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Math Help - Permutations question...

  1. #1
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    Permutations question...

    In how many ways can 12 basketball players be assigned to six double rooms?

    The answer is
    12!
    __
    (2!)^6

    But i dont understand why its done that way.
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  2. #2
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    The number of rearragements of the string "AABBCC" is \frac{6!}{(2!)^3}.
    Because there are six letters, but three letters are repeat twice.
    So we divide to account for that.
    So think about a string "AABBCCDDEEFF". That is six rooms A-F, two to a room.
    Then the team roster can correspond to any one of the arrangements of the 12-length string.
    Compare your given answer with this.
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  3. #3
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    Hello, Saibot!

    In how many ways can 12 basketball players be assigned to six double rooms?

    The answer is: . \frac{12!}{(2!)^6}

    But i don't understand why it's done that way.

    We could simply line up the twelve players: . ABCD{E}FGHIJKL
    . .
    There are: 12! permuations.

    Then partition them in pairs: . AB|CD|EF|GH|IJ|KL

    So that A,B go to room #1 . . . C,D go to room #2 . . . etc.

    But among the 12! possible permutations of the twelve players,
    . . there is a lot of duplication.


    With our game plan, another permuation would be: . BACD{E}FGHIJKL

    But this partitioning, BA|CD|EF|GH|IJ|KL
    . . is identical to the first room assignment.
    . . ( AB is in #1, CD is in #2, etc.)

    So if a pair is reversed, it results in a duplicated room assignment.


    How many reversals are there?
    Each of the six pairs can be reversed or not-reversed.
    So there are: . 2^6 possible reversals (each creating a duplication).

    To eliminate the duplications, we divide by 2^6


    Therefore, the number of room assignments is: . \frac{12!}{2^6}


    I hope all that is clear enough . . .

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  4. #4
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    Thankks

    very very gj

    now i jsut have one moer question :'(
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