<p>Find the total number of nine-digit numbers that can be formed using the digits 2, 2, 3, 3, 5, 5, 8, 8, 8 so that the odd digits occupy the even places.</p>
Step-by-Step Solution
Key Concept: Identify that odd digits (3, 3, 5, 5) must occupy even positions (2nd, 4th, 6th, 8th), and even digits (2, 2, 8, 8, 8) must occupy odd positions (1st, 3rd, 5th, 7th, 9th). Use permutations with repetition for each group separately.
<p><strong>Step 1:</strong> Identify positions. In a 9-digit number, even positions are: 2, 4, 6, 8 (total = 4 positions). Odd positions are: 1, 3, 5, 7, 9 (total = 5 positions).</p><p><strong>Step 2:</strong> Count available digits. Odd digits: 3, 3, 5, 5 (4 digits, 2 of one kind and 2 of another). Even digits: 2, 2, 8, 8, 8 (5 digits, 2 twos and 3 eights).</p><p><strong>Step 3:</strong> Arrange 4 odd digits (3, 3, 5, 5) in 4 even positions. Number of arrangements = 4!/(2!×2!) = 24/4 = 6.</p><p><strong>Step 4:</strong> Arrange 5 even digits (2, 2, 8, 8, 8) in 5 odd positions. Number of arrangements = 5!/(2!×3!) = 120/(2×6) = 120/12 = 10.</p><p><strong>Step 5:</strong> Apply multiplication principle. Total arrangements = 6 × 10 = 60.</p><p>∴ Answer: <strong>60</strong></p>
Correct Answer: 60