To find the number of odd numbers between 100 and 600 that have distinct digits, let's break down the problem step by step.

Step 1: Define the range and characteristics of the numbers

The numbers are three-digit numbers between 100 and 600. Therefore, the hundreds digit can be between 1 and 5. Additionally, since the number must be odd, the units digit must be one of the odd digits: 1, 3, 5, 7, or 9.

Step 2: Consider the hundreds digit

The hundreds digit determines the range:

• It can be 1, 2, 3, 4, or 5.

Step 3: Units digit (must be odd)

The units digit must be an odd number: 1, 3, 5, 7, or 9.

Step 4: Middle digit selection (tens digit)

The tens digit must be a distinct number that hasn't been used as the hundreds or units digit.

Step 5: Calculate the possible combinations

We will consider each of the cases based on the hundreds digit separately.

1. Hundreds digit = 1:

   • Units digit choices (odd and distinct from 1): 3, 5, 7, 9 (4 choices)
   • Tens digit choices (distinct from hundreds and units digit): 8 choices after the units digit is selected.

   Total combinations when hundreds digit is 1:
   $4 \times 8 = 32$

2. Hundreds digit = 2:

   • Units digit choices: 1, 3, 5, 7, 9 (5 choices)
   • Tens digit choices: 8 choices after the units digit is selected.

   Total combinations when hundreds digit is 2:
   $5 \times 8 = 40$

3. Hundreds digit = 3:

   • Units digit choices: 1, 5, 7, 9 (4 choices)
   • Tens digit choices: 8 choices.

   Total combinations when hundreds digit is 3:
   $4 \times 8 = 32$

4. Hundreds digit = 4:

   • Units digit choices: 1, 3, 5, 7, 9 (5 choices)
   • Tens digit choices: 8 choices.

   Total combinations when hundreds digit is 4:
   $5 \times 8 = 40$

5. Hundreds digit = 5:

   • Units digit choices: 1, 3, 7, 9 (4 choices since 5 cannot be repeated)
   • Tens digit choices: 8 choices.

   Total combinations when hundreds digit is 5:
   $4 \times 8 = 32$

Step 6: Add all the combinations

Total distinct three-digit odd numbers between 100 and 600:

$32 + 40 + 32 + 40 + 32 = 176$

Therefore, there are 176 odd numbers between 100 and 600 with distinct digits.

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Would you like further clarification on any part of this solution? Here are some related questions:

1. How would the count change if the digits didn't have to be distinct?
2. What if the number had to be even instead of odd?
3. How many three-digit numbers have all distinct digits without any restrictions on odd or even?
4. What if the range was between 500 and 700?
5. How can we generalize this method for different ranges or digit conditions?

Tip: When counting combinations, always start by setting constraints for each digit position and work through each scenario systematically!