Question

How do you simplify `(n!)/((n-2)!)`?

Answer 1

I got: `n(n-1)`

Explanation:

We can write it as:
`(n*(n-1)(n-2)(n-3)!)/((n-2)(n-3)!)=`
where you used the fact that `n! =n(n-1)!`

and so:

`(n*(n-1)cancel((n-2))cancel((n-3)!))/(cancel((n-2))cancel((n-3)!))=n(n-1)`

Check it with `n=4`
`(n!)/((n-2)!) =(4!)/(2!) =12`
`n(n-1)=4*3=12`