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Math Help - Distinguishable permutations of AAABBCC. Help!

  1. #1
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    Unhappy Distinguishable permutations of AAABBCC. Help!

    Find the number of distinguishable permutations of the given letters "AAABBCC''.

    First I thought I could just do this :

    A1A2A3B1B2C1C2
    """"""""""""" C2C1
    """""""""B2B1C1C2
    """"""""""""" C2C1
    A1A3A2B1B2C1C2...
    so 2 start with A1, 2 with A2, and 2 with A2
    then for every set of A there are two sets of B
    and for every set of B there are two more sets of C
    But apparently it's wrong.
    Can anyone tell me why and more importantly where does this formula 7!/(3!2!2!) come from.
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  2. #2
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    Re: Distinguishable permutations of AAABBCC. Help!

    Quote Originally Posted by SillyMath View Post
    Find the number of distinguishable permutations of the given letters "AAABBCC''.
    First I thought I could just do this :

    A1A2A3B1B2C1C2
    """"""""""""" C2C1
    """""""""B2B1C1C2
    """"""""""""" C2C1 if
    A1A3A2B1B2C1C2...
    so 2 start with A1, 2 with A2, and 2 with A2
    then for every set of A there are two sets of B
    and for every set of B there are two more sets of C
    But apparently it's wrong.
    Can anyone tell me why and more importantly where does this formula 7!/(3!2!2!) come from.
    In North America this is known as the Mississippi problem
    Consider "MISSISSIPPI" Put subscripts an the repeated letters: MI_1S_1S_2I_2S_3S_4I_3P_1P_2I_4
    There are 11! ways to rearrange those eleven distinct letters.
    Now we drop the subscripts we must divide by 2!\cdot 4!\cdot 4!.
    Thus the number of ways to rearrange the word MISSISSIPPI is \frac{11!}{2!\cdot 4!\cdot 4!}
    You can generalize that.
    Thanks from SillyMath
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  3. #3
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    Re: Distinguishable permutations of AAABBCC. Help!

    Oh so I'm right! Good! Silly webworks! Thanx!
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