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Math Help - Choosing different committees from a class of students

  1. #1
    Junior Member Shadow236's Avatar
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    Choosing different committees from a class of students

    The question that I am working on follows:

    A class has 10 male students and 7 female students. How many ways can we choose a committee consisting of

    (a) 2 students from the class
    (b) 2 males and 1 female from the class
    (c) All females from the class
    (d) No Females

    I found this solution that is kind of similar, but I am having trouble understanding it. How do I answer these questions?

    Thank you!
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  2. #2
    MHF Contributor

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    Re: Choosing different committees from a class of students

    These are pretty straight forward using a little common sense and the "fundamental rule of counting": If one thing can happen in n ways and another can happen in m ways they can happen together in nm ways.

    Quote Originally Posted by Shadow236 View Post
    The question that I am working on follows:

    A class has 10 male students and 7 female students. How many ways can we choose a committee consisting of

    (a) 2 students from the class
    There are a total of 17 students. You have 17 choices for the first student. Then there remain 16 choices for the second student. So 17(16) ways to do this.

    (b) 2 males and 1 female from the class
    If we were to choose the two males first, there would be 10 choices for the first male and 9 choices for the second, then 7 choices for the female so 10(9)(7). However, we do not have to choose in that order. There are 3(2)(1)= 6 orders in which to choose the two males and one female (writing M1 for the one of the males, M2 for the other, and F for the female those 6 orders would be M1M2F, M1FM2, M2M1F, M2FM1, FM1M2, FM2M1) so there are 10(9)(7)(6) ways to do this.

    (c) All females from the class
    ??? How large is the committee? If you mean "form a committee consisting of all the seven females in the class, there would, of course, be one such committee. But I suspect you mean "How many ways to form a committee of "n" people, chosen from the 7 females". To answer that we need to know what "n" is.

    (d) No Females
    This is the same as (c) but with males replacing females. How large is the committee to be?

    I found this solution that is kind of similar, but I am having trouble understanding it. How do I answer these questions?

    Thank you!
    Thanks from Shadow236
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  3. #3
    Junior Member Shadow236's Avatar
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    Re: Choosing different committees from a class of students

    First of all, thank you very much. I knew that the solution would be rather simple, but I just couldn't find a good example online.

    For (c) and (d) it doesn't specify a committee size. Therefore, I'm guessing it just means form a committee with all of the females in the class, and then one with all of the males. With that being said, does that mean that (c) = 7, and (d) = 10.
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  4. #4
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    Re: Choosing different committees from a class of students

    To go over some of the fine points of what HallsofIvy stated, for (a): You there are two orders to choose the males, but both orders create the same committee. Hence, 17*16 counts each committee twice. Therefore, there are $\dfrac{17\cdot 16}{2}$ ways to choose the committee.

    For (b), you must choose males and females separately. Suppose you choose the males first. None of the committees chosen will have any females yet (since you have not yet chosen the females). If you choose the females first, none of the committees chosen will have any males yet (since you have not yet chosen the males). So, the order that you choose the gender doesn't matter. However, while choosing the males, after you choose the first male, there is already a male in the committee, so when you choose the second male, the order you chose the two males does cause overcounting. Hence, there are $\dfrac{17\cdot 16}{2}$ ways to choose the males and $7$ ways to choose a female. This gives a total of $\dfrac{17\cdot 16\cdot 7}{2}$ ways to choose the committee.

    For (c), since you want all females in the class, the committee must contain all seven of those females. Then, from each male in the class, the male is either in the committee or is not. Hence, there are two choices for each male. If a specific male is chosen (or not), it does not affect the choosing of a different male, so these choices are independent. Hence, there are $2^{10} = 1024$ such committees.

    For (d), since you want no females, the problem is extremely similar to part (c). However, of the possible committees, one of them will not contain any members. If a committee must contain at least one member, then there are $2^{10}-1 = 1023$ committees with no females (and at least one member).
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