## Precalculus (10th Edition)

$80000.$
According to the Multiplication Principle, if we have $p$ selections for the first choice, $q$ selections for the second choice, then we have $p\cdot q$ different ways of selections. (similarly for more than two choices). For the first digit we have $8$ selections and for the rest we have $10$, hence $8\cdot10\cdot10\cdot10\cdot10=80000.$