2. Which three-digit numbers are more numerous: those in which all digits have the same parity, or those in which adjacent digits have different parity
Answer: equally

Instructions. Method 1. We will consider the numbers of interest by tens. In each ten, the hundreds and units digits are the same (they are of the same parity). There are exactly 10 tens: five of each parity, and thus, from this ten, five numbers will be added to each type. Therefore, the numbers of each type are the same.

Method 2. Let's count how many three-digit numbers have digits of the same parity. Even: the hundreds digit can be chosen in 4 ways (the number cannot start with zero), and the tens and units digits can be chosen in 5 ways each. In total, $4 \cdot 5 \cdot 5 = 100$ ways. Odd: the hundreds digit can be chosen in 5 ways, as well as the tens and units digits. In total, $5 \cdot 5 \cdot 5 = 125$ ways. Therefore, the number of numbers with digits of the same parity is $100 + 125 = 225$.

Let's count how many three-digit numbers have adjacent digits of different parity. EOE: the hundreds digit can be chosen in 4 ways (the number cannot start with zero), and the tens and units digits can be chosen in 5 ways each. In total, $4 \cdot 5 \cdot 5 = 100$ ways. OEO: the hundreds digit can be chosen in 5 ways, as well as the tens and units digits. In total, $5 \cdot 5 \cdot 5 = 125$ ways. Therefore, the number of numbers with adjacent digits of different parity is $100 + 125 = 225$.

There are 225 of each, and thus, they are equally numerous.

Criteria. Answer without justification: 0 points.