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Math Problem
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Thread: Math Problem

  1. #1
    Member vcuber13's Avatar
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    Lightbulb Math Problem

    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:
    Spoiler:

    The answer is 70.


    My thoughts:
    Spoiler:

    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.
    Last edited by vcuber13; 04-27-2010 at 05:34 PM.
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  2. #2
    Member rjohnson_8ball's Avatar
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    Default

    Spoiler:
    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?
    Last edited by rjohnson_8ball; 04-27-2010 at 05:46 PM.

  3. #3
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    Default My Idea

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

    Spoiler:
    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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  5. #5

    Default

    Spoiler:
    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:math.png

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