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Thread: Combination question about forming mixed hockey

  1. #1
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    Combination question about forming mixed hockey

    Question: In a game of mixed hockey there are ten married couples and 2 spinsters playing. In how many ways can the two teams be made up, if no husband may play against his wife?

    case 1:
    team 1 (include 2 spinsters)

    2 spinsters ; 2C2 =1
    there are 4 spaces to fill, we need 2 couples to fill it; 10C2 =45
    2C2 x 10 C2 = 45

    team 2;
    there are 6 spaces ( 3 couples to fill it)
    8C3 = 56


    case 2:
    team 1 ( no spinsters)
    we need 3 couples to fill it;
    10C3 = 120

    team 2
    there are 2 options include 2 spinsters or not

    if we include 2 spinsters; 2C2 =1
    and we are left with 4 spaces to fill it; 7C2 = 21

    if we do not include spinsters;

    then we need 3 couples; 7C3 = 35

    sum ; 45+120+56+35+21 = 277

    there are 277 ways but the book gives an answer of 252 please help

    I do not know where I went wrong?
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    Re: Combination question about forming mixed hockey

    Quote Originally Posted by bigmansouf View Post
    Question: In a game of mixed hockey there are ten married couples and 2 spinsters playing. In how many ways can the two teams be made up, if no husband may play against his wife?
    I am clueless as to anything about hockey. Please tell us how one forms a hockey team? How many on a team?
    Moreover, can a husband & wife be on the same team?
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    Re: Combination question about forming mixed hockey

    Ice hockey has 6 people per team, Field hockey has 11 people per team.

    Given the wording of the question, i.e. "the two teams", I think they are referring to field hockey. This would use all the people mentioned.

    A team then must be comprised of 5 couples and 1 spinster. Otherwise some couple would be playing against each other.

    given this the answer should be simply

    $N = \dbinom{10}{5}\cdot \dbinom{2}{1} = 504$

    This is precisely double the answer given. I can't account for the discrepancy. I think it quite possible the answer given is wrong.
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    Re: Combination question about forming mixed hockey

    Quote Originally Posted by Plato View Post
    I am clueless as to anything about hockey. Please tell us how one forms a hockey team? How many on a team?
    Moreover, can a husband & wife be on the same team?
    i thought that it was about ice hockey but it is about field hockey
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    Re: Combination question about forming mixed hockey

    Quote Originally Posted by romsek View Post
    Ice hockey has 6 people per team, Field hockey has 11 people per team.

    Given the wording of the question, i.e. "the two teams", I think they are referring to field hockey. This would use all the people mentioned.

    A team then must be comprised of 5 couples and 1 spinster. Otherwise some couple would be playing against each other.

    given this the answer should be simply

    $N = \dbinom{10}{5}\cdot \dbinom{2}{1} = 504$

    This is precisely double the answer given. I can't account for the discrepancy. I think it quite possible the answer given is wrong.
    thank you very much i am having problems with this kind of questions
    such as; "How many mixed hockey teams may be made from 6 married couples, 1 bachelor and 3 spinsters, if no wife will play without her husband? "
    for this question;
    by following your example
    i realise that there are two ways to do this 4 couples and 3 individuals and 5 couples and 1 individual


    4 couples and 3 individuals;
    6C4 = 15 4C3 = 4 15 x 4 = 60

    5 couples and 1 individual
    6C5 =6 4C1= 4 6 x 4 = 24


    the answer is 60 + 24 = 84 but the book gives an answer of 876 but i keep getting the answer wrong

    please can you help me and how do i improve for these kind of questions
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    Re: Combination question about forming mixed hockey

    Quote Originally Posted by bigmansouf View Post
    thank you very much i am having problems with this kind of questions
    such as; "How many mixed hockey teams may be made from 6 married couples, 1 bachelor and 3 spinsters, if no wife will play without her husband? "
    for this question;
    by following your example
    i realise that there are two ways to do this 4 couples and 3 individuals and 5 couples and 1 individual


    4 couples and 3 individuals;
    6C4 = 15 4C3 = 4 15 x 4 = 60

    5 couples and 1 individual
    6C5 =6 4C1= 4 6 x 4 = 24


    the answer is 60 + 24 = 84 but the book gives an answer of 876 but i keep getting the answer wrong

    please can you help me and how do i improve for these kind of questions
    One thing you've neglected is that single husbands can play on the team w/o their wives.

    $\left(
    \begin{array}{ccc}
    H& C & S \\
    0 & 5 & 1 \\
    1 & 5 & 0 \\
    0 & 4 & 3 \\
    1 & 4 & 2 \\
    2 & 4 & 1 \\
    1 & 3 & 4 \\
    2 & 3 & 3 \\
    3 & 3 & 2 \\
    4 & 2 & 3 \\
    3 & 2 & 4 \\
    5 & 1 & 4 \\
    \end{array}
    \right)$

    For each triple compute $\dbinom{6}{H}\dbinom{6-H}{C}\dbinom{4}{S}$ and sum them all up.
    This sums to 876.
    Last edited by romsek; Jun 30th 2019 at 06:27 PM.
    Thanks from bigmansouf
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    Re: Combination question about forming mixed hockey

    Quote Originally Posted by romsek View Post
    Ice hockey has 6 people per team, Field hockey has 11 people per team.

    Given the wording of the question, i.e. "the two teams", I think they are referring to field hockey. This would use all the people mentioned.

    A team then must be comprised of 5 couples and 1 spinster. Otherwise some couple would be playing against each other.

    given this the answer should be simply

    $N = \dbinom{10}{5}\cdot \dbinom{2}{1} = 504$

    This is precisely double the answer given. I can't account for the discrepancy. I think it quite possible the answer given is wrong.
    Thinking about this I guess the factor of two comes from the fact that they don't care about the order the two teams are formed in.
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    Re: Combination question about forming mixed hockey

    Quote Originally Posted by romsek View Post
    One thing you've neglected is that single husbands can play on the team w/o their wives.

    $\left(
    \begin{array}{ccc}
    H& C & S \\
    0 & 5 & 1 \\
    1 & 5 & 0 \\
    0 & 4 & 3 \\
    1 & 4 & 2 \\
    2 & 4 & 1 \\
    1 & 3 & 4 \\
    2 & 3 & 3 \\
    3 & 3 & 2 \\
    4 & 2 & 3 \\
    3 & 2 & 4 \\
    5 & 1 & 4 \\
    \end{array}
    \right)$

    For each triple compute $\dbinom{6}{H}\dbinom{6-H}{C}\dbinom{4}{S}$ and sum them all up.
    This sums to 876.
    thank you very much

    for to be a bother but can you explain what C stands for in the table
    H- Husband
    C- ?
    S- spinsters

    is C for Bachelors
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    Re: Combination question about forming mixed hockey

    Quote Originally Posted by bigmansouf View Post
    thank you very much

    for to be a bother but can you explain what C stands for in the table
    H- Husband
    C- ?
    S- spinsters

    is C for Bachelors
    H is a husband taken by himself
    C is an entire couple
    S is for anyone not married (the spinsters and the bachelor)
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