I have the following problem:
$_xC_6$ = $_xC_4$
I expand both sides to: $$\frac{x!}{[(x-6)!]6!} = \frac{x!}{[(x-4)]!4!}$$
Next I multiply both sides by the denominator of the right-hand expression to get:
$$\frac{x![(x-4)!]4!}{[(x-6)!]6!}=x!$$
At this point things start to become a mess, so I'm wondering if these first few steps are correct.
EDIT: Similar, but different, problem:
$$_{12}C_4 = _xC_8$$ translates to: $$\frac{12!}{8!4!}=\frac{x!}{(x-8)!8!}$$
First, I multiply both sides by $8!$ to get:
$$ \frac{12!}{4!}=\frac{x!}{(x-8)!} $$
Then I reduce/simplify the RHS to get: $$ \frac{12!}{4!}=(x-7)! $$ (I'm not completely confident this step is correct)
Here's where I get stuck. The LHS=19,958,400 but I can't figure out how to manipulate the factorial on the RHS to get to just $x$.
Thanks in advance!
Your steps are correct. However, more simply, from your first equation $$ \frac{{x!}}{{(x - 6)!6!}} = \frac{{x!}}{{(x - 4)!4!}}, $$ you can see (by dividing both sides by $x!$) that $$ \frac{1}{{(x - 6)!6!}} = \frac{1}{{(x - 4)!4!}}, $$ hence $$ \frac{{(x - 4)!}}{{(x - 6)!}} = \frac{{6!}}{{4!}}. $$ The rest is easy.