Question about a Question: Simplifying Fractions

Question

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?

Answer 1

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!)

Answer 2

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