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Math Help - Combinations

  1. #1
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    Combinations

    solved
    Last edited by john-1; November 30th 2009 at 11:11 AM.
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  2. #2
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    I'm assuming in all of these that when you say 'starting lineup', you are referring to the final team and not just the effect the problem has on it.

    A) \binom{2}{1}\binom{4}{2}\binom{6}{1}\binom{5}{1} = 2*6*6*5 = 360

    One center out of 2 possible, 2 guards out of 4 possible, then you have 1 power forward out of 6 possible, and 1 small forward out of the remaining 5 forwards.

    B) \binom{1}{1}\binom{5}{2}\binom{6}{2} = 1*10*15 = 150

    You only have 1 center, then you have 5 possible forwards, of which you need 2. What's left is 3 guards and 3 forwards, of which you need 2.

    C) The wording here is rather poor, I take it to mean 'is the same result achieved in problem B if the guards are free to change positions rather than the forwards?'. As it does not specify if your players are sick or not, I may be doing the wrong problem.

    \binom{1}{1}\binom{3}{2}\binom{6}{2} = 1*3*15 = 45

    You have your one center, then 3 choices to pick from for 2 guards. You have 1 guard remaining to toss into the forward position, so 6 players to pick 2 from. Therefor: No, the same result is not achieved.
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