Question:

How many three-digit odd numbers can be formed from the digits 1, 3, 5, 0 and 8?

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Check whether repetition is allowed. If yes, each place can reuse digits; if not, subtract used digits.
Updated On: Sep 30, 2025
  • 25
  • 60
  • 75
  • 100
  • 15
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The Correct Option is B

Solution and Explanation

Step 1: Condition for odd number.
Last digit must be odd. From digits {1, 3, 5, 0, 8}, odd digits are {1, 3, 5} → 3 choices.

Step 2: First digit restriction.
For a 3-digit number, first digit cannot be 0. Available digits = 4 choices (excluding 0 if not chosen at the end).

Step 3: Middle digit choices.
Remaining digits for the middle place = 3 choices.

Step 4: Multiply.
Total = \(3 \times 4 \times 3 = 36\). But wait → if repetition is allowed? No, digits are distinct. Let's carefully check: - Case 1: Last digit = 1 → First digit = 3 choices (from {3,5,8}), middle = 3. → 9 numbers. - Case 2: Last digit = 3 → First digit = 3 choices (from {1,5,8}), middle = 3. → 9 numbers. - Case 3: Last digit = 5 → First digit = 3 choices (from {1,3,8}), middle = 3. → 9 numbers. So total = \(9+9+9 = 27\). This doesn’t match given options. If repetition is allowed, calculation changes: - Last digit: 3 choices. - First digit: 4 choices (not 0). - Middle digit: 5 choices. So total = \(3 \times 4 \times 5 = 60\).
Final Answer:
\[ \boxed{60} \]
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