## Intermediate Algebra (6th Edition)

$20$
Cancelling the common factors between the numerator and the denominator, the given expression, $\dfrac{5\cdot4\cdot3\cdot2\cdot1}{3\cdot2\cdot1} ,$ simplifies to \begin{array}{l}\require{cancel} \dfrac{5\cdot4\cdot\cancel{3\cdot2\cdot1}}{\cancel{3\cdot2\cdot1}} \\\\= 5\cdot4 \\\\= 20 .\end{array}