$\left[\begin{array}{l}[\text { Classical combinatorics (other) })] \\ {[\quad \text { Rule of product }}\end{array}\right]$

a) Among the first thousand integers (including 1000), which are there more of: those in whose notation there is a one, or the others?

b) Which are there more of: seven-digit numbers in whose notation there is a one, or the others?
a) There are $8^{3}=512$ three-digit numbers that do not contain 1 and 0. This is already more than half of the numbers in the first thousand.

b) Let's count the number of numbers that do not contain the digit 1. The first digit can be any of the eight digits (not 0 and not 1), and each of the remaining digits can be any of the nine digits different from 1. In total, we get $8 \cdot 9^{6}$ numbers, which is less than half of the total $9 \cdot 10^{6}$ seven-digit numbers (see problem $\underline{60336}$ ).

## Answer

There are more numbers in which a) there is no digit 1; b) there is a digit 1.