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?
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?
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!)
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 $\frac{(x-4)!}{(x-6)!}=\frac{6!}{4!}$