7.3. How many five-digit natural numbers exist, each of which has adjacent digits with different parity
Answer: 5625. Solution. If the first 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 $(1,3,5,7,9$ - for the second and fourth digits and $0,2,4,6,8$ - for the third and fifth). In the end (by the rule of product), we have a total of $4 \cdot 5^{4}=2500$ numbers. Similarly, in the case of an odd first digit, we get $5^{5}=3125$ numbers. Thus, the total number of numbers is 5625. Remark. The same result can be obtained if we immediately use the rule of product. The first (most significant) 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: $9 \cdot 5^{4}$.