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Math Help - Simple Factoral Problem

  1. #1
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    Simple Factoral Problem

    Hi everyone, I've been trying to figure out this factoral problem and wondered if you could help. This is in a statistics class but figured the thread should be here.

    \frac {\left(\begin{array}{cc}q\\x\end{array}\right)\lef  t(\begin{array}{cc}8-q\\2-x\end{array}\right)}<br />
{\left(\begin{array}{cc}8\\2\end{array}\right)}

    where x=0

    I have gotten to

    \frac {(8-q)!6!2!}{2!(6-q)!8!}

    q is an unknown variable, the answer I should get is

    \frac{(8-q)(7-q)}{7*8}

    and from there I can do the rest, any help?
    Last edited by Raezputin; September 14th 2009 at 10:22 AM. Reason: sorry the d was just wrong =)
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  2. #2
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    Quote Originally Posted by Raezputin View Post
    \frac {\left(\begin{array}{cc}d\\x\end{array}\right)\lef  t(\begin{array}{cc}8-d\\2-x\end{array}\right)}<br />
{\left(\begin{array}{cc}8\\2\end{array}\right)}
    where x=0
    Where and why q?
    \frac {\left(\begin{array}{cc}d\\x\end{array}\right)\lef  t(\begin{array}{cc}8-d\\2-x\end{array}\right)}<br />
{\left(\begin{array}{cc}8\\2\end{array}\right)} =\frac{\displaystyle\frac{(8-d)(8-d-1)}{2\cdot 1}}{\displaystyle\frac{8\cdot 7}{2\cdot 1}}
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  3. #3
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    Hey Plato thanks for the response, sorry about the d and q mixup. Im still a bit confused about what you are doing, in stats we were taught that

    \left(\begin{array}{cc}n\\r\end{array}\right) = \frac {n!}{r!(n-r)!}

    is this the rule you are using or another? Thanks for helping me =)
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  4. #4
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    Quote Originally Posted by Raezputin View Post
    Hey Plato thanks for the response, sorry about the d and q mixup. Im still a bit confused about what you are doing, in stats we were taught that

    \left(\begin{array}{cc}n\\r\end{array}\right) = \frac {n!}{r!(n-r)!}
    Yes of course that is the standard use. It is also an ole-timers trick.
    \binom{8}{2}=\frac{8\cdot 7}{2\cdot 1} note that there are the same numbers of factors in numerator as in denominator.

    So if I want \binom{20}{4} I start the denominator \frac{}{4\cdot 3\cdot 2\cdot 1} and fill out the numerator \frac{20\cdot 19\cdot 18\cdot 17}{4\cdot 3\cdot 2\cdot 1} so they match.
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