Results 1 to 7 of 7

Math Help - Comb application 2

  1. #1
    Member
    Joined
    Jun 2009
    Posts
    108

    Comb application 2

    in how many ways can a committee of five be selected from 8 women and 5 men.

    a) without restriction
    b) if there must be three women on the committee
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,404
    Thanks
    1293
    Quote Originally Posted by johnsy123 View Post
    in how many ways can a committee of five be selected from 8 women and 5 men.

    a) without restriction
    b) if there must be three women on the committee
    a) There are 13 people to choose from and you must choose 5.

    So the number of combinations of 5 people from 13 is \displaystyle {13\choose{5}} = \frac{13!}{5!(13 - 5)!}.

    For part b) does that mean there must be EXACTLY three women or AT LEAST three women?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Jun 2009
    Posts
    108
    never mind guys, i worked it out.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,404
    Thanks
    1293
    You're welcome :P
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Jun 2009
    Posts
    108
    Your working out is wrong, because you can't simply work out the number of combinations as there are independent variables, so what i did is i worked out the combinations of which you can variate 4 men and 8 women into a committee of 5, there were 5.

    -If i knew how to use the symbols on MHF, i would show you what i done.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,404
    Thanks
    1293
    Quote Originally Posted by johnsy123 View Post
    Your working out is wrong, because you can't simply work out the number of combinations as there are independent variables, so what i did is i worked out the combinations of which you can variate 4 men and 8 women into a committee of 5, there were 5.

    -If i knew how to use the symbols on MHF, i would show you what i done.
    But there are 5 men... And since you're choosing any 5 people out of those 13, it doesn't matter how many men and women there are. It's just the number of ways you can pick 5 people from 13, i.e. \displaystyle {13\choose{5}}.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by Prove It View Post
    But there are 5 men... And since you're choosing any 5 people out of those 13, it doesn't matter how many men and women there are. It's just the number of ways you can pick 5 people from 13, i.e. \displaystyle {13\choose{5}}.
    I agree.

    I assume the OP no longer needs (or no longer thinks s/he needs) help with part (b) ....?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. comb or perm?
    Posted in the Discrete Math Forum
    Replies: 4
    Last Post: August 5th 2010, 07:16 AM
  2. comb or perm?
    Posted in the Discrete Math Forum
    Replies: 6
    Last Post: August 3rd 2010, 04:23 AM
  3. probility and comb and perms
    Posted in the Advanced Statistics Forum
    Replies: 0
    Last Post: April 22nd 2010, 03:03 AM
  4. Non periodic Dirac Comb (Shah) Function
    Posted in the Advanced Math Topics Forum
    Replies: 0
    Last Post: April 14th 2010, 03:49 AM
  5. Perm/Comb/Pascal Math/Geometry- Please help!
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: May 19th 2008, 08:56 AM

Search Tags


/mathhelpforum @mathhelpforum