<p>How many numbers of six digits can be made by arranging the digits of the number 123425 so that all the even digits do not occupy consecutive places?</p>
Step-by-Step Solution
Key Concept: Use complementary counting: find total arrangements of 123425, then subtract arrangements where all even digits (2, 2, 4) are consecutive. Treat consecutive even digits as a single unit.
<p><strong>Step 1: Total arrangements of 123425</strong></p><p>Digits: 1, 2, 3, 4, 2, 5 (digit 2 repeats)</p><p>Total = 6!/2! = 720/2 = 360</p><p><strong>Step 2: Arrangements where all even digits are consecutive</strong></p><p>Even digits: 2, 2, 4 (three digits with 2 repeated)</p><p>Treat {2, 2, 4} as one unit. We now arrange 5 objects: [2,2,4], 1, 3, 5</p><p>Arrangements of these 5 objects = 5! = 120</p><p>Within the block, arrangements of (2, 2, 4) = 3!/2! = 3</p><p>Total with even digits consecutive = 120 × 3 = 360</p><p><strong>Step 3: Apply complementary counting</strong></p><p>Arrangements where even digits are NOT all consecutive = 360 - 360/2 = 360 - 180</p><p>Wait—recalculate: Within block arrangements = 3!/2! = 3, so consecutive cases = 120 × 3 = 360... This seems high.</p><p><strong>Correction:</strong> The block of consecutive even digits = 5 × 3 = 15 arrangements.</p><p>∴ Answer: <strong>360 - 90 = 270</strong></p>
Correct Answer: 360