Results 1 to 6 of 6

Math Help - Ways to create 2 distinguishable committees

  1. #1
    Newbie
    Joined
    Sep 2010
    Posts
    3

    Ways to create 2 distinguishable committees

    From a group of 11 people, how many ways are to create 2 distinguishable committees in which at least one person is in both committees?


    I'm having a hard time understanding what to do with the second committee. I think for the first one, since it does not specify any number of people that have to be in either committee, there should be 2^11 ways to create it, minus 1 because there can't be an empty committee (because there must be at least one person in both). I'm not sure where to go from here to find how many ways there are to create the second committee.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,793
    Thanks
    1688
    Awards
    1
    Quote Originally Posted by IronicMan View Post
    From a group of 11 people, how many ways are to create 2 distinguishable committees in which at least one person is in both committees?
    As far I can see, there is absolutely no way to know what exactly that question means. for all the reasons you say and much more.
    Did you copy the question exactly? Can you get any clarification from the source of this question?

    If you must answer the question, then if I were you I would makes some written assumptions and proceed from there.

    This assumption makes the question interesting.
    Assume that all eleven people must be on at least one of the committees and there are equal numbers on each committee.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Sep 2010
    Posts
    3
    Quote Originally Posted by Plato View Post
    As far I can see, there is absolutely no way to know what exactly that question means. for all the reasons you say and much more.
    Did you copy the question exactly? Can you get any clarification from the source of this question?

    If you must answer the question, then if I were you I would makes some written assumptions and proceed from there.

    This assumption makes the question interesting.
    Assume that all eleven people must be on at least one of the committees and there are equal numbers on each committee.

    From what I understand, there can be any number of people in either committee, and some people do not have to be in either. So you could have a committee of 4 and another committee of 5, and three or more people could be left over (since at least one of those people in the two committees is in both). I don't see how you could count the ways to create the second group.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,793
    Thanks
    1688
    Awards
    1
    Quote Originally Posted by IronicMan View Post
    From what I understand, there can be any number of people in either committee, and some people do not have to be in either. So you could have a committee of 4 and another committee of 5, and three or more people could be left over (since at least one of those people in the two committees is in both). I don't see how you could count the ways to create the second group.
    Well, once again you have made my point.
    I would really like to know what the author meant by the question.

    Several years ago I edited a contest that had this question.
    There are an odd number of people (eleven will do). How many ways can we form two teams of equal numbers so that exactly one person is on both teams to serve as the coach of both teams?

    That is an interesting question.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member
    Joined
    Apr 2009
    Posts
    677
    @Plato - Is the answer to your question
    Let there be 2n+1 people then the ways to form two teams are
    (2n+1)*(2n)! / (2*n!*n!)
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,793
    Thanks
    1688
    Awards
    1
    Quote Originally Posted by aman_cc View Post
    @Plato - Is the answer to your question
    Let there be 2n+1 people then the ways to form two teams are
    (2n+1)*(2n)! / (2*n!*n!)
    It is indeed.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. how many distinguishable words
    Posted in the Discrete Math Forum
    Replies: 4
    Last Post: December 23rd 2010, 08:46 PM
  2. Committees with 12 men and 8 women
    Posted in the Discrete Math Forum
    Replies: 2
    Last Post: May 16th 2010, 04:40 AM
  3. Arrangements and committees.
    Posted in the Discrete Math Forum
    Replies: 0
    Last Post: September 6th 2009, 08:20 AM
  4. Problem - distinguishable permuations
    Posted in the Discrete Math Forum
    Replies: 4
    Last Post: August 15th 2009, 01:25 PM
  5. Just can't do it - COMMITTEES
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: July 18th 2009, 07:46 AM

Search Tags


/mathhelpforum @mathhelpforum