Math CentralQuandaries & Queries


Question from Brad, a parent:

I am organizing a golf trip with 12 golfers. We will play 5 rounds (3 foursomes each round)
total and play two 9 holes matches per round (18holes). So two matches are created within each foursome. What are the possible combinations of foursomes so that everyone plays each other at least once with the least number of repeat matches?
For example: Here are the foursomes for round 1:(1,2,3,4) (5,6,7,8) (9,10,11,12)
match 1 round 1: 1vs2, 3vs4, 5vs6, 7vs8, 9vs10,11vs12
we can not split foursomes up for the second match (9 hole matches):
So match 2 round 1: 1vs3, 2vs4,5vs7,6vs8,9vs11,10vs12

Hopefully this makes sense! Any help would be much appreciated!!


There are 66 possible two-player matches with a group 12 players. Each 18 hole round accounts for 6 matches in the first 9 holes, and 6 more in the second 9 holes. With five 18 hole rounds you can play at most 60 matches. That is, the schedule you want does not exist. It seems unlikely that a schedule with 60 matches exists either, although the reason why would be more technical. Can you revise your plans a bit?


About Math Central


Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.
Quandaries & Queries page Home page University of Regina PIMS