<p>Five different digits from the set of numbers \(\{1, 2, 3, 4, 5, 6, 7\}\) are written in random order. Find the probability that the five-digit number thus formed is divisible by 9.</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. We need to find which 5-digit subsets from {1,2,3,4,5,6,7} have digit sums divisible by 9, then count favorable arrangements.
<p><strong>Step 1:</strong> Find the sum of all digits: 1+2+3+4+5+6+7 = 28</p><p><strong>Step 2:</strong> When selecting 5 digits from 7, we exclude 2 digits. If we exclude digits with sum S, the remaining 5 digits have sum (28 - S).</p><p><strong>Step 3:</strong> For the 5-digit number to be divisible by 9, we need: 28 - S ≡ 0 (mod 9), which means S ≡ 28 ≡ 1 (mod 9)</p><p><strong>Step 4:</strong> Find all pairs of digits from {1,2,3,4,5,6,7} whose sum ≡ 1 (mod 9):</p><p>Possible sums: 1, 10. Check pairs:</p><p>• Sum = 10: (3,7), (4,6) → 2 pairs</p><p>• Sum = 1: Not possible with distinct positive digits</p><p><strong>Step 5:</strong> Total ways to choose and arrange 5 digits from 7: P(7,5) = 7!/(2!) = 2520</p><p><strong>Step 6:</strong> Favorable outcomes (2 valid pairs × arrangements of 5 digits): 2 × 5! = 2 × 120 = 240</p><p><strong>Step 7:</strong> Probability = 240/2520 = 2/21</p><p>∴ Answer: <strong>2/21</strong></p>
Correct Answer: A