<p>How many numbers of five digits can be made with at least one repeated digit?</p>
Step-by-Step Solution
Key Concept: Use complementary counting: Total 5-digit numbers with at least one repeated digit = (All 5-digit numbers) - (5-digit numbers with all distinct digits). The first digit cannot be 0, which requires careful case analysis.
<p><strong>Step 1:</strong> Count total 5-digit numbers (first digit ≠ 0): 9 × 10^4</p><p><strong>Step 2:</strong> Count 5-digit numbers with all distinct digits using permutation: 10P5 = 10!/(10-5)! = 30,240</p><p><strong>Step 3:</strong> This 10P5 includes invalid cases where first digit is 0. When first digit is 0, we need 4 more distinct digits from remaining 9 digits: 9P4 = 9!/(9-4)! = 3,024</p><p><strong>Step 4:</strong> Valid 5-digit numbers with all distinct digits = 10P5 - 9P4</p><p><strong>Step 5:</strong> Numbers with at least one repeated digit = (Total 5-digit numbers) - (All distinct digits) = 9 × 10^4 - (10P5 - 9P4) = 9 × 10^4 - 10P5 + 9P4</p><p>∴ Answer: <i>9 × 10^4 - </i>^10P_5 <i>+ </i>^9P_4</p>
Correct Answer: \(9 \times 10^4 - {}^{10}P_5 + {}^{9}P_4\)