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

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Answer 1 · 2016-11-02T03:29:55

This can be simplified to #n(n - 1) = n^2- n#.

Expand the factorial in the numerator, using the definition that #n! = (n)(n - 1)(n - 2)(n - 3)...#.

#(n!)/((n - 2)!)#

#=>((n)(n - 1)(n - 2)!)/((n -2)!)#

#=>n(n - 1)#

#=>n^2- n#

Hopefully this helps!

Answer 2 · 2016-11-02T03:34:47

Please see the explanation to understand how it simplifies to #n(n -1)#

#(n!)/((n -2)!) = ((n)(n-1)(n-2)!)/((n-2)!) =#

Cancel:

#((n)(n-1)cancel(n-2)!)/(cancel(n-2)!) = n(n -1)#