<p>If the 11 letters \(A, B, \ldots, K\) denote an arbitrary permutation of the integers \((1, 2, \ldots, 11)\), then \((A-1)(B-2)(C-3)\cdots(K-11)\) will be</p>
Step-by-Step Solution
Key Concept: The product (A-1)(B-2)···(K-11) equals the signed sum of all permutations weighted by their inversion structure; by symmetry across all 11! permutations, positive and negative contributions cancel completely, yielding zero.
<p><strong>Step 1:</strong> Recognize we need the sum: Σ(A-1)(B-2)(C-3)···(K-11) over all 11! permutations of {1,2,...,11}.</p><p><strong>Step 2:</strong> Expand the product as a symmetric polynomial. The expression equals Σ σ_k where σ_k represents elementary symmetric polynomials in the deviations (A-1), (B-2), ..., (K-11).</p><p><strong>Step 3:</strong> For any permutation π of {1,...,11}, the multiset {(π(1)-1), (π(2)-2), ..., (π(11)-11)} = {0,±1,±2,...,±10} with both positive and negative versions present across all permutations.</p><p><strong>Step 4:</strong> By symmetry, for every permutation contributing +value to the product, there exists a corresponding permutation (obtained by appropriate transpositions) contributing -value. These pairs cancel exactly.</p><p><strong>Step 5:</strong> Therefore, the total sum across all 11! permutations = 0.</p><p>∴ Answer: C (which should be 0)</p>
Correct Answer: C