<p>The number of numbers formed by nine non-zero digits such that all the digits in the first four places are less than the digit in the middle and all the digits in the last four places are greater than that in the middle is</p>
Step-by-Step Solution
Key Concept: The middle (5th) digit acts as a pivot: choose it first, then select 4 digits less than it for positions 1-4 and 4 digits greater than it for positions 6-9, accounting for all arrangements.
<p><strong>Step 1:</strong> We have 9 non-zero digits {1, 2, 3, ..., 9}. Let the middle digit (5th position) be d.</p><p><strong>Step 2:</strong> We need exactly 4 digits less than d for positions 1-4, and exactly 4 digits greater than d for positions 6-9. This means d must have exactly 4 digits smaller than it and 4 digits larger than it.</p><p><strong>Step 3:</strong> Only d = 5 satisfies this condition (digits 1,2,3,4 are less than 5 and digits 6,7,8,9 are greater than 5).</p><p><strong>Step 4:</strong> For d = 5: We must use {1,2,3,4} in first four positions and {6,7,8,9} in last four positions.</p><p><strong>Step 5:</strong> Number of ways to arrange {1,2,3,4} in positions 1-4 = 4!</p><p><strong>Step 6:</strong> Number of ways to arrange {6,7,8,9} in positions 6-9 = 4!</p><p><strong>Step 7:</strong> Total numbers = 4! × 4! = 24 × 24 = <strong>576</strong></p><p>∴ Answer: C</p>
Correct Answer: C