1
$\begingroup$

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!

  • 0
    Glad it helped.2011-08-14

1 Answers 1

1

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.

  • 0
    If you had $1/((x-6)(x-5)=30$, it would just be a quadratic equation: $ (x-6)(x-5) = \frac{1}{30} $ $ x^2 - 11x + 30 - \frac{1}{30} = 0 $2011-07-24