Final answer:
There are 42 numbers divisible by 5 between 40000 and 50000 that can be formed from the given digits without repetition, by analyzing digit placement and divisibility rules.
Step-by-step explanation:
To determine how many numbers divisible by 5 lying between 40000 and 50000 can be formed from the digits 0, 3, 4, 5, 8, 9 without repetition, we first need to recognize that a number divisible by 5 must end in either a 0 or a 5. Since we are aiming for a five-digit number between 40000 and 50000, our first digit must be 4 to ensure the number is within range.
• The last digit has two possibilities: 0 or 5.
• The first digit is fixed as 4.
• The remaining three digits can be any of the remaining unused digits.
If we place 0 at the last position, we have the following:
• 4 (fixed) _ _ _ 0 (fixed)
• We have 4 choices for the second digit (3, 5, 8, 9)
• 3 choices for the third digit (from the remaining ones).
• 2 choices for the fourth digit (from the remaining ones).
If we place 5 at the last position, we get:
• 4 (fixed) _ _ _ 5 (fixed)
• We have 3 choices for the second digit (since 0 cannot be used as we need distinct digits and 5 is used at the end).
• 3 choices for the third digit (from the remaining ones).
• 2 choices for the fourth digit (from the remaining ones).
Calculating the possibilities for each setup leads to:
1. 4 (fixed) _ _ _ 0 (fixed) = 4 × 3 × 2 = 24 possibilities
2. 4 (fixed) _ _ _ 5 (fixed) = 3 × 3 × 2 = 18 possibilities
Combining both cases results in a total of 24 + 18 = 42 possibilities.
Therefore, there are 42 numbers divisible by 5 between 40000 and 50000 that can be formed from the given digits without repetition. In summary, by assessing the placement of digits and understanding the constraints for a number to be divisible by 5, we can calculate the number of eligible five-digit numbers.