



 
David, It turns out that the schedule you want (no duplicate pairs) does not exist, though that is not so easy to prove. It is possible to come fairly close. If you look here: http://mathcentral.uregina.ca/QQ/database/QQ.09.08/h/bill4.html, you will find a schedule for 16 golfers in 4 groups of 4 for 5 days; every pair are together exactly once. Delete players (teams) d, e, f, g wherever they occur. That will give you 4 weeks of four groups of three, and one week with three groups of 4. Still, every pair is together exactly once. In the week with three groups of four, choose any three of the four in from each group, and put the remaining three into a group. That’s week 5. At this point the only duplication is among that final group of 3. Each of those three has not been paired with three teams: the other three in the 4 it came from when making round 5. In your pairings, make sure that each of these three is paired with a team they have not faced before. The other three pairs will be duplicates. Overall there three teams who have each not faced two others, and six teams who have faced the same opponent twice. All other pairs of teams have been together exactly once. Hope this makes sense.  


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