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Math Help - Permutations Involving Places

  1. #1
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    Permutations Involving Places

    Hello Everyone!

    Lets say we have 6 places: A_1, A_2, B_1, B_2, C_1, C_2 and we want to pass by these places in a way that we always start by A_1 and end by A_2 and that X_1 always comes after X_2. I'm thinking of 4! / 4 = 6, I know it's 6, why I used this formula is because it works for 4 places: A_1, B_1, A_2, B_2...

    (1) What if each time, our trip starts by B or C, we get a total of 3*6=18?

    (2) What about when there are 8 places, is it 6!/6 = 120?

    Thanks for the help!
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  2. #2
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    Quote Originally Posted by rebghb View Post
    Hello Everyone!

    Lets say we have 6 places: A_1, A_2, B_1, B_2, C_1, C_2 and we want to pass by these places in a way that we always start by A_1 and end by A_2 and that X_1 always comes after X_2. I'm thinking of 4! / 4 = 6, I know it's 6, why I used this formula is because it works for 4 places: A_1, B_1, A_2, B_2...
    We can disregard A_1 and A_2 because they are fixed at the beginning and end, respectively. There are 4! ways to arrange the Bs and Cs without any restrictions; in 1/2 of these B_1 comes before B_2, and in half of the ones with B_1 before B_2 we also have C_1 before C_2. So the total number of arrangements is

    \frac{1}{2} \cdot \frac{1}{2} \cdot 4! = 6.

    (1) What if each time, our trip starts by B or C, we get a total of 3*6=18?
    I'm not sure what you mean. If you start with a B, do you have to end with a B? And do you have to start with B_1?

    (2) What about when there are 8 places, is it 6!/6 = 120?

    Thanks for the help!
    No, \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} \cdot 6!.
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  3. #3
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    I'm not sure what you mean. If you start with a B, do you have to end with a B? And do you have to start with ?

    Yes, I mean B1 ... B2 and C1 ... C2

    Regards,
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  4. #4
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    It has taken me quite sometime to begin to understand this question.
    If n\in \mathbb{Z}^+ and we have n pairs so there are 2n places.
    If we begin and end a string with members of the same pair, we can choose that pair in n ways.
    The other (n-1) pairs can be arranged in \dfrac{(2n-2)!}{2^{n-1}}

    So the total ways is n\cdot\dfrac{(2n-2)!}{2^{n-1}}

    Have I understood what you meant?
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  5. #5
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    Thank you
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