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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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  • $\begingroup$ 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 ${}_{12} C_4$. $\endgroup$
    – J. M. ain't a mathematician
    Aug 14, 2011 at 15:46
  • $\begingroup$ @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? $\endgroup$
    – Nick
    Aug 14, 2011 at 15:49
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    $\begingroup$ $\frac{{x!}}{{(x - 8)!}} \ne (x - 7)!$, 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). $\endgroup$
    – Shai Covo
    Aug 14, 2011 at 15:50
  • $\begingroup$ In that case, the best you can do is note that $12=4+8$ $\endgroup$
    – J. M. ain't a mathematician
    Aug 14, 2011 at 15:52
  • $\begingroup$ @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? $\endgroup$
    – Nick
    Aug 14, 2011 at 15:54

1 Answer 1

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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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  • $\begingroup$ 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... $\endgroup$
    – Nick
    Jul 24, 2011 at 4:16
  • $\begingroup$ Nick: $\frac{{(x - 4)!}}{{(x - 6)!}} = \frac{{(x - 6)!(x - 5)(x - 4)}}{{(x - 6)!}} = (x - 5)(x - 4)$. So you need to solve $(x - 5)(x - 4) = 30$. $\endgroup$
    – Shai Covo
    Jul 24, 2011 at 4:20
  • $\begingroup$ 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 $(10-6)!=4!$). $\endgroup$
    – Shai Covo
    Jul 24, 2011 at 4:25
  • $\begingroup$ 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! $\endgroup$
    – Nick
    Jul 24, 2011 at 4:47
  • $\begingroup$ If you had $1/((x-6)(x-5)=30$, it would just be a quadratic equation: $$ (x-6)(x-5) = \frac{1}{30} $$ $$ x^2 - 11x + 30 - \frac{1}{30} = 0 $$ $\endgroup$
    – Michael Hardy
    Jul 24, 2011 at 12:41

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