## Precalculus (10th Edition)

$132,860$
Order matters so the question is involves permutation. We know that the number of permutations for $n$ objects taken $r$ at a time is given by the formula $P(n,r)=\dfrac{n!}{(n-r)!}.$ Here we have $n=365$ and $r=2$. Hence, $P(365,2)=\dfrac{365!}{(365-2)!}=\dfrac{365\cdot364\cdot363!}{363!}=365\cdot364=132860$.