Let the eleven letters A, B, dots, K denote a | vedclass

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(C) The set of integers is $S = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11\}$.In this set, there are $6$ odd numbers $\{1, 3, 5, 7, 9, 11\}$ and $5$ even num

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Always even

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(C) The set of integers is $S = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11\}$.In this set, there are $6$ odd numbers $\{1, 3, 5, 7, 9, 11\}$ and $5$ even numbers $\{2, 4, 6, 8, 10\}$.Let the product be $P = \prod_{i=1}^{11} (X_i - i)$, where $X_i$ is a permutation of $S$.The sum of all terms $(X_i - i)$ is $\sum_{i=1}^{11} (X_i - i) = \sum X_i - \sum i = 0$.For the sum of $11$ integers to be $0$ (an even number), there must be an even number of odd terms in the sum.Since $11$ is odd, if all terms were odd, the sum would be odd. Thus, there must be at least one even term in the product.Alternatively, consider the parity: the number of odd integers in the set is $6$. In the product, we subtract $11$ integers from $11$ integers. By the Pigeonhole Principle, since there are $6$ odd numbers being subtracted, at least one odd number must be subtracted from another odd number, resulting in an even difference.Therefore, at least one factor $(X_i - i)$ must be even, making the entire product even.

Let the eleven letters $A, B, \dots, K$ denote an arbitrary permutation of the integers $(1, 2, \dots, 11)$. Then, the product $(A - 1)(B - 2)(C - 3) \dots (K - 11)$ is:

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