Results 1 to 3 of 3

Math Help - problem

  1. #1
    Newbie
    Joined
    Oct 2009
    Posts
    13

    problem

    A team of 7 people is to be chosen from 12
    couples. How many ways can a team of three men and four
    women be chosen? How many ways can a team containing
    no couple be chosen? How many ways can a team
    containing exactly two couples be chosen?

    my guess is for part 1 is: (12 choose 7) * (6 choose 3) * (6 choose 4)?

    cant figure out the rest!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,685
    Thanks
    616
    Hello, dannyshox!

    Your counting is off . . .


    A team of 7 people is to be chosen from 12 couples.

    (a) How many ways can a team of 3 men and 4 women be chosen?
    There are 12 couples: 12 men and 12 women.

    Choose 3 men: . {12\choose3} \:=\:\frac{12!}{3!\,9!} \:=\:220 ways.

    Choose 4 women: . {12\choose4} \:=\:\frac{12!}{4!\,8!} \:=\:495 ways.


    Therefore: . n(\text{3M, 4W}) \:=\:220\cdot 495 \:=\:108,\!900 ways.




    (b) How many ways can a team containing no couples be chosen?
    I found no neat formula for this . . . I made an exhaustive list.


    7 Men, 0 Women
    . . . n(\text{7M}) \;=\;{12\choose7} ways.


    6 Men, 1 Woman
    Choose 6 men: . {12\choose6} ways.

    The 1 woman must be chosen from the 6 women
    . . who are not wives of the 6 men: . {6\choose1} ways.
    . . n(\text{6M, 1W}) \:=\:{12\choose6}{6\choose1} ways.


    5 Men, 2 Women
    Choose 5 men: . {12\choose5} ways.

    The 2 women must be chosen from the 7 women
    . . who are not wives of the 5 men: . {7\choose2} ways.
    . . n(\text{5M, 2W}) \:=\:{12\choose5}{7\choose2} ways.


    4 Men, 3 Women
    Choose 4 men: . {12\choose4} ways.

    The 3 women must be chosen from the 8 women
    . . who are not wives of the 4 men: . {8\choose3} ways.
    . . n(\text{4M, 3W}) \:=\:{12\choose4}{8\choose3} ways.


    3 Men, 4 Women
    Choose 3 men: . {12\choose3} ways.

    The 4 women must be chosen from the 9 women
    . . who are not wives of the 3 men: . {9\choose4} ways.
    . . n(\text{3M, 4W}) \:=\:{12\choose3}{9\choose4} ways.


    2 Men, 5 Women
    Choose 2 men: . {12\choose2} ways.

    The 5 women must be chosen from the 10 women
    . . who are not wives of the 2 men: . {10\choose5} ways.
    . . n(\text{2M, 5W}) \:=\:{12\choose2}{10\choose5} ways.


    1 Man, 6 Women
    Choose 1 man: . {12\choose1} ways.

    The 6 women must be chosen from the 11 women
    . . who are not wives of the 2 men: . {11\choose6} ways.
    . . n(\text{1M, 6W}) \:=\:{12\choose1}{11\choose6} ways.


    0 Man, 7 Women
    Choose 7 women: . {12\choose7} ways.


    Now add them . . .




    (c) How many ways can a team containing exactly two couples be chosen?

    Choose 2 of the 12 men to be on the team: . {12\choose2} \:=\:66 ways.

    There is 1 way that their wives are also on the team.

    We have 4 people on the team.: 2M, 2W.
    We must choose 3 more people from the other 4 men and 4 women
    . . and include no couples.


    There are 4 cases:

    . . 3M, 0W: . {4\choose3} \:=\:4

    . . 2M, 1W: . {4\choose2}{2\choose1} \:=\:12

    . . 1M, 3W: . {4\choose1}{3\choose2} \:=\:12

    . . 0M, 3W: . {4\choose3} \:=\:4

    There are: . 4 + 12 + 12 + 4 \:=\:32 ways.


    Therefore: . n(\text{exactly 2 couples}) \;=\;66\cdot32 \;=\;2112 ways.

    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    Apr 2009
    Posts
    677
    Quote Originally Posted by Soroban View Post
    Hello, dannyshox!

    Your counting is off . . .


    There are 12 couples: 12 men and 12 women.

    Choose 3 men: . {12\choose3} \:=\:\frac{12!}{3!\,9!} \:=\:220 ways.

    Choose 4 women: . {12\choose4} \:=\:\frac{12!}{4!\,8!} \:=\:495 ways.


    Therefore: . n(\text{3M, 4W}) \:=\:220\cdot 495 \:=\:108,\!900 ways.




    I found no neat formula for this . . . I made an exhaustive list.


    7 Men, 0 Women
    . . . n(\text{7M}) \;=\;{12\choose7} ways.


    6 Men, 1 Woman
    Choose 6 men: . {12\choose6} ways.

    The 1 woman must be chosen from the 6 women
    . . who are not wives of the 6 men: . {6\choose1} ways.
    . . n(\text{6M, 1W}) \:=\:{12\choose6}{6\choose1} ways.


    5 Men, 2 Women
    Choose 5 men: . {12\choose5} ways.

    The 2 women must be chosen from the 7 women
    . . who are not wives of the 5 men: . {7\choose2} ways.
    . . n(\text{5M, 2W}) \:=\:{12\choose5}{7\choose2} ways.


    4 Men, 3 Women
    Choose 4 men: . {12\choose4} ways.

    The 3 women must be chosen from the 8 women
    . . who are not wives of the 4 men: . {8\choose3} ways.
    . . n(\text{4M, 3W}) \:=\:{12\choose4}{8\choose3} ways.


    3 Men, 4 Women
    Choose 3 men: . {12\choose3} ways.

    The 4 women must be chosen from the 9 women
    . . who are not wives of the 3 men: . {9\choose4} ways.
    . . n(\text{3M, 4W}) \:=\:{12\choose3}{9\choose4} ways.


    2 Men, 5 Women
    Choose 2 men: . {12\choose2} ways.

    The 5 women must be chosen from the 10 women
    . . who are not wives of the 2 men: . {10\choose5} ways.
    . . n(\text{2M, 5W}) \:=\:{12\choose2}{10\choose5} ways.


    1 Man, 6 Women
    Choose 1 man: . {12\choose1} ways.

    The 6 women must be chosen from the 11 women
    . . who are not wives of the 2 men: . {11\choose6} ways.
    . . n(\text{1M, 6W}) \:=\:{12\choose1}{11\choose6} ways.


    0 Man, 7 Women
    Choose 7 women: . {12\choose7} ways.


    Now add them . . .





    Choose 2 of the 12 men to be on the team: . {12\choose2} \:=\:66 ways.

    There is 1 way that their wives are also on the team.

    We have 4 people on the team.: 2M, 2W.
    We must choose 3 more people from the other 4 men and 4 women
    . . and include no couples.


    There are 4 cases:

    . . 3M, 0W: . {4\choose3} \:=\:4

    . . 2M, 1W: . {4\choose2}{2\choose1} \:=\:12

    . . 1M, 3W: . {4\choose1}{3\choose2} \:=\:12

    . . 0M, 3W: . {4\choose3} \:=\:4

    There are: . 4 + 12 + 12 + 4 \:=\:32 ways.


    Therefore: . n(\text{exactly 2 couples}) \;=\;66\cdot32 \;=\;2112 ways.

    @Soroban - hi
    For b)
    An easy way - as there can't be a couple so first select 7couples from 12 couples. 12C7.
    Now that you have 7 couples select either of the partner. i.e. 2^7ways.
    So we have 12C7.2^7

    solution to c) also appears wrong at following places plz

    We must choose 3 more people from the other 4 men and 4 women
    . . and include no couples.

    1M, 3W: {4\choose1}{3\choose2} \:=\:12
    Follow Math Help Forum on Facebook and Google+

Search Tags


/mathhelpforum @mathhelpforum