# Factorials

• Apr 27th 2008, 07:30 AM
mathgeek777
Factorials
I understand how to find them when numbers are involved. However, my problem is I can not figure out how to do it with n terms.

Example:

$\displaystyle \frac{(2n)!}{(2n-2)!}$

Any help will be greatly appreciated.
• Apr 27th 2008, 07:38 AM
Isomorphism
Quote:

Originally Posted by mathgeek777
I understand how to find them when numbers are involved. However, my problem is I can not figure out how to do it with n terms.

Example:

$\displaystyle \frac{(2n)!}{(2n-2)!}$

Any help will be greatly appreciated.

Hint: $\displaystyle \frac{(2n)!}{(2n-2)!} = \frac{(2n)\times(2n-1)\times(2n-2)!}{(2n-2)!}$
• Apr 27th 2008, 07:38 AM
Kai
Hmm, try using the fact that n!= n(n-1)!=n(n-1)(n-2)!
• Apr 27th 2008, 07:41 AM
elizsimca
Quote:

Originally Posted by mathgeek777
I understand how to find them when numbers are involved. However, my problem is I can not figure out how to do it with n terms.

Example:

$\displaystyle \frac{(2n)!}{(2n-2)!}$

Any help will be greatly appreciated.

$\displaystyle \frac{(2n)!}{(2n-2)!}$= $\displaystyle \frac{1*2*...*(n-2)*(n-1)*n*(n+1)*(n+2)*...*(2n-2)*(2n-1)*(2n)}{1*2*...*(n-2)*(n-1)*n*(n+1)*(n+2)*...*(2n-2)}$

Do you see where you can cancel and reduce the expression?
• Apr 27th 2008, 07:44 AM
finch41
i dont know how sound this logic is since i haven't done factorials in a long time but maybe it will steer you in some sort of direction

(2n)!
------
(2n-2)!

gives:

1x2x3...x(2n-3) x(2n-2) x (2n-1) x (2n)
--------------------------------------
1x2x3...x((2n-2)-1) x (2n-2)

this leaves (2n-1)(2n)
• Apr 27th 2008, 07:45 AM
mathgeek777