<p>The number of numbers between 2,000 and 5,000 that can be formed with the digits 0, 1, 2, 3, 4 (repetition of digits is not allowed) and are multiple of 3 is</p>
Step-by-Step Solution
Key Concept: A number is divisible by 3 if the sum of its digits is divisible by 3. For 4-digit numbers between 2000-5000, the first digit must be 2, 3, or 4. Identify which 4-digit combinations from {0,1,2,3,4} have digit sums divisible by 3, then count valid arrangements.
<p><strong>Step 1: Identify constraints</strong></p><p>Numbers are between 2000-5000, so first digit ∈ {2, 3, 4}. We use 4 digits from {0,1,2,3,4} without repetition.</p><p><strong>Step 2: Find digit sum of all 5 digits</strong></p><p>0+1+2+3+4 = 10 ≡ 1 (mod 3)</p><p><strong>Step 3: Determine valid 4-digit combinations</strong></p><p>For sum ≡ 0 (mod 3), we exclude one digit whose value ≡ 1 (mod 3):</p><p>• Exclude 1: {0,2,3,4}, sum = 9 ✓</p><p>• Exclude 4: {0,1,2,3}, sum = 6 ✓</p><p><strong>Step 4: Count arrangements for {0,2,3,4}</strong></p><p>First digit ∈ {2,3,4}:</p><p>• First digit = 2: arrange {0,3,4} in 3 positions = 3! = 6</p><p>• First digit = 3: arrange {0,2,4} in 3 positions = 3! = 6</p><p>• First digit = 4: arrange {0,2,3} in 3 positions = 3! = 6</p><p>Subtotal = 18</p><p><strong>Step 5: Count arrangements for {0,1,2,3}</strong></p><p>First digit ∈ {2,3}:</p><p>• First digit = 2: arrange {0,1,3} in 3 positions = 3! = 6</p><p>• First digit = 3: arrange {0,1,2} in 3 positions = 3! = 6</p><p>Subtotal = 12</p><p><strong>Step 6: Total count</strong></p><p>18 + 12 = 30</p><p>∴ Answer: C</p>
Correct Answer: C