Problem 3. How many odd three-digit numbers exist where the digit in the tens place is 4?
Solution. The unit digit can be 1, 3, 5, 7 or 9, which means we have 5 possibilities. The tens digit is 4, which means we have 1 possibility. The hundreds digit can be: $1,2,3,4,5,6,7,8$ or 9, which means we have 9 possibilities.

Finally, the number of desired numbers is $5 \cdot 1 \cdot 9=45$.