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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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