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Math Help - seating arrangement variation

  1. #1
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    seating arrangement variation

    I would like to extend our old circular permutation problem to 10 boys and 4 girls.

    That is, "how many arrangements are possible with 10 boys and 4 girls if no 2 girls can set together?".

    Anyone wanna tackle this one. It's a little rougher than the 5 boys and 3 girls.

    We know there are 13! possible arrangements, altogether. Big number.
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    Quote Originally Posted by galactus
    I would like to extend our old circular permutation problem to 10 boys and 4 girls.

    That is, "how many arrangements are possible with 10 boys and 4 girls if no 2 girls can set together?".

    Anyone wanna tackle this one. It's a little rougher than the 5 boys and 3 girls.

    We know there are 13! possible arrangements, altogether. Big number.
    Extending the concept of counting orbits you have,
    \frac{1}{14}S\cdot 4!\cdot 10!
    Where, S
    The the number of different position without rotations, for example,
    Code:
    G*G*G*G******
    G*G*G**G*****
    G**G**G**G***
    AND SO ON
    The problem reduces to finding S.
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  3. #3
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    My I ask, what exactly is an Orbit?.
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    "Where, S (is) the the number of different position without rotations"
    But that is the whole point of this new question. Finding S is the hard part.
    With 5 & 3, it was easy by listing. But in this case it is more difficult.
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    Quote Originally Posted by galactus
    My I ask, what exactly is an Orbit?.
    If you studied group theory I would be happy to give you a lecture on the concept of G-sets.
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    Thanks for the offer, PH, but I haven't studied much group theory. Wish I could say I had. Fields, rings and the occasional -morphism.

    Could you recommend a good introductory text on group theory?. I have a nice abstract algebra text.

    You don't need that to figure this problem up, though, but it is an interesting approach.
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    BUT the point is that ‘group theory’ cannot solve this counting problem.
    Given 10 blue beads and four red beads, how many ways can one form a ring of the beads no the red beads are adjacent?
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    Quote Originally Posted by Plato
    BUT the point is that ‘group theory’ cannot solve this counting problem.
    Given 10 blue beads and four red beads, how many ways can one form a ring of the beads no the red beads are adjacent?
    Yes group theory can solve this problem. You need to consider flips as well as rotations. This gives you a full dihidrel group D_{14} having order of 2(14)=28
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    Quote Originally Posted by ThePerfectHacker
    Yes group theory can solve this problem. You need to consider flips as well as rotations. This gives you a full dihidrel group D_{14} having order of 2(14)=28
    You are once again wrong!
    The answer is 21.
    What did you did you do?
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    Quote Originally Posted by Plato
    You are once again wrong!
    How?
    You simply take the total of all premutations and divide it by 28
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    10\cdot{9}\cdot{4}\cdot{3!}\cdot{8!}\cdot{21}
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    Quote Originally Posted by galactus
    10\cdot{9}\cdot{4}\cdot{3!}\cdot{8!}\cdot{21}
    What is that?
    It cannot be "S".
    Because "S" represents the different possible seating positions.
    Since there are 14 different slots for the girls. The number of different assigments is,
    S={14 \choose 4}, S cannot exceed this number.
    ---
    Here is my method of Finding S,
    Find the total possible positions which is given above by the combination. Then subtract the undesired. For example, subtract when all girls are together. There are 11 such instances. Then subtract 3 girls together 1 seperate.... This method which I posted before might be slightly time consuming because you need to count S but S is not such a large number.
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  13. #13
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    Another way:

    C(10,4)9!4!=1,828,915,200
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