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| Math Central | Quandaries & Queries |
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Question from AL, a parent: I have a golf tourney where we have fourteen players and are playing five times/rounds of golf. I am trying to pair up two threesomes and two foursomes so we don't play with the same person twice or the least amount of times. |
Al,
You can play with everyone exactly once if, on one of the days, you play in 3 foursomes and a twosome. It isn't possible to do it using two foursomes and two threesomes on each day: you need to "cover" 91 pairs of players, but a collection of rounds like that only covers 90 (each foursome covers 6 pairs and each threesome covers 3 pairs).
What's shown below is a schedule for 16 players to play each other exactly once over five rounds. If you delete players O and P everywhere they occur in the schedule, then you get the schedule that includes a twosome on one of the days. If you fix that up to always have two foursomes and two threesomes by adjusting the last day, then there are 3 pairs that will never play together, and two that will play together twice.
16 player schedule:
AEIM BFJN CGKO DHLP, AFKP BELO CHIN DGJM, AGLN BHKM CEJP DFIO,
AHJO BGIP CFLM DEKN, ABCD EFGH IJKL MNOP
Have fun!
Victoria
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