Permutations & Combinations
Permutations with Restrictions
Grade 11
Question:
<p>The number of integers greater than 6,000 that can be formed, using the digits 3, 5, 6, 7 and 8, without repetition, is:</p>
<p>(A) 72</p>
<p>(B) 216</p>
<p>(C) 192</p>
<p>(D) 120</p>
Step-by-Step Solution
Key Concept: Count 4-digit numbers ≥ 6000 (by first digit 6,7,8) and all 5-digit permutations separately, then add.
<p><strong>Concept:</strong> Count all numbers using digits {3, 5, 6, 7, 8} without repetition that exceed 6,000.</p><p><strong>Step 1:</strong> Numbers greater than 6,000 include:</p><p>• 4-digit numbers ≥ 6,000</p><p>• All 5-digit numbers</p><p><strong>Step 2 (4-digit numbers ≥ 6,000):</strong> First digit must be 6, 7, or 8.</p><p>Case 1: First digit = 6. Remaining 3 positions filled by 4 remaining digits: \(P(4,3) = 4 \times 3 \times 2 = 24\)</p><p>Case 2: First digit = 7. Remaining 3 positions: \(P(4,3) = 24\)</p><p>Case 3: First digit = 8. Remaining 3 positions: \(P(4,3) = 24\)</p><p>Total 4-digit: \(24 + 24 + 24 = 72\)</p><p><strong>Step 3 (5-digit numbers):</strong> All permutations of 5 digits without repetition: \(5! = 120\)</p><p><strong>Step 4:</strong> Total = \(72 + 120 = 192\)</p><p>∴ Answer is C.</p>
Correct Answer: C