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Number of ways to form a committee with men and women

  1. Mar 26, 2013 #1
    1. The problem statement, all variables and given/known data
    From a group of 5 women and 7 men, how many different committees con-
    sisting of 2 women and 3 men can be formed? What if 2 of the men are
    feuding and refuse to serve on the committee together?

    Part 1:
    my attempt:
    for men, we have 7 choose 3
    for woman, we have 5 choose 2.
    multiply together, and we get 350 possible committees

    Part 2 - what if 2 of the men are feuding.
    I'm trying to grasp the concept of that and what it means and how to build an equation for it. So we have 7 men and any 2 can be feuding, so if A and B are feuding, A can still serve with C, D, E, F, or G. So that means 5 ways for A. B can do the same, so 5 ways for B, 5 ways for C, 5 ways for D, 5 ways for E which is 30 ways. The men can be swapped around in 30 ways.
    Women can still be grouped in to 5 choose 2, so 30 x 10 = 300.

    I was able to work out the solution manually thinking it out, but I want to know how the author did it for part 2.

    author solution:

    There are C(5; 2)C(7; 3) = 350 possible committees consisting of 2 women
    and 3 men. Now, if we suppose that 2 men are feuding and refuse to serve
    together then the number of committees that do not include the two men
    is C(7; 3) - C(2; 2)C(5; 1) = 30 possible groups. Because there are still
    C(5; 2) = 10 possible ways to choose the 2 women, it follows that there are
    30 10 = 300 possible committees

    what is C(2; 2) and C(5; 1)?? how did he come up with that?
     
  2. jcsd
  3. Mar 26, 2013 #2

    Ibix

    User Avatar
    Science Advisor

    He's using C(n;r) to mean n-choose-r, sometimes written nCr, nCr, or Cnr.

    He has divided the men into two groups - the two feuders and the other five. If C(7;3) is all groups of three men, and C(7;3)-C(2;2)C(5;1) is all groups not including both feuders, what must C(2;2)C(5;1) be?
     
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