Let the eleven letters A, B, ....., K denote | vedclass

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(C) Let $P = (A - 1)(B - 2)(C - 3) ..... (K - 11)$.Consider the sum of the factors: $S = (A - 1) + (B - 2) + (C - 3) + ..... + (K - 11)$.$S = (A + B +

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Always even

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(C) Let $P = (A - 1)(B - 2)(C - 3) ..... (K - 11)$.Consider the sum of the factors: $S = (A - 1) + (B - 2) + (C - 3) + ..... + (K - 11)$.$S = (A + B + C + ..... + K) - (1 + 2 + 3 + ..... + 11)$.Since $(A, B, ....., K)$ is a permutation of $(1, 2, ....., 11)$,the sum $(A + B + C + ..... + K) = (1 + 2 + 3 + ..... + 11) = 55 + 1 = 66$.Thus,$S = 66 - 66 = 0$.In any set of integers,the sum of the differences $(A_i - i)$ is zero. 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 there are $11$ terms in total (an odd number),at least one term $(A_i - i)$ must be even.If at least one factor in a product of integers is even,the entire product must be even.Therefore,the product is always even.

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

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