1. [3 points] Find the number of eight-digit numbers, the product of the digits of each of which is equal to 3375. The answer should be presented as an integer.
Answer: 1680.

Solution. Since $3375=3^{3} \cdot 5^{3}$, the sought numbers can consist of the following digits: (a) three threes, three fives, and two ones, or (b) one three, one nine, three fives, and three ones. We will calculate the number of variants in each case.

(a) First, we choose three places out of eight for the threes $\left(C_{8}^{3}=\frac{8!}{3!5!}\right.$ ways), then three places out of the remaining five for the fives ( $C_{5}^{3}=\frac{5!}{3!2!}$ ways). Finally, the remaining places are occupied by ones. By the multiplication rule, we get $C_{8}^{3} \cdot C_{5}^{3}=\frac{8!}{3!3!2!}=560$ ways.

(b) Reasoning similarly, we find that the number of ways in this case is $\frac{8!}{3!3!}=1120$.

Finally, we get $560+1120=1680$ ways.