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Math Help - Permatations

  1. #1
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    Permatations

    OK... I am in the procsess of building a spreadsheet to help me with route planning but I think I may have stumbled on a problem that someone might be able to help me with.

    To put it simply, I have 14 objects and need to list the different combinations that they can be ordered into.

    e.g 1,2,3,4,5,6,7,8,910,11,12,13,14
    or
    1,3,2,4,5,6,7,8,9,10,11,12,13,14


    Am I right in saying that there are over eighty seven billion results / combinations or am I working this out wrong?
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  2. #2
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    Re: Permatations

    There are 14! ways to arrange the numbers 1-14

    This turns out to be about 8.72 \cdot 10^{10}
    Last edited by e^(i*pi); August 3rd 2011 at 11:28 AM. Reason: inserting the word "about" since Plato gave a more accurate answer
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  3. #3
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    Re: Permatations

    14!=87178291200
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    Re: Permatations

    OK.

    Well is there any way to find out the number of results under certain rules?

    1 has to be before 2
    3 has to be before 4
    5 has to be before 6
    7 has to be before 8
    9 has to be before 10
    11 has to be before 12
    13 has to be before 14

    By before I mean to the left of. so it can't start with 2,1 etc.

    Is there anyway I can find out how many that leaves me with?
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  5. #5
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    Re: Permatations

    Quote Originally Posted by thegooser123 View Post
    Well is there any way to find out the number of results under certain rules?
    1 has to be before 2
    3 has to be before 4
    5 has to be before 6
    7 has to be before 8
    9 has to be before 10
    11 has to be before 12
    13 has to be before 14
    Is there anyway I can find out how many that leaves me with?
    As I understand it, we could have: 1,13,2,11,5,7,3,14,6,9,8,12,4,10.
    Is that correct.
    If so, there are \frac{14!}{2^7}=681080400 possible such strings.
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    Re: Permatations

    Could you tell me what the results would be with the same criteria but with numbers 1 t0 8 and also numbers 1 to 6?
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  7. #7
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    Re: Permatations

    for 8 it would be 2520,and for 6 it would be 90.
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    Re: Permatations

    Thanks for all your help.
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  9. #9
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    Re: Permatations

    Quote Originally Posted by thegooser123 View Post
    Could you tell me what the results would be with the same criteria but with numbers 1 t0 8 and also numbers 1 to 6?
    If we go with these rules
    1 has to be before 2
    3 has to be before 4
    5 has to be before 6
    7 has to be before 8
    then \frac{8!}{2^4}

    In general for \{1,2,\dots,2n\}, \frac{(2n)!}{2^n}.
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  10. #10
    MHF Contributor Also sprach Zarathustra's Avatar
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    Re: Permatations

    Quote Originally Posted by Plato View Post
    If we go with these rules
    1 has to be before 2
    3 has to be before 4
    5 has to be before 6
    7 has to be before 8
    then \frac{8!}{2^4}

    In general for \{1,2,\dots,2n\}, \frac{(2n)!}{2^n}.

    Plato, could you maybe explain how you came to that formula?


    Thanks!
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  11. #11
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    Re: Permatations

    Quote Originally Posted by Also sprach Zarathustra View Post
    Plato, could you maybe explain how you came to that formula?
    If n=5 then there are 10! ways to arrange the 2(5) numbers.
    If half of them the 1 comes before the 2. Also in half the 3 comes before the 4. Just continue that pattern.
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