<p>The number of five-digit numbers that contain 7 exactly once is</p>
Step-by-Step Solution
Key Concept: Partition the problem: first choose 1 position (out of 5) for the digit 7, then fill remaining 4 positions with digits from {0,1,2,3,4,5,6,8,9} while ensuring the first digit is never 0.
<p><strong>Step 1:</strong> Identify that we need exactly one 7 in a five-digit number. Choose 1 position out of 5 for the 7: this splits into two cases.</p><p><strong>Step 2 (Case 1 - 7 in first position):</strong> If 7 is in position 1, the remaining 4 positions can be filled with any digit from {0,1,2,3,4,5,6,8,9} (9 choices each). Count = 1 × 9⁴ = 6561</p><p><strong>Step 3 (Case 2 - 7 in positions 2, 3, 4, or 5):</strong> Choose 1 of these 4 positions for 7: 4 ways. First position must be non-zero and not 7, so 8 choices {1,2,3,4,5,6,8,9}. Remaining 3 positions can have any digit except 7, so 9 choices each. Count = 4 × 8 × 9³ = 4 × 8 × 729 = 23,328</p><p><strong>Step 4:</strong> Total = 6561 + 23,328 = 29,889</p><p>∴ Answer: A</p>
Correct Answer: A