Results 1 to 4 of 4

Math Help - Combinations with restrictions.

  1. #1
    Senior Member
    Joined
    Oct 2009
    Posts
    295
    Thanks
    9

    Combinations with restrictions.

    Four married couples have bought 8 seats in the same row for a concert. In how many different ways can they be seated
    (a) with no restrictions?
    8!=40320

    (b) if each couple is to sit together?
    2(4!)=48

    (c) if all the men sit together to the right of all the women?
    4!+4!=48


    I just want to make sure that they're right. Thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5

    Re: Quick combination question

    Quote Originally Posted by downthesun01 View Post
    Four married couples have bought 8 seats in the same row for a concert. In how many different ways can they be seated
    (a) with no restrictions?
    8!=40320

    (b) if each couple is to sit together?
    2(4!)=48

    (c) if all the men sit together to the right of all the women?
    4!+4!=48


    I just want to make sure that they're right. Thanks
    I think the + should be a x in your answer to part (c), and therefore the answer should be .....
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,649
    Thanks
    1597
    Awards
    1

    Re: Combinations with restrictions.

    Quote Originally Posted by downthesun01 View Post
    Four married couples have bought 8 seats in the same row for a concert. In how many different ways can they be seated
    (a) with no restrictions?
    8!=40320

    (b) if each couple is to sit together?
    2(4!)=48

    (c) if all the men sit together to the right of all the women?
    4!+4!=48
    Part (b) is a gross under-count. The answer should be 2^4(4!). Now you tell us why that is correct.

    As Mr F points out, part (c) is multiplicative and is missing one factor.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member
    Joined
    Oct 2009
    Posts
    295
    Thanks
    9

    Re: Combinations with restrictions.

    I see why b and c are wrong now. For B, I thought it was enough to multiply by 2, but after thinking about it, it makes sense that I have to multiply by 2 four times (once of each couple).

    C was more of a slight mental error on my part.

    Thank you for the help
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Counting combinations of tuples with restrictions
    Posted in the Discrete Math Forum
    Replies: 0
    Last Post: December 14th 2009, 04:59 PM
  2. Restrictions for Max./Min. Problem
    Posted in the Calculus Forum
    Replies: 1
    Last Post: September 5th 2009, 12:43 AM
  3. Combinations seating order restrictions
    Posted in the Statistics Forum
    Replies: 1
    Last Post: January 26th 2009, 05:33 AM
  4. restrictions
    Posted in the Algebra Forum
    Replies: 2
    Last Post: December 4th 2008, 10:11 AM
  5. restrictions for x
    Posted in the Algebra Forum
    Replies: 2
    Last Post: July 27th 2008, 12:33 PM

/mathhelpforum @mathhelpforum