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| Math Central | Quandaries & Queries |
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Question from Steve: We are having a golf tournament. We have 10 golfers and four rounds of golf in two days. We are going to break the group of 10 into two flights. Flight A and Flight B. We would like all members in Flight A to play one nine hole match against each other. All players in Flight B to play one nine hole match against each other. Could we play in groups of 4,3 and 3? What is the best way to make the pairings work? |
Steve,
Each flight needs ten pairings, for a total of 20 required:
AB AC AD AE BC BD BE CD CE DE ab ac ad ae bc bd be cd ce de
A foursome and two threesomes provide twelve pairings:
EG: ABCD Eab cde -> AB AC AD BC BD CD Ea Eb bc cd ce de
so the task looks fairly easy. Three rounds will more than do it:
ABCD Eab cde ; ABE CDe abcd ' CDE ABab abe
You cannot do even one flight in two rounds: with a threesome & a pair within a flight at most four pairs are covered, so a foursome within the flight is needed: eg, ABCD. But then E must play four others in the next round which cannot be achieved even within a foursome.
Good Hunting!
RD
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