How many positive integer tuples $ (x_1,x_2,\dots, x_{13})$ are there satisfying the inequality $x_1+x_2+\dots + x_{13}\leq 2006$?

$ 
\textbf{(A)}\ \frac{2006!}{13!1993!}
\qquad\textbf{(B)}\ \frac{2006!}{14!1992!}
\qquad\textbf{(C)}\ \frac{1993!}{12!1981!}
\qquad\textbf{(D)}\ \frac{1993!}{13!1980!}
\qquad\textbf{(E)}\ \text{None of above}
$
1. We start with the inequality \( x_1 + x_2 + \dots + x_{13} \leq 2006 \). To convert this into an equation, we introduce a new variable \( x_{14} \) such that:
   \[
   x_1 + x_2 + \dots + x_{13} + x_{14} = 2007
   \]
   where \( x_{14} \) is a positive integer.

2. Since \( x_1, x_2, \dots, x_{13} \) are positive integers, we need \( x_{14} \) to also be a positive integer. This transforms our problem into finding the number of positive integer solutions to the equation:
   \[
   x_1 + x_2 + \dots + x_{14} = 2007
   \]

3. To solve this, we use the stars and bars theorem. The stars and bars theorem states that the number of ways to distribute \( n \) indistinguishable objects (stars) into \( k \) distinguishable bins (variables) is given by:
   \[
   \binom{n-1}{k-1}
   \]
   Here, \( n = 2007 \) (the total sum) and \( k = 14 \) (the number of variables including \( x_{14} \)).

4. Applying the stars and bars theorem, we get:
   \[
   \binom{2007-1}{14-1} = \binom{2006}{13}
   \]

5. The binomial coefficient \( \binom{2006}{13} \) is calculated as:
   \[
   \binom{2006}{13} = \frac{2006!}{13!(2006-13)!} = \frac{2006!}{13!1993!}
   \]

Thus, the number of positive integer tuples \( (x_1, x_2, \dots, x_{13}) \) satisfying the inequality \( x_1 + x_2 + \dots + x_{13} \leq 2006 \) is \( \frac{2006!}{13!1993!} \).

The final answer is \( \boxed{\frac{2006!}{13!1993!}} \)