Question:

<p><strong>Step 1: Identify available digits and their frequencies.</strong></p><p>Available: 0(1), 1(1), 2(2), 3(3). We need to form 3-digit numbers where the first digit ≠ 0.</p><p><strong>Step 2: Organize by cases based on the digits used in the three-digit number.</strong></p><p>Since we have limited copies, we consider which multisets of 3 digits can be selected from our available pool.</p><p><strong>Case 1: All three digits are the same</strong></p><p>Only 3 appears 3 times → {3,3,3}</p><p>Number of arrangements = 1</p><p><strong>Case 2: Two digits are the same, one different</strong></p><p>• {2,2,x} where x ∈ {0,1,3}: We have 3 such sets</p><p> - {2,2,0}: arrangements = 3!/2! = 3, but first digit ≠ 0 → all 3 valid ✓</p><p> - {2,2,1}: arrangements = 3!/2! = 3, all valid ✓</p><p> - {2,2,3}: arrangements = 3!/2! = 3, all valid ✓</p><p> Subtotal: 9</p><p>• {3,3,x} where x ∈ {0,1,2}: We have 3 such sets</p><p> - {3,3,0}: arrangements = 3!/2! = 3, first digit ≠ 0 → 2 valid ✓</p><p> - {3,3,1}: arrangements = 3!/2! = 3, all valid ✓</p><p> - {3,3,2}: arrangements = 3!/2! = 3, all valid ✓</p><p> Subtotal: 8</p><p><strong>Case 3: All three digits are different</strong></p><p>Select 3 different digits from {0,1,2,3}: C(4,3) = 4 sets</p><p>• {0,1,2}: arrangements = 3! = 6, first digit ≠ 0 → 4 valid ✓</p><p>• {0,1,3}: arrangements = 3! = 6, first digit ≠ 0 → 4 valid ✓</p><p>• {0,2,3}: arrangements = 3! = 6, first digit ≠ 0 → 4 valid ✓</p><p>• {1,2,3}: arrangements = 3! = 6, all valid ✓</p><p>Subtotal: 16</p><p><strong>Step 3: Add all valid numbers.</strong></p><p>Total = 1 + 9 + 8 + 16 = 34</p><p><strong>∴ Answer: 34</strong></p>