## Factorial Equation Problem

Complex numbers, rational functions, logarithms, sequences and series, matrix operations, etc.
maroonblazer
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### Factorial Equation Problem

I'm struggling to solve for x in the following:

$\frac{1}{[(x-8)!]8!} = \frac{1}{[(x-7)!]7!}$

Do I multiply each side by $[(x-7)!]7!$

Thanks in advance,
mb
Last edited by maroonblazer on Sun Jul 24, 2011 1:49 pm, edited 1 time in total.

nona.m.nona
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### Re: Factorial Equation Problem

$\frac{1}{[x-8]!8!} = \frac{1}{[x-7]!7!}$

Do I multiply each side by $[x-7]!7!$
You could do that, or multiply through by the larger denominator in order to get a linear equation.

Do the brackets indicate the "floor" or "ceiling" function, or something else? Thank you.

maroonblazer
Posts: 51
Joined: Thu Aug 12, 2010 11:16 am
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### Re: Factorial Equation Problem

You could do that, or multiply through by the larger denominator in order to get a linear equation.
Thanks for your reply! I was using the brackets to separate the factorials. I've edited the question so it's clearer.

So following that approach I get:

$\frac{[(x-8)!]8!}{[(x-7)!]7!}=1$

$\frac{[(x-8)]8!}{7!}=1$

$(x-8)8=1$...I've done something wrong by this point because the answer is supposed to be 15 and this isn't headed in that direction.

Help?

nona.m.nona
Posts: 288
Joined: Sun Dec 14, 2008 11:07 pm
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### Re: Factorial Equation Problem

$(x-8)8=1$...I've done something wrong by this point because the answer is supposed to be 15 and this isn't headed in that direction.
My apologies: For factorials, of course (x - 7)! is larger than (x - 8)!, because (x - 7)! = 1*2*3*...*(x - 9)*(x - 8)*(x - 7), while (x - 8)! = 1*2*3*...*(x - 9)(x - 8).

Start from your first equation in your last post, noting that (x - 8)! / (x - 7)! = 1/(x - 7).

maroonblazer
Posts: 51
Joined: Thu Aug 12, 2010 11:16 am
Contact:

### Re: Factorial Equation Problem

$(x-8)8=1$...I've done something wrong by this point because the answer is supposed to be 15 and this isn't headed in that direction.
My apologies: For factorials, of course (x - 7)! is larger than (x - 8)!, because (x - 7)! = 1*2*3*...*(x - 9)*(x - 8)*(x - 7), while (x - 8)! = 1*2*3*...*(x - 9)(x - 8).

Start from your first equation in your last post, noting that (x - 8)! / (x - 7)! = 1/(x - 7).
Got it! Thanks very much!!

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