<p>How many six-digit odd numbers, greater than 6,00,000, can be formed from the digits 5, 6, 7, 8, 9, and 0 if<br>(a) repetition of digits is allowed?</p>
Step-by-Step Solution
Key Concept: Break the constraint into independent conditions: first digit must be 6-9 (to exceed 6,00,000), last digit must be odd (5, 7, or 9), and middle positions can be any of 6 digits. Use multiplication principle treating each position separately.
<p><strong>Step 1:</strong> Identify constraints for a 6-digit odd number > 6,00,000</p><ul><li>Position 1 (first digit): Must be from {6, 7, 8, 9} to ensure number ≥ 6,00,000 → <strong>4 choices</strong></li><li>Position 6 (last digit): Must be odd from {5, 7, 9} → <strong>3 choices</strong></li><li>Positions 2, 3, 4, 5 (middle digits): Can be any digit from {5, 6, 7, 8, 9, 0} → <strong>6 choices each</strong></li></ul><p><strong>Step 2:</strong> Apply multiplication principle (digits are independent since repetition is allowed)</p><p>Total = (Choices for position 1) × (Choices for position 2) × (Choices for position 3) × (Choices for position 4) × (Choices for position 5) × (Choices for position 6)</p><p>Total = 4 × 6 × 6 × 6 × 6 × 3</p><p><strong>Step 3:</strong> Calculate</p><p>= 4 × 3 × 6⁴</p><p>= 12 × 1296</p><p>= <strong>15,552</strong></p><p>∴ Answer: <strong>15552</strong></p>
Correct Answer: 15552