To solve the problem, we need to evaluate the expression \((A-1)(B-2)(C-3)\ldots(K-11)\), where \(A, B, C, \ldots, K\) are an arbitrary permutation of the integers \(1, 2, \ldots, 11\).
### Step-by-Step Solution:
1. **Understanding the Expression**:
The expression \((A-1)(B-2)(C-3)\ldots(K-11)\) consists of 11 terms. Each letter \(A, B, C, \ldots, K\) corresponds to a unique integer from \(1\) to \(11\).
2. **Substituting the Values**:
Since \(A, B, C, \ldots, K\) are a permutation of \(1, 2, \ldots, 11\), we can express each letter in terms of its position in the permutation. For example, if \(A = 2\), then \(A - 1 = 1\).
3. **Analyzing Each Term**:
Let's denote the permutation as \(A_1, A_2, A_3, \ldots, A_{11}\) where \(A_i\) is the \(i\)-th element of the permutation. The expression can be rewritten as:
\[
(A_1 - 1)(A_2 - 2)(A_3 - 3)\ldots(A_{11} - 11)
\]
4. **Finding the Value of Each Term**:
Each term \(A_i - i\) can take values from \(-10\) to \(10\) depending on the permutation. However, since \(A_i\) is always one of the integers from \(1\) to \(11\), the maximum value of \(A_i - i\) is \(10\) (when \(A_i = 11\) and \(i = 1\)) and the minimum is \(-10\) (when \(A_i = 1\) and \(i = 11\)).
5. **Calculating the Product**:
The product will include terms that can be both positive and negative. However, we can observe that for every \(i\), \(A_i - i\) will yield a value that is symmetric around zero. This means that for every positive term, there is a corresponding negative term.
6. **Conclusion**:
Since the product involves terms that can be both positive and negative, and given that there are 11 terms (an odd number), the overall product will always yield an even number. This is because the product of an odd number of negative terms results in a negative product, while the product of any positive terms will not change the parity.
### Final Answer:
The value of \((A-1)(B-2)(C-3)\ldots(K-11)\) will always be an even number.
