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Permutations matches in basketball league

  1. Mar 4, 2008 #1
    1. The problem statement, all variables and given/known data
    Suppose that a basketball league has 32 teams, split into two conferences of 16 teams each. Each conference is split into three divisions. Suppose that the North Central Division has five teams. Each of the teams in the North Central Division plays four games agains each of the other teams in this division, three games against each of the 11 remaining teams in the conference, and two games against each of the 16 teams in the other conference. In how many different orders can the games of one of the teams in the North Central DIvision be scheduled?



    2. Relevant equations
    [tex]\frac{n!}{n_1n_2 \cdots n_k}[/tex]


    3. The attempt at a solution
    Well, I think it would be [tex]\frac{81!}{4!^4 \times 11!^3 \times 16!^2}[/tex]
    But, it is a hard stuff to compute this if we do not use calculator. And, I believe it has implementation with how to distribute [tex]82[/tex] objects into [tex]n[/tex] boxes. Unfortunately, I do not know how to find [tex]n[/tex].
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Mar 5, 2008 #2
    I don't think you are understanding what a permutation is exactly. The definition of a permutation is:

    [tex]_nP_k = \frac{n!}{(n-k)!}[/tex]

    The reason for this is because, in the beginning, you have n objects from which to choose. For the second object, you have n-1 choices. Third object, you have n-2 choices. Kth object, you have n-k+1 choices. This multiplication is equivalent to the permutation definition above.

    In your case, how many total games are there? Are the independent of each other--that is, does playing one game affect whether you can play any of the other games? Think about how many total choices you have for the first game, how many total choices you have for the second game, etc.
     
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