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 Question from Mike, a teacher: Dear Math Central: I have a combination problem where I need to identify 1 EXACT combination out of the possible thousands that will occur. So here's my problem: I need to schedule 18 high school teams to play each other in a "Jeopardy!"-like academic competition. Teams compete against each other 6 at a time (simultaneously) in 3 different rooms. What will the minimum amount of matches (I believe 6 ?) we would have to play to ensure that each team plays every other team exactly twice? And more importantly: what would that schedule look like? The best I've come up with is having most teams playing twice, but with 2 pairs of teams having to play either only once or sometimes 3 times. Thanks in advance for any help you can give! :-) Mike

Mike,

You are going to need at east 7 rounds and even then I can't guarantee that things will work out the way you want.

There are 18 choose 2 pairs to be accommodated twice, making $9 \times 17 \times 2 = 306$ pairs in all.
Each round accommodates 6 choose 2 in each room times 3 rooms. 6 choose 2 is 15 so you have 45 pairs per round. 6 rounds will therefore account for only 270 pairs. With 7 rounds there are 315 pairs so necessarily some pairs will play together 3 times. My guess is that no matter how clever the schedule, even with 7 rounds there will remain some pairs playing together only once.

Chris

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