<p>8-digit numbers are formed using the digits 1, 1, 2, 2, 2, 3, 4, 4. The number of such numbers in which the odd digits do not occupy odd places, is</p>
Step-by-Step Solution
Key Concept: Use complementary counting: find total arrangements, then subtract arrangements where at least one odd digit occupies an odd position. Alternatively, count directly by placing odd digits (1,1,3) only in even positions (4 available), then arrange remaining digits in remaining positions.
<p><strong>Step 1: Identify the digits and positions</strong></p><p>Digits: 1, 1, 2, 2, 2, 3, 4, 4 (odd digits: 1, 1, 3; even digits: 2, 2, 2, 4, 4)</p><p>In an 8-digit number: Odd positions (1st, 3rd, 5th, 7th) = 4 positions; Even positions (2nd, 4th, 6th, 8th) = 4 positions</p><p><strong>Step 2: Constraint interpretation</strong></p><p>"Odd digits do not occupy odd places" means all odd digits (1, 1, 3) must go in the 4 even positions only.</p><p><strong>Step 3: Place odd digits in even positions</strong></p><p>We have 3 odd digits (1, 1, 3) and 4 even positions. Choose 3 of 4 even positions:</p><p>Ways = C(4,3) × (3!/2!) = 4 × 3 = 12</p><p><strong>Step 4: Arrange remaining digits</strong></p><p>Remaining digits (2, 2, 2, 4, 4) must fill 4 odd positions + 1 remaining even position = 5 positions.</p><p>Arrangements = 5!/(3! × 2!) = 120/(6 × 2) = 10</p><p><strong>Step 5: Apply multiplication principle</strong></p><p>Total arrangements = 12 × 10 = 120</p><p>∴ Answer: C</p>
Correct Answer: C