If the 11 letters A-K denote the arbitrary permutations of the integers (1,2,3,.....11) Then $(A-1)(B-2)(C-3).........(K-11)$ is necessarily: 
a) even b) odd c) zero
I didn't get the question itself. If those integers could be arranged in $11!$ ways... then should $A+B+\dots+K$ be 11! ?
Or am I approaching it incorrectly?
 A: The correct answer is A) even.
To see why this is, first notice that for the product of the 11 terms to be odd, every term must be odd ($A-1, B-2, ..., K-11$ must all be odd) since having a single even term would make the whole product even.
Next, observe that there are 6 odd numbers and 5 even numbers in 1 to 11, and A through K must also contain 6 odd letters and 5 even letters. By the pigeonhole principle, there must be at least one letter that is odd and is paired with an odd number in any permutation. (Basically, since there are 5 even numbers and 6 odd letters, at least one odd letter is leftover and has to be matched with an odd number.)
Now since there is always a pair that has an odd number and an odd letter, at least one of the terms in the product must be even (the one with an odd number and odd letter of course). By the first point, that means the entire product ends up being even.
To show that it isn't option C) zero, just let $A=2, B=3, ..., J=11, K=1.$
A: Suppose it is odd. Then each factor should be odd. Since there is odd nuber of factors their sum should be odd. But it is 0! A contradiction. 
So the product is even.
