



 
Gordon, There are 50 choose 3 = 19,600 different triples of numbers from among 1, 2, ..., 50. There are 5 choose 3 = 10 triples of numbers on each ticket. Thus, at least 1960 tickets are required. I think it is possible to prove that, in this case, there is no collection of 1960 tickets that does the job. In that case I doubt that the actual minimum number of tickets is known. Problems like this are remarkably difficult. Victoria  


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