How do you simplify #(n!)/((n-3)!)#?

1 Answer
Shwetank Mauria
Mar 17, 2018

#(n!)/((n-3)!)=n^3-3n^2+2n#

Explanation:

#(n!)/((n-3)!)#

= #(n(n-1)(n-2)color(blue)((n-3)(n-4).....3*2*1))/((n-3)(n- 4).....3*2*1)#

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

= #n(n^2-3n+2)#

= #n^3-3n^2+2n#

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