7.4. How many six-digit natural numbers exist, each of which has adjacent digits with different parity
Answer: 28125. Solution. If the first (most significant) digit is even, then it can be chosen in four ways $(2,4,6,8)$, and all subsequent ones can be chosen in five ways (possible candidates for the second, fourth, and sixth digits are $1,3,5,7,9$, and for the third and fifth - $0,2,4,6,8$). In the end, by the rule of product, we will have a total of $4 \cdot 5^{5}=12500$ numbers with the first even digit. Similarly, in the case of an odd first digit, we get $5 \cdot 5^{5}=15625$ numbers. Thus, the total number of the desired numbers is 28125. Remark. The same result can be obtained if we immediately use the rule of product. The first digit can be chosen in nine ways (taking any digit except 0), after which the second digit can be chosen in five ways (taking any digit with a different parity from the first digit), and so on: the following digits can be chosen in five ways (and the number of ways does not depend on the previous digits). Therefore, by the rule of product, we get the result: $9 \cdot 5^{5}$.