Question about a Question: Simplifying Fractions In a question I asked several weeks ago an interim step reached was a.):
$$\frac{1}{(x-6)!6!}=\frac{1}{(x-4)!4!}$$
hence b.):
$$ \frac{(x-4)!}{(x-6)!}=\frac{6!}{4!}$$
I'm not following how we got from a.) to b.)
Help?
 A: Cross multiply.
Multiplying both sides by $6!$ you get
$$\begin{align*}
\frac{1}{(x-6)!6!} &= \frac{1}{(x-4)!4!}\\
\frac{6!}{(x-6)!6!} &= \frac{6!}{(x-4)!4!}.
\end{align*}$$
Now the $6!$ factor in the numerator and denominator on the left hand side cancel, and you get
$$\frac{1}{(x-6)!} = \frac{6!}{(x-4)!4!}.$$
Now multiply both sides by $(x-4)!$ to get
$$\frac{(x-4)!}{(x-6)!} = \frac{(x-4)!6!}{(x-4)!4!}.$$
Again, you have a factor of $(x-4)!$ in both the numerator and denominator of the right hand side, so these cancel. You end up with
$$\frac{(x-4)!}{(x-6)!} = \frac{6!}{4!},$$
as desired.
P.S. It would have made more sense to follow-up that answer with a query in comments (and even more sense not to accept the answer until you understood all the steps!)
A: Multiply both sides by (x-4)! and 6! and you will have (b)
$$\frac{1}{(x-6)!6!}=\frac{1}{(x-4)!4!} \Leftrightarrow \frac{(x-4)!6!}{(x-6)!6!}=\frac{(x-4)!6!}{(x-4)!4!}$$
as $\frac{6!}{6!} = 1$ and $\frac{(x-4)}{(x-4)} = 1$, it simplifies to $$\frac{(x-4)!}{(x-6)!}=\frac{6!}{4!}$$
