represent the players, 1 through 11. The numbers represent the group that player is in. For example, on Day 1, the groups are 1, 2, 5, 9; 3, 6, 7, 10; 4, 8, 11.

Because of the way the possibilities are generates, a flaw arises when there are groups of unequal size. In the schedule above, players 1 and 2 are never in the threesome.

Hope this works for you.
--Victoria

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