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Math Help - Permutation of the word STATISTIC

  1. #1
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    Permutation of the word STATISTIC

    Q: In how many ways can the letters in the word STATISTIC be arranged?

    Now, I am not interested in the actual answer but rather the method whereby the problem is solved. We can begin by computing the number of arrangements while neglecting the repeated letters, which yields a total number of 9! ways of arrangements.

    Now here's is where I am confused: Why do we then followingly divide by the number of arrangements that can be made of the repeated letters? The right answer is obtainable by dividing by the 2! (permutations of S), 3! (permutations of T) and again 2! (permutations of I).

    Are we not supposed to take the overcounted arrangements and successively subtract the arrangements of the repeated letters, instead of dividing?
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    Re: Permutation of the word STATISTIC

    Quote Originally Posted by MathCrusader View Post
    Q: In how many ways can the letters in the word STATISTIC be arranged?

    Now, I am not interested in the actual answer but rather the method whereby the problem is solved. We can begin by computing the number of arrangements while neglecting the repeated letters, which yields a total number of 9! ways of arrangements.

    Now here's is where I am confused: Why do we then followingly divide by the number of arrangements that can be made of the repeated letters? The right answer is obtainable by dividing by the 2! (permutations of S), 3! (permutations of T) and again 2! (permutations of I).

    Are we not supposed to take the overcounted arrangements and successively subtract the arrangements of the repeated letters, instead of dividing?
    The best way to see is to use a simple example.
    Consider ABBBCD for our 'word'. Put subscripts in.
    AB_1B_2B_3CD now we have six different letters that can be rearranged in 6! ways.

    Here are six, 3!, of them:
    B_1ACB_3DB_2
    B_1ACB_2DB_3
    B_2ACB_1DB_3
    B_2ACB_3DB_1
    B_3ACB_2DB_1
    B_3ACB_1DB_2

    Notice that the B's are in the same relative position, only the subscripts changed.
    If we remove the subscripts then all six of those would be identical.

    Thus the total is divided by 3! to account for duplication.

    Does that help.
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    Re: Permutation of the word STATISTIC

    Quote Originally Posted by Plato View Post
    The best way to see is to use a simple example.
    Consider ABBBCD for our 'word'. Put subscripts in.
    AB_1B_2B_3CD now we have six different letters that can be rearranged in 6! ways.

    Here are six, 3!, of them:
    B_1ACB_3DB_2
    B_1ACB_2DB_3
    B_2ACB_1DB_3
    B_2ACB_3DB_1
    B_3ACB_2DB_1
    B_3ACB_1DB_2

    Notice that the B's are in the same relative position, only the subscripts changed.
    If we remove the subscripts then all six of those would be identical.

    Thus the total is divided by 3! to account for duplication.

    Does that help.
    I understand your example, however I still can't we see division before subtraction is justified. We're taking a huge number 6!, i.e. the total number of arrangements, and dividing it by the arrangements of the B's i.e. 3!. We have 3! superfluous arrangements, so if one wishes to get rid of these extraneous arrangements from total number of arrangements, one would naturally want to subtract, but evidently no. Why?
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    Re: Permutation of the word STATISTIC

    Quote Originally Posted by MathCrusader View Post
    I understand your example, however I still can't we see division before subtraction is justified. We're taking a huge number 6!, i.e. the total number of arrangements, and dividing it by the arrangements of the B's i.e. 3!. We have 3! superfluous arrangements, so if one wishes to get rid of these extraneous arrangements from total number of arrangements, one would naturally want to subtract, but evidently no. Why?

    Don't you understand that there is no subtraction whatsoever in these problems.

    The number of ways to rearrange the word MISSISSIPPI is
    \frac{11!}{4!\cdot 4!\cdot 2!}. We divide to eliminate duplication.

    There is no subtraction to it.
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    Re: Permutation of the word STATISTIC

    Quote Originally Posted by Plato View Post
    Don't you understand that there is no subtraction whatsoever in these problems.

    The number of ways to rearrange the word MISSISSIPPI is
    \frac{11!}{4!\cdot 4!\cdot 2!}. We divide to eliminate duplication.

    There is no subtraction to it.
    I do understand that, but I am still asking why since I want to deepen my understanding of this concept.
    There must be rationale behind dividing in order to reduce the extraneous arrangements, as opposed to subtracting the extraneous arrangements.
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    Re: Permutation of the word STATISTIC

    Quote Originally Posted by mathcrusader View Post
    i am still asking why since i want to deepen my understanding of this concept. There must be rationale behind dividing in order to reduce the extraneous arrangements, as opposed to subtracting the extraneous arrangements.


    MI_1S_1S_2I_2S_3S_4I_3P_1P_2I_4
    That is eleven different letters. They can be rearranged in 11!

    In any one of those arrangements
    The I's represent 4! ways.
    The S's represent 4! ways.
    The P's represent 2! ways.

    So we have over-countered by 4!\cdot 4!\cdot 2! times.
    So divide.
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