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.)


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    $\begingroup$ Multiply both sides by $\ 6!\:(x-4)!$ $\endgroup$ – Bill Dubuque Aug 14 '11 at 2:00
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    $\begingroup$ Err, why wait so long to ask for clarification? $\endgroup$ – anon Aug 14 '11 at 2:03

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!)

  • $\begingroup$ Thanks. I thought I did understand it a few weeks ago however looking back on it today I was struggling to recall the steps. I also thought posing the question in the comments, so long after the question had been asked, would not have received a response. $\endgroup$ – Nick Aug 14 '11 at 2:39
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    $\begingroup$ @Nick: When you post the comment, the author of the original answer gets a notification, calling his attention to your query; you could also have edited the question adding your new confusion, which would immediately "bump" the question up to the top. $\endgroup$ – Arturo Magidin Aug 14 '11 at 2:42
  • $\begingroup$ Yikes, I didn't realize that, thanks. In that case my apologies for the clutter and sloppy reading of StackExchange mechanics. I've done some digging on Meta to see if there's a way for me to consolidate the two questions or if a Mod needs to do it. $\endgroup$ – Nick Aug 14 '11 at 14:40

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


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