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Thread: 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: $\displaystyle MI_1S_1S_2I_2S_3S_4I_3P_1P_2I_4$
    There are $\displaystyle 11!$ ways to rearrange those eleven distinct letters.
    Now we drop the subscripts we must divide by $\displaystyle 2!\cdot 4!\cdot 4!$.
    Thus the number of ways to rearrange the word MISSISSIPPI is $\displaystyle \frac{11!}{2!\cdot 4!\cdot 4!}$
    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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