<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>(b) repetition of digits is not allowed?</p>
Step-by-Step Solution
Key Concept: For a six-digit odd number greater than 6,00,000: the first digit must be 6, 7, 8, or 9 (and ≥6); the last digit must be odd (5, 7, or 9); all digits must be distinct from {5, 6, 7, 8, 9, 0}. Use conditional counting by cases based on first and last digit constraints.
<p><strong>Step 1: Identify constraints</strong></p><p>• Six-digit number using digits {0, 5, 6, 7, 8, 9} without repetition<br>• First digit ≥ 6 (so number > 6,00,000): must be from {6, 7, 8, 9}<br>• Last digit must be odd: must be from {5, 7, 9}<br>• No repetition allowed</p><p><strong>Step 2: Case-by-case analysis</strong></p><p><strong>Case 1: First digit = 6</strong><br>Last digit ∈ {5, 7, 9} → 3 choices<br>Middle 4 digits: arrange 4 remaining digits in 4! = 24 ways<br>Subtotal: 3 × 24 = 72</p><p><strong>Case 2: First digit = 7</strong><br>Last digit ∈ {5, 9} (cannot be 7) → 2 choices<br>Middle 4 digits: arrange 4 remaining digits in 4! = 24 ways<br>Subtotal: 2 × 24 = 48</p><p><strong>Case 3: First digit = 8</strong><br>Last digit ∈ {5, 7, 9} → 3 choices<br>Middle 4 digits: arrange 4 remaining digits in 4! = 24 ways<br>Subtotal: 3 × 24 = 72</p><p><strong>Case 4: First digit = 9</strong><br>Last digit ∈ {5, 7} (cannot be 9) → 2 choices<br>Middle 4 digits: arrange 4 remaining digits in 4! = 24 ways<br>Subtotal: 2 × 24 = 48</p><p><strong>Step 3: Sum all cases</strong><br>Total = 72 + 48 + 72 + 48 = <strong>240</strong></p>
Correct Answer: 240