How many six-digit odd numbers can be formed from the digits 0, 2, 3, 5, 6, 7, 8 and 9 if no digit is repeated in any number?

Introduction / Context:
This question asks for the count of six-digit odd numbers formed from a specific set of digits where repetition is not allowed. It tests understanding of both place value restrictions (odd number means last digit must be odd) and the basic permutation counting rules with exclusions on leading zeros. Paying attention to which digits can occupy the units and thousands places is essential.

Given Data / Assumptions:
- Available digits: 0, 2, 3, 5, 6, 7, 8, 9 (8 distinct digits in total).
- We must form six-digit numbers (first digit cannot be zero).
- Each digit can appear at most once (no repetition allowed).
- The number must be odd, so the last digit must be one of the available odd digits.
- Odd digits available: 3, 5, 7, 9 (4 digits).

Concept / Approach:
We first determine the possible choices for the last digit because of the odd-number condition. After fixing the last digit, we then choose the first digit, remembering that it cannot be zero. The remaining four middle positions can then be filled with any of the remaining digits without repetition. We proceed step by step, counting the number of options at each position and multiplying using the rule of product.

Step-by-Step Solution:
Step 1: Choose the units (last) digit. Since the number must be odd, the last digit must be 3, 5, 7, or 9.Step 2: There are 4 choices for the last digit.Step 3: Once the last digit is chosen, it cannot be used again. There remain 7 unused digits, including zero.Step 4: Choose the first (leftmost) digit from these 7 remaining digits, but it cannot be zero because we need a six-digit number.Step 5: Out of the 7 remaining digits, exactly 1 is zero, so there are 7 - 1 = 6 valid choices for the first digit.Step 6: After fixing the first and last digit, there are 6 unused digits left to fill the 4 middle positions.Step 7: The middle four positions can be filled by these 6 digits without repetition in 6P4 ways (permutations of 6 taken 4 at a time).Step 8: Compute 6P4 = 6 * 5 * 4 * 3 = 360.Step 9: Total number of valid numbers = (choices for last digit) * (choices for first digit) * (arrangements of middle digits) = 4 * 6 * 360.Step 10: Compute 4 * 6 = 24; 24 * 360 = 8640.

Verification / Alternative check:
We can confirm the reasoning by considering the total six-digit numbers that can be formed without the oddness restriction and then restricting the last digit to the 4 odd ones. The step-by-step approach above ensures we never assign zero to the first place and never reuse digits. The final multiplication 4 * 6 * 360 is straightforward, and recomputing 6P4 as 6 * 5 * 4 * 3 = 360 confirms no arithmetic mistake has been made.

Why Other Options Are Wrong:
- 720 is far too small and might come from counting only 6! permutations, ignoring the restrictions.

- 3620 and 4512 are intermediate-like numbers but do not match any natural product of 4, 6 and a permutation term; they likely reflect mis-handling of the zero constraint or incomplete multiplication.

Common Pitfalls:
Students often forget that the first digit of a six-digit number cannot be zero, leading them to count some invalid numbers. Another error is to treat the middle four positions as combinations instead of permutations, using 6C4 instead of 6P4, which ignores the importance of order in forming different numbers. Ensuring that each positional choice respects both non-repetition and the number structure avoids such mistakes.

Final Answer:
There are 8640 six-digit odd numbers that can be formed from the given digits without repetition.