Results 1 to 5 of 5

Thread: Permutations Involving Places

  1. #1
    Member
    Joined
    Jan 2010
    Posts
    133

    Permutations Involving Places

    Hello Everyone!

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

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

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

    Thanks for the help!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Mar 2008
    Posts
    934
    Thanks
    33
    Awards
    1
    Quote Originally Posted by rebghb View Post
    Hello Everyone!

    Lets say we have 6 places: $\displaystyle A_1, A_2, B_1, B_2, C_1, C_2$ and we want to pass by these places in a way that we always start by $\displaystyle A_1$ and end by $\displaystyle A_2$ and that $\displaystyle X_1$ always comes after $\displaystyle X_2$. I'm thinking of $\displaystyle 4! / 4 = 6$, I know it's 6, why I used this formula is because it works for 4 places: $\displaystyle A_1, B_1, A_2, B_2$...
    We can disregard $\displaystyle A_1$ and $\displaystyle A_2$ 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 $\displaystyle B_1$ comes before $\displaystyle B_2$, and in half of the ones with $\displaystyle B_1$ before $\displaystyle B_2$ we also have $\displaystyle C_1$ before $\displaystyle C_2$. So the total number of arrangements is

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

    (1) What if each time, our trip starts by $\displaystyle B$ or $\displaystyle C$, we get a total of $\displaystyle 3*6=18$?
    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 $\displaystyle B_1$?

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

    Thanks for the help!
    No, $\displaystyle \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} \cdot 6!$.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Jan 2010
    Posts
    133
    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,
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Aug 2006
    Posts
    21,737
    Thanks
    2812
    Awards
    1
    It has taken me quite sometime to begin to understand this question.
    If $\displaystyle n\in \mathbb{Z}^+$ and we have $\displaystyle n$ pairs so there are $\displaystyle 2n$ places.
    If we begin and end a string with members of the same pair, we can choose that pair in $\displaystyle n$ ways.
    The other $\displaystyle (n-1)$ pairs can be arranged in $\displaystyle \dfrac{(2n-2)!}{2^{n-1}}$

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

    Have I understood what you meant?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Jan 2010
    Posts
    133
    Thank you
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 4
    Last Post: Mar 6th 2012, 02:05 PM
  2. Replies: 3
    Last Post: Mar 5th 2012, 04:53 PM
  3. Permutations involving playing cards
    Posted in the Advanced Statistics Forum
    Replies: 2
    Last Post: Aug 12th 2011, 04:06 AM
  4. Problem involving combinations and permutations
    Posted in the Discrete Math Forum
    Replies: 5
    Last Post: Jun 1st 2011, 11:36 PM
  5. Proof involving permutations
    Posted in the Advanced Math Topics Forum
    Replies: 3
    Last Post: Feb 10th 2010, 01:03 AM

Search Tags


/mathhelpforum @mathhelpforum