108. Bus tickets have numbers from 000001 to 999999. A number is considered lucky if the first three digits are odd and different, the second three digits are even, and 7 and 8 do not stand next to each other. How many different lucky numbers exist?
Solution. The number of numbers satisfying the first two conditions of the problem (but perhaps not satisfying the third condition) can be calculated as follows: the first digit of such a number can be chosen in five ways, the second in four ways, the third in three ways, and each subsequent digit in five ways, so the total number of such tickets is 7500 $(5 \cdot 4 \cdot 3 \cdot 5 \cdot 5 \cdot 5=7500)$.

Among the considered numbers, those that do not satisfy the third condition of the problem are the numbers where the third digit is 7 and the fourth digit is 8. The number of such numbers is calculated in the same way as above: it is equal to $300(4 \cdot 3 \cdot 5 \cdot 5=300)$.

Therefore, the number of different lucky numbers is $7200(7500-300=7200)$.

Answer: 7200.