



 
Tom, The following scheme has every pair playing together at least once and at most twice. To do this I took the solution to a previous problem which was for 16 golfers playing 5 rounds. I numbered the golfers from 0 to 15 and for your problem I discarded numbers 0 and 15. Round 2 in the previous solution had 0 and 15 playing together and I modified this round in two ways to get your required 6 rounds. These are day 1 and day 6 in the list below. For day 1 I took player 1 from his 4some and put him with 5 and 10 and for day 6 I put player 3 with 5 and 10, and placed 1 back in his foursome. Day 1 Day 2 Day 3 Day 4 Day 5 Day 6 Chris  


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