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4 March, 16:24

There are girls and boys in a chess club. The club holds a round-robin tournament in which every player plays against every other player exactly once. What fraction of the games are boy-versus-boy? Enter your answer as a fraction in simplified form.

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  1. 4 March, 16:31
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    Note: We need the number of girls and boys to calculate this question.

    This is the application of combinations;

    Suppose we have 6 boys and 6 girls participating in the tournament;

    take C to represent the combination:

    The total number of games = 12 C 2 = 66

    only boys versus boys games = 6 C 2=15

    only girls versus girls = 6 C 2=15

    One girl versus one boy game=66-15-15=36

    the fraction of the boys versus boys game = (total number of boys versus boys games) / (total number of games)

    =15/66

    =5/22

    This methodology can be used to solve similar problems with different number of boys and girls participating in the tournament
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