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Thread: Combinatorics: Word Arrangement Problem

  1. #1
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    Combinatorics: Word Arrangement Problem

    A screenshot to my problem: http://prntscr.com/md3hhc

    Combinatorics: Word Arrangement Problem-mathematics.png

    I found a Chegg answer to the word WISCONSIN, and I understand that you separate each letter into its appearance but afterwards they didn't really explain much about how to solve it. They just solved it with no explanation and I can't really follow along.

    Thanks for the help!
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  2. #2
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    Re: Combinatorics: Word Arrangement Problem

    Quote Originally Posted by Psych View Post
    A screenshot to my problem: http://prntscr.com/md3hhc
    Click image for larger version. 

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    There are five letters that do not repeat: $\_\_\_H\_\_\_E\_\_\_I\_\_\_C\_\_\_S\_\_\_$ we use those to separate those that do repeat.
    Note that the five create six spaces for the repeaters: $MMAATT$
    The separators can be arranged in $5!$ ways and the repeaters can be arranged in $\dfrac{6!}{(2!)^3}$ ways.
    So what is the answer?
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  3. #3
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    Re: Combinatorics: Word Arrangement Problem

    Here's how I would do that. Imagine labeling each of the repeated letters with a subscript so you can tell them apart: $\displaystyle WI_1S_1CON_1S_2I_2N_2$. Then there are 9 distinct letters so 9! ways to arrange them. But two of those 9! arrangements are, for example $\displaystyle WI_1S_1CON_1S_2I_2N_2$ and $\displaystyle WI_2S_1CON_1S_2I_1N_2$, exactly the same except that the two "I"s have been swapped. That is, for every one of the 9! ways to arrange WISCONSIN, there is another that has only the "I"s swapped that we don't want to count as "different". To allow for that, divide by 2!= 2, the number of ways to arrange, or swap, the two "I"s. Similarly with the two "S"s and two "N"s. The number of distinct arrangements is $\displaystyle \frac{9!}{2!2!2!}= \frac{362880}{8}= 45360$.

    More generally, if a word has $\displaystyle m$ letters, $\displaystyle n_1$ of which are the same, $\displaystyle n_2$ of which are the same but different from the first, … to $\displaystyle n_i$ of which are the same, then the number of distinct arrangements is $\displaystyle \frac{m!}{n_1!n_2!\cdot\cdot\cdot n_i!}$. Notice that, since 1!= 1, we can include "n"s that are equal to 1 requiring that $\displaystyle n_1+ n_2+ \cdot\cdot\cdot+ n_i= m$.
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