# Combination Problem with a Variable

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 $x$.

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It might help you to remember that the binomial coefficients ${}_p C_q$ are coefficients of the binomial expansion, and then see that they satisfy a symmetry, e.g. $(x+1)^4=x^4+4x^3+6x^2+4x+1$, from which you can see, e.g. that ${}_4 C_1={}_4 C_3$... try to do something similar with ${}_{12} C_4$. –  Ｊ. Ｍ. Aug 14 '11 at 15:46
@J. M. Thanks for your reply. The textbook I'm working from hasn't covered binomial coefficients yet. Is there an alternative way to look at it? –  Nick Aug 14 '11 at 15:49
$\frac{{x!}}{{(x - 8)!}} \ne (x - 7)!$, hence the equation $\frac{{12!}}{{4!}} = (x - 7)!$ is wrong. Rather, from $\frac{{12!}}{{4!}} = \frac{{x!}}{{(x - 8)!}}$, it is readily seen that $x=12$ is a solution: indeed, $12!=12!$ and $4!=(12-8)!$ (compare numerators and denominators). –  Shai Covo Aug 14 '11 at 15:50
In that case, the best you can do is note that $12=4+8$ –  Ｊ. Ｍ. Aug 14 '11 at 15:52
@Shai Covo Thanks. I now see, if not so readily, that is in fact what's happening. I'm trying to work through the algebra as much, if not more, than simply getting to the answer. What is the correct simplification of the RHS? –  Nick Aug 14 '11 at 15:54

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.
Nick: $\frac{{(x - 4)!}}{{(x - 6)!}} = \frac{{(x - 6)!(x - 5)(x - 4)}}{{(x - 6)!}} = (x - 5)(x - 4)$. So you need to solve $(x - 5)(x - 4) = 30$. –  Shai Covo Jul 24 '11 at 4:20
Nick: Moreover, it is readily seen that $x=10$ is a solution of $\frac{{(x - 4)!}}{{(x - 6)!}} = \frac{{6!}}{{4!}}$ (indeed, $10-4=6$ and $10-6=4$, hence $(10-4)!=6!$ and $(10-6)!=4!$). –  Shai Covo Jul 24 '11 at 4:25