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Math Help - a prob ques

  1. #1
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    a prob ques

    question:

    a football team consists of 20 offensive and 20 defensive players. the players are to be paired in grps of 2 for the purpose of determining roomates. if the pairing is done at random, what is the prob that there are no offensive-defensive roommate pairs?

    the ans goes like this: the denominator of the prob is (40)!/((2^20)(20)!) and the numerator is ((20)!/((2^10)(10)!))^2.

    why is there the (20)! at the bottom of the denominator and (10)! at the bottom of the numerator.

    the book kinda explained that its 'cos its unordered pair, but i don't understand why have to divide by (20)! & (10)!?
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    Quote Originally Posted by bitzz View Post
    a football team consists of 20 offensive and 20 defensive players. the players are to be paired in grps of 2 for the purpose of determining roomates. if the pairing is done at random, what is the prob that there are no offensive-defensive roommate pairs?
    the ans goes like this: the denominator of the prob is (40)!/((2^20)(20)!) and the numerator is ((20)!/((2^10)(10)!))^2. why is there the (20)! at the bottom of the denominator and (10)! at the bottom of the numerator. the book kinda explained that its 'cos its unordered pair, but i don't understand why have to divide by (20)! & (10)!?
    Because the pairs are not labeled there is no way to distinguish them one from another. If we had the pairs labeled A to T then there would be not need for the for the (20!). If Pete and John are paired together it makes no difference when the two are chosen. However it may make a difference if the pair is assigned to room Q and not to room A. The first is an unordered partition whereas the second is an ordered partition.
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