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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!

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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 ${}_{12} C_4$.2011-08-14
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    @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?2011-08-14
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    $\frac{{x!}}{{(x - 8)!}} \ne (x - 7)!$, 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).2011-08-14
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    In that case, the best you can do is note that $12=4+8$2011-08-14
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    @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?2011-08-14
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    You repeatedly apply the relation $n!=n\cdot(n-1)!=n(n-1)\cdot(n-2)!=\cdots$2011-08-14
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    For the RHS, it holds $$\frac{{x!}}{{(x - 8)!}} = \frac{{1 \cdot 2 \cdots (x - 8)(x - 7)(x - 6) \cdots x}}{{1 \cdot 2 \cdots (x - 8)}} = (x - 7)(x - 6) \cdots x$$.2011-08-14
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    Thus, for example, $$\frac{{12!}}{{(12 - 8)!}} = (12 - 7)(12 - 6) \cdots 12 = 5 \cdot 6 \cdots 12 = {\rm 19958400}.$$2011-08-14
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    @Shai Covo Got it - thank you!2011-08-14
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    Glad it helped.2011-08-14

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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.

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    Of course - dividing by x! makes perfect sense. Then the rest would be: 1/(x-6)(x-5)=30? Now I start to get lost again...2011-07-24
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    Nick: $\frac{{(x - 4)!}}{{(x - 6)!}} = \frac{{(x - 6)!(x - 5)(x - 4)}}{{(x - 6)!}} = (x - 5)(x - 4)$. So you need to solve $(x - 5)(x - 4) = 30$.2011-07-24
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    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 $(10-6)!=4!$).2011-07-24
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    Yes, I could see it given the numbers are small. However I wanted to ensure I could perform the computation to get to the answer. Thanks again for your help!2011-07-24
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    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