<p>Among the \(8!\) permutations of the digits 1, 2, 3, ..., 8, consider those arrangements which have the following property. If we take any five consecutive positions, the product of the digits in these positions is divisible by 5. The number of such arrangements is equal to</p>
Step-by-Step Solution
Key Concept: For the product of any 5 consecutive digits to be divisible by 5, the digit 5 must appear in every window of 5 consecutive positions. This forces 5 to be positioned such that no gap of 5 or more consecutive positions exists without it.
<p><strong>Step 1: Identify the constraint.</strong> For any 5 consecutive positions to have product divisible by 5, the digit 5 must appear in every such window.</p><p><strong>Step 2: Determine valid positions for digit 5.</strong> In an 8-digit arrangement, the 5 consecutive position windows are: (1,2,3,4,5), (2,3,4,5,6), (3,4,5,6,7), (4,5,6,7,8). For 5 to appear in ALL these windows, it must be in the intersection of all windows: positions {2,3,4,5} ∩ all windows = only position 4 works universally, but checking: if 5 is at position 4, windows (1,2,3,4,5) ✓, (2,3,4,5,6) ✓, (3,4,5,6,7) ✓, (4,5,6,7,8) ✓. Actually, 5 can be in positions 4 ONLY ensures all windows contain it.</p><p><strong>Step 3: Verify constraint more carefully.</strong> If 5 is at position k, then windows [i, i+4] for i=1 to 4 all contain position k iff k ∈ [i, i+4] for all such i. This means k must be in [4,4], so position 4 only. However, rechecking: position 4 is in windows (1-5), (2-6), (3-7), (4-8) ✓.</p><p><strong>Step 4: Count arrangements.</strong> With 5 fixed at position 4, we arrange the remaining 7 digits (1,2,3,4,6,7,8) in the remaining 7 positions: 7! = 5040.</p><p>∴ Answer: B</p>
Correct Answer: B