How many 3-digit numbers can be composed by the digits 234568 if the 3-digit numbers were to be odd and repetion of a number is not allowed?

To form a 3-digit odd number using the digits 2, 3, 4, 5, 6, and 8 without repetition, the last digit must be odd. The available odd digits are 3 and 5.

If we choose 3 as the last digit, we can use any of the remaining 5 digits (2, 4, 5, 6, 8) for the first digit, and then any of the remaining 4 digits for the second digit, giving us (5 \times 4 = 20) combinations.

If we choose 5 as the last digit, we again have 5 choices for the first digit (2, 3, 4, 6, 8) and then 4 choices for the second digit, resulting in another (5 \times 4 = 20) combinations.

Adding both cases together, we have (20 + 20 = 40) possible 3-digit odd numbers.