# Math Problem



## vcuber13 (Apr 28, 2010)

I did this math contest and the answers are now posted and this one question I can't figure out, and I was hoping someone here could help me.

Six soccer teams are competing in a tournament in Waterloo. Every team is to play three games, each against a different team. (Note that not every pair of teams plays a game together.) Judene is in charge of pairing up the teams to create a schedule of games that will be played. Ignoring the order and times of the games, how many different schedules are possible?

The answer is:


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The answer is 70.



My thoughts:


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Whenever I tried I was getting either 120 or 60, but I think I was going about it wrong.

What I was doing was \( 6!/3!=120 \) since there was 6 teams and 3 games each.


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## rjohnson_8ball (Apr 28, 2010)

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There are (6*5)/2 possible pairings for the first game. (Each team can play one of the remaining teams. Divide by 2 because A vs. B is the same as B vs. A.) Once that game has been played there are (6*4)/2 pairings for the 2nd game. Finally, there are (6*3)/2 pairings for the final game. Finally, the order of the 3 games should not matter, so divide the result by 3!. So I get 6*5*6*4*6*3/(2*2*2*3*2) = 270. Are you sure it should be 70? Is my logic wrong? If so, why, and what is the correct logic?


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## CuBeOrDiE (Apr 28, 2010)

*My Idea*

I know it's wrong, but here's my idea: 



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Each team can play 5 other teams for the first game, 4 for the second, and 3 for the third. So each team can play 5x4x3= 60 ways. BUT, you have to divide by 3! since order of games doesn't matter. So we have 10. Since there are 6 teams, there are 6x10= 60 combinations. BUT, since Avs.B is the same as Bvs.A, we divide by two to get the answer, 30.


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## Meep (Apr 28, 2010)

Detailed solutions: http://www.cemc.uwaterloo.ca/contests/past_contests/2010/2010PascalSolution.pdf


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## Stefan (Apr 28, 2010)

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Start with 3 3 3 3 3 3, pick a team and assign its opponents, 10 ways to do that all resulting in a 3 3 2 2 2 situation. The rest as picture:


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