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

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Answer 1 · 2015-11-18T19:03:56

I got: $n(n-1)$

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 $n=4$
$(n!)/((n-2)!) =(4!)/(2!) =12$
$n(n-1)=4*3=12$

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