<p>If the probability of a six-digit number \(N\) whose six digits are 1, 2, 3, 4, 5, 6 written as random order is divisible by 6 is \(p\), then the value of \(1/p\) is ________.</p>
Step-by-Step Solution
Key Concept: A number is divisible by 6 iff it's divisible by both 2 and 3. Since digits 1,2,3,4,5,6 sum to 21 (divisible by 3), ALL arrangements are divisible by 3. Divisibility by 6 depends only on the last digit being even (2, 4, or 6).
<p><strong>Step 1:</strong> Find the total number of six-digit numbers using digits 1, 2, 3, 4, 5, 6 in random order: Total = 6! = 720</p><p><strong>Step 2:</strong> Check divisibility by 3: Sum of digits = 1+2+3+4+5+6 = 21, which is divisible by 3. Therefore, EVERY arrangement of these digits gives a number divisible by 3.</p><p><strong>Step 3:</strong> Check divisibility by 2: A number is divisible by 2 iff its last digit is even. Among {1,2,3,4,5,6}, the even digits are {2, 4, 6} (3 choices).</p><p><strong>Step 4:</strong> Count favorable outcomes: If the last digit is even (3 choices), the remaining 5 digits can be arranged in 5! = 120 ways. Favorable outcomes = 3 × 120 = 360</p><p><strong>Step 5:</strong> Calculate probability: p = 360/720 = 1/2</p><p><strong>Step 6:</strong> Calculate 1/p: 1/p = 2/1 = 2</p><p>∴ Answer: 2</p>
Correct Answer: 3