



 
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!  


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