Thread: Simple Factoral Problem

1. 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.

$\frac {\left(\begin{array}{cc}q\\x\end{array}\right)\lef t(\begin{array}{cc}8-q\\2-x\end{array}\right)}
{\left(\begin{array}{cc}8\\2\end{array}\right)}$

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?

2. Originally Posted by Raezputin
$\frac {\left(\begin{array}{cc}d\\x\end{array}\right)\lef t(\begin{array}{cc}8-d\\2-x\end{array}\right)}
{\left(\begin{array}{cc}8\\2\end{array}\right)}$

where x=0
Where and why q?
$\frac {\left(\begin{array}{cc}d\\x\end{array}\right)\lef t(\begin{array}{cc}8-d\\2-x\end{array}\right)}
{\left(\begin{array}{cc}8\\2\end{array}\right)}$
$=\frac{\displaystyle\frac{(8-d)(8-d-1)}{2\cdot 1}}{\displaystyle\frac{8\cdot 7}{2\cdot 1}}$

3. 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 =)

4. Originally Posted by Raezputin
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.