Ten prime-digit numbers with digit product 100000

Count the ten-digit numbers in which every digit is a prime number and the product of all ten digits equals $100000$.

$252$. Since $100000=2^5\cdot5^5$, the digits can only be $2$s and $5$s, with five of each to use all ten places. Choosing which five positions hold the $2$s gives $\binom{10}{5}=252$.