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

1 Answer
Nov 18, 2015

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#