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Question from Richard:

Hi

I have looked through the examples on your site but can't find one that suits. I am fiddling with one to see if I can get a result; but not confident.
I have 17 players playing 6 rounds in 3 groups of 4 and 1 group of 5. Two players will not be included in any 5 ball group. Can you help me please.

Richard,

This may not work perfectly, but it will be ok.

There is a schedule for 16 players to meet each other exactly once over five rounds (four foursomes). Try adding player 17 to the last foursome in each of these days, and then delete two players from the fivesome in as balanced a way as possible. For the last day, try to form the groups from the players who haven't played together in the same group yet -- there will be quite a few such pairs.

Good luck!
--Victoria

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