Results 1 to 4 of 4
Like Tree1Thanks
  • 1 Post By Plato

Math Help - Another simple probability question

  1. #1
    MHF Contributor alexmahone's Avatar
    Joined
    Oct 2008
    Posts
    1,074
    Thanks
    7

    Another simple probability question

    If n people are seated in a random manner in a row containing 2n seats, what is the probability that no two people will occupy adjacent seats?


    My solution:


    There are 2n! favourable arrangements.


    Total no. of arrangements:


    The 1st person can sit on any of the 2n seats, the 2nd person can sit on any of the other 2n-1 seats and so on. So, the number of ways to seat n people on 2n seats is \displaystyle 2n(2n-1)\cdots (n+1)=\frac{(2n)!}{n!}.


    \displaystyle P=\frac{2(n!)^2}{(2n)!}


    But the answer given in the book is \displaystyle P=\frac{(n+1)(n!)^2}{(2n)!}. Where have I gone wrong?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,617
    Thanks
    1581
    Awards
    1

    Re: Another simple probability question

    Quote Originally Posted by alexmahone View Post
    If n people are seated in a random manner in a row containing 2n seats, what is the probability that no two people will occupy adjacent seats? the answer given in the book is \displaystyle P=\frac{(n+1)(n!)^2}{(2n)!}.
    Think of a string of n zeros and n ones.
    There are \frac{(2n)!}{(n!)^2} ways to arrange that string.
    The n zeros create n+1 spaces to put the ones so that no two ones are together.
    EXAMPLE: n=4, "_0_0_0_0_" that is five places.

    Thus \binom{n+1}{n}=n+1 places seat the n people no two together.

    \frac{(n+1)}{\dfrac{(2n)!}{(n!)^2}} which is the book's answer.
    Thanks from alexmahone
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member
    Joined
    Mar 2012
    From
    Sheffield England
    Posts
    440
    Thanks
    76

    Re: Another simple probability question

    The first person can sit on any of 2n seats but the second person cant sit in next seat so doesnt have a choice of 2n-1 seats.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor alexmahone's Avatar
    Joined
    Oct 2008
    Posts
    1,074
    Thanks
    7

    Re: Another simple probability question

    Quote Originally Posted by biffboy View Post
    The first person can sit on any of 2n seats but the second person cant sit in next seat so doesnt have a choice of 2n-1 seats.
    Actually, I was counting the total number of arrangements.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Simple probability question
    Posted in the Statistics Forum
    Replies: 11
    Last Post: March 30th 2012, 05:13 PM
  2. simple probability question
    Posted in the Statistics Forum
    Replies: 1
    Last Post: November 2nd 2010, 12:11 AM
  3. simple probability question
    Posted in the Statistics Forum
    Replies: 8
    Last Post: September 19th 2009, 01:22 PM
  4. Simple probability question...
    Posted in the Advanced Statistics Forum
    Replies: 6
    Last Post: April 22nd 2008, 03:04 AM
  5. simple probability question
    Posted in the Advanced Statistics Forum
    Replies: 2
    Last Post: April 12th 2008, 05:23 PM

Search Tags


/mathhelpforum @mathhelpforum