# factorial!

• Jan 18th 2010, 11:47 AM
needuhelp
factorial!
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, 11:54 AM
VonNemo19
Quote:

Originally Posted by needuhelp
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! :)

For the second one $\frac{(n+2)!}{n!}=30$ means that $\frac{(n+2)(n+1)(n)(n-1)(n-2)...}{n(n-1)(n-2)...}=30$.

So by cancelling we get

$(n+2)(n+1)=30$

I'm sure you can finish up. The first is similar, just more messy.
• Jan 18th 2010, 03:37 PM
needuhelp
could someone post the work involved with the first question, i'm lost lol
• Jan 18th 2010, 03:41 PM
Jhevon
Quote:

Originally Posted by needuhelp
could someone post the work involved with the first question, i'm lost lol

post what you tried first
• Jan 18th 2010, 04:23 PM
needuhelp
i have..

2!(n-1)(n-2)(n-3)(n-4) / 4! = (n-1)(n-2)(n-3)
• Jan 18th 2010, 04:58 PM
mr fantastic
Quote:

Originally Posted by needuhelp
okay, i'm stuck on a couple questions... :/

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

I assume the red factorial was meant to be there.

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

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.