1. [4 points] Find the number of eight-digit numbers, the product of whose digits equals 4900. The answer should be presented as an integer.
Answer: 4200.

Solution. Since $4900=7^{2} \cdot 2^{2} \cdot 5^{2}$, the sought numbers can consist of the following digits: (a) two twos, two fives, two sevens, and two ones, or (b) a four, two fives, two sevens, and three ones. We will calculate the number of variants in each case.

(a) First, we choose two places out of eight for the twos ( $C_{8}^{2}=\frac{8!}{2!6!}$ ways), then two places out of the remaining six for the fives ( $C_{6}^{2}=\frac{6!}{4!2!}$ ways), and then two places out of the remaining four for the sevens ( $C_{4}^{2}=\frac{4!}{2!2!}$ ways). Finally, the remaining places are occupied by ones. By the rule of product, we get $C_{8}^{2} \cdot C_{6}^{2} \cdot C_{4}^{2}=\frac{8!}{2!2!2!2!} \cdot 4=2520$ ways.

(b) Reasoning similarly, we find that the number of ways in this case is $\frac{8!}{2!2!3!}=1680$. Finally, we get $2520+1680=4200$ ways.