Results 1 to 11 of 11

Math Help - permutation and combinations

  1. #1
    Super Member
    Joined
    Dec 2009
    Posts
    755

    permutation and combinations

    3 men and 4 women are seated at a round table. Find the number of possible arrangements if one particular woman must sit between 2 men.

    My answer: (5-1)!2! but answer is 144
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Nov 2011
    From
    Crna Gora
    Posts
    420
    Thanks
    64

    Re: permutation and combinations

    Quote Originally Posted by Punch View Post
    3 men and 4 women are seated at a round table. Find the number of possible arrangements if one particular woman must sit between 2 men.

    My answer: (5-1)!2! but answer is 144
    Since table is round answer is :

    6\cdot 4!
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    Dec 2009
    Posts
    755

    Re: permutation and combinations

    Quote Originally Posted by princeps View Post
    Since table is round answer is :

    6\cdot 4!
    Yes, the answer is right but I dont understand why it is 6x4
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    Joined
    Dec 2009
    Posts
    755

    Re: permutation and combinations

    In addition, what is wrong with 4!2!? I don't see how it is wrong...
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member
    Joined
    Nov 2011
    From
    Crna Gora
    Posts
    420
    Thanks
    64

    Re: permutation and combinations

    Quote Originally Posted by Punch View Post
    Yes, the answer is right but I dont understand why it is 6x4
    Actually correct formula is :

    \frac{3!}{(3-2)!}\cdot 4!
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member
    Joined
    Dec 2009
    Posts
    755

    Re: permutation and combinations

    Quote Originally Posted by princeps View Post
    Actually correct formula is :

    \frac{3!}{(3-2)!}\cdot 4!
    I see, I understand the 4! part, but I still dont get the 3!/(3-2)! part.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Senior Member
    Joined
    Nov 2011
    From
    Crna Gora
    Posts
    420
    Thanks
    64

    Re: permutation and combinations

    Quote Originally Posted by Punch View Post
    I see, I understand the 4! part, but I still dont get the 3!/(3-2)! part.
    That part represents number of permutations with no repetition .

    You have 3 men and 2 seats . Do you understand now ?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Super Member
    Joined
    Dec 2009
    Posts
    755

    Re: permutation and combinations

    Quote Originally Posted by princeps View Post
    That part represents number of permutations with no repetition .

    You have 3 men and 2 seats . Do you understand now ?
    So there are 3 men vying for 2 seats, but I still dont understand how it is worked out
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,605
    Thanks
    1573
    Awards
    1

    Re: permutation and combinations

    Quote Originally Posted by Punch View Post
    So there are 3 men vying for 2 seats, but I still dont understand how it is worked out
    There are three ways to select the two men who will sit on either side of that woman. There are two ways the seat those three. That gives six. Put that group at the table. Now there are 4! ways to seat the other four.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Super Member
    Joined
    Dec 2009
    Posts
    755

    Re: permutation and combinations

    Quote Originally Posted by princeps View Post
    That part represents number of permutations with no repetition .

    You have 3 men and 2 seats . Do you understand now ?
    So that is 3P2 whereby 2 man are selected to sit beside the woman. Why is there no 4C1 included to select 1 woman from the 4 to sit in between the men?

    Which gives, (choose 2men from 3)*(permute the 2men)*(choose 1 woman from 4)*(Permute in a circle)=3C2*2!*4C1*4!=576
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,659
    Thanks
    600

    Re: permutation and combinations

    Hello, Punch!

    3 men and 4 women are seated at a round table.
    Find the number of possible arrangements
    if one particular woman must sit between 2 men.

    Since we have a round table, that woman can sit anywhere.

    She will be flanked by two men.
    There are 3 choices for the man on her right, 2 choices for the man on her left.

    The remaining 4 people (one man, 3 women) can be seated in 4! ways.


    Therefore, there are: . 3\cdot 2\cdot 4! \:=\:144 arrangements.

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Counting, Permutation, and Combinations
    Posted in the Statistics Forum
    Replies: 6
    Last Post: March 27th 2010, 12:14 PM
  2. Permutation and Combinations
    Posted in the Discrete Math Forum
    Replies: 3
    Last Post: October 14th 2009, 08:13 AM
  3. permutation
    Posted in the Math Topics Forum
    Replies: 1
    Last Post: August 9th 2009, 08:54 AM
  4. Permutation and Combinations
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: March 22nd 2009, 01:38 PM
  5. Combinations/ Permutation
    Posted in the Statistics Forum
    Replies: 5
    Last Post: December 5th 2007, 03:06 PM

Search Tags


/mathhelpforum @mathhelpforum