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Math Help - Clarifying the first combination example of this text.

  1. #1
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    Clarifying the first combination example of this text.

    From a group of 5 women and 7 men, how many different committees consisting 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?

    For the second question the book says that there are 5 total invalid combinations, \binom{2}{2}\binom{5}{1}=5, but then provides no explanation as to how it arrived at that conclusion.

    Could someone explain how this poorly executed textbook arrived at this conclusion for me?
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    Re: Clarifying the first combination example of this text.

    Quote Originally Posted by bkbowser View Post
    From a group of 5 women and 7 men, how many different committees consisting 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?
    For the second question the book says that there are 5 total invalid combinations, \binom{2}{2}\binom{5}{1}=5, but then provides no explanation as to how it arrived at that conclusion.
    With no restrictions at all there are \binom{5}{2}\binom{7}{3} ways to form the committee.

    Put two feuding men altogether there are \binom{5}{2}\binom{5}{1} ways to form the committee.

    Now subtract the second from the first.
    Last edited by Plato; January 18th 2014 at 09:29 AM.
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    Re: Clarifying the first combination example of this text.

    Quote Originally Posted by Plato View Post
    With no restrictions at all there are \binom{5}{2}\binom{7}{3} ways to form the committee.

    Leaving the two feuding men off altogether there are \binom{5}{2}\binom{5}{3} ways to form the committee.

    Now subtract the second from the first.
    So \binom{2}{2}\binom{5}{1} is a simplification of \binom{5}{2}\binom{5}{3}?

    I'm getting;

    \binom{5}{2}\binom{5}{3}=\frac{5!}{(5-2)!2!}\frac{5!}{(5-3)!3!}=2\frac{5!}{3!2!}=20\neq5

    ?
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    Re: Clarifying the first combination example of this text.

    Op sorry there's an edit.
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    Re: Clarifying the first combination example of this text.

    Still getting something wierd.
    \binom{5}{2}\binom{5}{1}=\frac{5!}{(5-2)!2!}\frac{5!}{(5-1)!1!}=\frac{5!5!}{3!2!4!}=100\neq5
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    Re: Clarifying the first combination example of this text.

    I think you may be misunderstanding the book.
    An "invalid" committee is one in which those two specific men serve together. Treat those two men as a single unit. There is one way, \begin{pmatrix}2 \\ 2 \end{pmatrix} to choose that unit and \begin{pmatrix}5 \\ 1\end{pmatrix} ways to choose the third man. So there are \begin{pmatrix}2 \\ 2 \end{pmatrix}\begin{pmatrix}5 \\ 1\end{pmatrix} to choose the men for an "invalid" committee. But there will still be \begin{pmatrix}5 \\ 2 \end{pmatrix} ways to choose the women and so \begin{pmatrix}5 \\ 2\end{pmatrix}\begin{pmatrix}2 \\ 2 \end{pmatrix}\begin{pmatrix}5 \\ 1\end{pmatrix} "invalid" committees.
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    Re: Clarifying the first combination example of this text.

    Quote Originally Posted by bkbowser View Post
    Still getting something wierd.
    \binom{5}{2}\binom{5}{1}=\frac{5!}{(5-2)!2!}\frac{5!}{(5-1)!1!}=\frac{5!5!}{3!2!4!}=100\neq5
    Sorry that should be 50 not 100
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    Re: Clarifying the first combination example of this text.

    Quote Originally Posted by HallsofIvy View Post
    I think you may be misunderstanding the book.
    An "invalid" committee is one in which those two specific men serve together. Treat those two men as a single unit. There is one way, \begin{pmatrix}2 \\ 2 \end{pmatrix} to choose that unit and \begin{pmatrix}5 \\ 1\end{pmatrix} ways to choose the third man. So there are \begin{pmatrix}2 \\ 2 \end{pmatrix}\begin{pmatrix}5 \\ 1\end{pmatrix} to choose the men for an "invalid" committee. But there will still be \begin{pmatrix}5 \\ 2 \end{pmatrix} ways to choose the women and so \begin{pmatrix}5 \\ 2\end{pmatrix}\begin{pmatrix}2 \\ 2 \end{pmatrix}\begin{pmatrix}5 \\ 1\end{pmatrix} "invalid" committees.
    OK, that does make sense to me.

    Thanks.
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