1. ## 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!

2. 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.

3. could someone post the work involved with the first question, i'm lost lol

4. Originally Posted by needuhelp
could someone post the work involved with the first question, i'm lost lol
post what you tried first

5. i have..

2!(n-1)(n-2)(n-3)(n-4) / 4! = (n-1)(n-2)(n-3)

6. 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.