okay, i'm stuck on a couple questions... :/

2!n! / 4!(n-5) = (n-1)! / (n-4)!

and..

(n+2)! / n! = 30

any help would be great, thanks! :)

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- Jan 18th 2010, 10:47 AMneeduhelpfactorial!
okay, i'm stuck on a couple questions... :/

2!n! / 4!(n-5) = (n-1)! / (n-4)!

and..

(n+2)! / n! = 30

any help would be great, thanks! :) - Jan 18th 2010, 10:54 AMVonNemo19
- Jan 18th 2010, 02:37 PMneeduhelp
could someone post the work involved with the first question, i'm lost lol

- Jan 18th 2010, 02:41 PMJhevon
- Jan 18th 2010, 03:23 PMneeduhelp
i have..

2!(n-1)(n-2)(n-3)(n-4) / 4! = (n-1)(n-2)(n-3) - Jan 18th 2010, 03:58 PMmr fantastic
I assume the red factorial was meant to be there.

This re-arranges into $\displaystyle \frac{2!}{4!} = \frac{(n-1)! (n-5)!}{n! (n-4)!}$.

The left hand side is 1/12. Simplify the right hand side using what you have learned from this thread (and what you should have learned from class too) and solve the resulting equation.