<p>A five-digit numbers divisible by 3 is to be formed using the numerals 0, 1, 2, 3, 4 and 5, without repetition. Find the total number of ways in which this can be done.</p>
Step-by-Step Solution
Key Concept: A number is divisible by 3 if and only if the sum of its digits is divisible by 3. From {0,1,2,3,4,5}, we must select 5 digits whose sum is divisible by 3, then arrange them ensuring 0 is not in the first position.
<p><strong>Step 1: Find valid 5-digit subsets</strong></p><p>Sum of all digits: 0+1+2+3+4+5 = 15 (divisible by 3)</p><p>When we exclude one digit, the remaining 5 must have sum ≡ 0 (mod 3):</p><p>• Exclude 0: sum = 15 ✓ (divisible by 3)</p><p>• Exclude 1: sum = 14 ✗</p><p>• Exclude 2: sum = 13 ✗</p><p>• Exclude 3: sum = 12 ✓ (divisible by 3)</p><p>• Exclude 4: sum = 11 ✗</p><p>• Exclude 5: sum = 10 ✗</p><p><strong>Step 2: Count arrangements for each valid subset</strong></p><p><strong>Case 1: Digits {1,2,3,4,5} (0 excluded)</strong></p><p>All 5 digits can start: 5! = 120 arrangements</p><p><strong>Case 3: Digits {0,1,2,3,4} (5 excluded)</strong></p><p>Total arrangements: 5! = 120</p><p>Arrangements with 0 first (invalid): 4! = 24</p><p>Valid arrangements: 120 - 24 = 96</p><p><strong>Step 3: Total</strong></p><p>120 + 96 = <strong>216</strong></p>
Correct Answer: 216