1. ## 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:

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

Any help will be greatly appreciated.

2. 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:

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

Any help will be greatly appreciated.
Hint: $\frac{(2n)!}{(2n-2)!} = \frac{(2n)\times(2n-1)\times(2n-2)!}{(2n-2)!}$

3. Hmm, try using the fact that n!= n(n-1)!=n(n-1)(n-2)!

4. 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:

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

Any help will be greatly appreciated.
$\frac{(2n)!}{(2n-2)!}$= $\frac{1*2*...*(n-2)*(n-1)*n*(n+1)*(n+2)*...*(2n-2)*(2n-1)*(2n)}{1*2*...*(n-2)*(n-1)*n*(n+1)*(n+2)*...*(2n-2)}$

Do you see where you can cancel and reduce the expression?

5. 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)

6. So the answer is 2n(2n-1)?

7. Originally Posted by mathgeek777