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Thread: Permutation of Letters problem - please help me think it through

  1. #1
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    Permutation of Letters problem - please help me think it through

    Problem: "How many distinguishable 11-letter 'words' can be formed using the letters in
    MISSISSIPPI?"

    Because "words" is in quotations, I interpret that to simply mean 11-letter permutations and not real language words.

    While each permutation must contain 11 letters, only 4 of those are unique: M (exactly 1 occurrence), I (exactly 4 occurrences), S (exactly 4 occurrences), and P (exactly 2 occurrences).

    11! cannot be the answer because that would produce many duplicate permutations (whereas the problem calls only for distinguishable ones). So the answer must be 11!-x where x is the number of duplicates.

    Here I get stuck. Any help will be appreciated.

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  2. #2
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    Re: Permutation of Letters problem - please help me think it through

    Google is well aware of this!
    Example:
    https://math.stackexchange.com/quest...ers-in-mississ
    Thanks from NicholasDeMaio
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  3. #3
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    Re: Permutation of Letters problem - please help me think it through

    Quote Originally Posted by NicholasDeMaio View Post
    Problem: "How many distinguishable 11-letter 'words' can be formed using the letters in
    MISSISSIPPI?"
    Because "words" is in quotations, I interpret that to simply mean 11-letter permutations and not real language words.
    While each permutation must contain 11 letters, only 4 of those are unique: M (exactly 1 occurrence), I (exactly 4 occurrences), S (exactly 4 occurrences), and P (exactly 2 occurrences).
    Consider these two 'word' $MISSISSIPPI$ and $MI_1S_1S_2I_2S_3S_4I_3P_1P_2I_4$
    I hope that you agree those are indeed different letter strings.
    And these are three different arangements:
    $MI_1S_4S_2I_2S_1S_3I_3P_1P_2I_4$, $MI_1S_2S_3I_2S_4S_1I_3P_1P_2I_4$, $MI_1S_4S_24_2S_3S_1I_3P_1P_2I_4$ (places are the same but S-subscripts different)

    To explain the point consider the string $"O_1UTDO_2O_3RS"$ made up of eight different characters.
    We can rearrange that string in $8!$ ways.
    How many ways rearrange the subscripted O's leaving all other letters fixed? Is it $3!~?$ WHY?
    In the list of the eight factorial rearrangements with subscripts, lets go through and delete all subscripts.
    Now in that list we would have $3!=6$ identical strings without subscripts in fixed positions.
    So we need count only one out of six.
    Thus, there are $\dfrac{8!}{3!}$ ways to rearrange the word 'OUTDOORS'
    There are $\dfrac{11!}{(4!)^2(2!)}$ ways to rearrange the word 'MISSISSIPPI'
    Last edited by Plato; Apr 21st 2017 at 01:05 PM.
    Thanks from HallsofIvy
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