# Thread: Combinations simplification, please revise and help

1. ## Combinations simplification, please revise and help

Hi, I'm trying to simplify this but I get stuck at some point.

it goes 1/n! * C(n,2) * P(n,n-2)

which yields:

1/n! * n!/2!(n-2)! * n!/(n - (n-2))!

which so far I have reduced down to

2! * (n-2)! * n! / (n - (n - 2))!

Is this correct? help with further simplification? Thank you!

2. Originally Posted by domain07
Hi, I'm trying to simplify this but I get stuck at some point.

it goes 1/n! * C(n,2) * P(n,n-2)

which yields:

1/n! * n!/2!(n-2)! * n!/(n - (n-2))!

which so far I have reduced down to

2! * (n-2)! * n! / (n - (n - 2))!

Is this correct? help with further simplification? Thank you!
Hmm, I get

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

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

and $\displaystyle C(n,2)=\frac{n!}{(n-2)!\cdot2!}$

Overall, $\displaystyle \left(\frac{1}{n!}\right)\cdot C(n,2)\cdot P(n,n-2)$

$\displaystyle =\left(\frac{1}{n!}\right)\left(\frac{n!}{2}\right )\left(\frac{n!}{(n-2)!\cdot2!}\right)$

$\displaystyle =\frac{n!}{(n-2)!\cdot4}$

$\displaystyle =\frac{n\cdot(n-1)\cdot(n-2)!}{(n-2)!\cdot4}$

$\displaystyle =\frac{n(n-1)}{4}$