<p>An eight digit number divisible by 9 is to be formed using digits from 0 to 9 without repeating the digits. The number of ways in which this can be done is</p>
Step-by-Step Solution
Key Concept: A number is divisible by 9 if and only if the sum of its digits is divisible by 9. Since we select 8 digits from {0,1,2,...,9}, we must exclude exactly 2 digits whose sum ≡ 0 (mod 9), then arrange the remaining 8 digits ensuring the first digit isn't 0.
<p><strong>Step 1:</strong> Find sum of all digits 0-9: 0+1+2+...+9 = 45 ≡ 0 (mod 9)</p><p><strong>Step 2:</strong> For an 8-digit number using 8 of these 10 digits, we exclude 2 digits. Their sum must satisfy: 45 - (sum of excluded) ≡ 0 (mod 9), so excluded digits must sum to 0, 9, or 18 (mod 9).</p><p><strong>Step 3:</strong> Find pairs (a,b) with a < b from {0,...,9} that sum to 0, 9, or 18:</p><p>• Sum = 0: (0,0) - invalid (need distinct digits)</p><p>• Sum = 9: (0,9), (1,8), (2,7), (3,6), (4,5) → 5 pairs</p><p>• Sum = 18: (9,9) - invalid</p><p><strong>Step 4:</strong> For each valid pair of excluded digits, we have 8 remaining digits to arrange as an 8-digit number (first digit ≠ 0).</p><p><strong>Case 1:</strong> If 0 is excluded (pair (0,9)): All 8 remaining digits can be arranged freely in first position. Arrangements = 8!</p><p><strong>Case 2:</strong> If 0 is not excluded (pairs (1,8), (2,7), (3,6), (4,5)): Total arrangements = 8!, but subtract those starting with 0. Valid arrangements = 8! - 7! = 7! × 7</p><p><strong>Step 5:</strong> Total ways = 1 × 8! + 4 × 7! × 7 = 8! + 28 × 7! = 40320 + 141120 = 181440 = 7! × (8 + 28) = 7! × 36</p><p><strong>Step 6:</strong> Alternatively: 1 × 8! + 4 × 7 × 7! = 8 × 7! + 28 × 7! = 36 × 7! = <strong>1,814,400</strong> or expressed as <strong>36 × 7!</strong></p><p>∴ Answer: D</p>
Correct Answer: D