## Intermediate Algebra for College Students (7th Edition)

$n^3-3n^2+2n$
Since,we have $\dfrac{n!}{(n-3)!}$ Thus, $\dfrac{n!}{(n-3)!}=\dfrac{n \cdot (n-1) \cdot (n-2) \cdot (n-3) !}{(n-3)!}$ or, $=n(n-1)(n-2)$ or, $=n(n^2-3n+2)$ or, $=n^3-3n^2+2n$