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Thread: combinatorics

  1. #1
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    combinatorics

    Hi guys ,
    I have some combinatoric problems. There are 6 couples. All the 6 wives are to sit together. Find the number of ways if the none of the wives are sitting next to her husband? Thank you very much.

    Regards
    Pang
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  2. #2
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    Re: combinatorics

    What does that mean? Is it a circular table? Are they sitting in a row? Are they just sitting randomly (wives in a gaggle, husbands spread out throughout the room randomly and just plop down when they feel like it)? The answer depends on how they are seated.
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  3. #3
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    Re: combinatorics

    Sorry , I din mention properly. They are sitting in a row. My answer is 2592000 but the answer given 2678400. Thanks for response!
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  4. #4
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    Re: combinatorics

    So, there are 12 seats in a row. You have six women sitting in a group. We have the following arrangements:
    wwwwwwhhhhhh
    hwwwwwwhhhhh
    hhwwwwwwhhhh
    hhhwwwwwwhhh
    hhhhwwwwwwhh
    hhhhhwwwwwwh
    hhhhhhwwwwww

    The first and last arrangement are similar. The middle 4 are similar. For the first and last case, we permute the wives. The only open seat that is next to a wife can go to any husband but her own. Then we permute the remaining husbands:
    $2\cdot 6!\cdot 5\cdot 5!$

    For the rest of the cases, there are two possibilities. The husband of the wife on the right is seated next to the wife on the left or he isn't. So,we permute the wives, there are 4 husbands that are not married to either wife "on the end". Then, then, there are 4 husbands not married to the wife on the right, then we permute the remaining 4 husbands. Or, the husband of the woman on the right sits next to the woman on the left. Then there we can simply permute the remaining 5 husbands. This gives:

    $5\left( 6!\cdot 4\cdot 4\cdot 4!+6!\cdot 5!\right) $

    Total:
    2,678,400
    Last edited by SlipEternal; Oct 8th 2017 at 08:07 AM.
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  5. #5
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    Re: combinatorics

    Thanks a million!
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