## College Algebra (10th Edition)

$151,200$
RECALL: $P(n, r)= \dfrac{n!}{(n-r)!}$ Using the definition above gives: $\require{cancel} P(10, 6)! \\= \dfrac{10!}{(10-6)!} \\=\dfrac{10!}{4!} \\=\dfrac{10(9)(8)(7)(6)(5)\cancel{(4!)}}{\cancel{4!}} \\=151,200$