What  is  the  highest  possible  probability  of  getting  $12$  of  these  $20$  multiple  choice  questions  correct, given  that  you  don't  know  how  to  work  any  of  them  and  are  forced  to  blindly  guess  on  each  one?

[b]Note: The first question has 1 choice, the second question has 2 choices, and so on.[/b]

$\text{(A) }\frac{1}{6!}\qquad
\text{(B) }\frac{1}{7!}\qquad
\text{(C) }\frac{1}{8!}\qquad
\text{(D) }\frac{1}{9!}\qquad
\text{(E) }\frac{1}{10!}\qquad
\text{(F) }\frac{1}{11!}\qquad\\ \\
\text{(G) }\frac{1}{12!}\qquad
\text{(H) }\frac{2}{8!}\qquad
\text{(I) }\frac{2}{10!}\qquad
\text{(J) }\frac{2}{12!}\qquad
\text{(K) }\frac{1}{20!}\qquad
\text{(L) }\text{none of the above}\qquad$
1. To determine the highest possible probability of getting exactly 12 out of 20 multiple choice questions correct by guessing, we need to consider the probability of guessing each question correctly. 

2. According to the problem, the first question has 1 choice, the second question has 2 choices, the third question has 3 choices, and so on up to the 20th question which has 20 choices.

3. The probability of guessing the first question correctly is \( \frac{1}{1} \), the probability of guessing the second question correctly is \( \frac{1}{2} \), the probability of guessing the third question correctly is \( \frac{1}{3} \), and so on.

4. To find the probability of getting exactly 12 questions correct, we need to consider the product of the probabilities of guessing each of these 12 questions correctly. 

5. The probability of guessing 12 specific questions correctly is:
   \[
   \frac{1}{1} \cdot \frac{1}{2} \cdot \frac{1}{3} \cdot \frac{1}{4} \cdot \frac{1}{5} \cdot \frac{1}{6} \cdot \frac{1}{7} \cdot \frac{1}{8} \cdot \frac{1}{9} \cdot \frac{1}{10} \cdot \frac{1}{11} \cdot \frac{1}{12}
   \]

6. This product simplifies to:
   \[
   \frac{1}{12!}
   \]

7. Therefore, the highest possible probability of getting exactly 12 out of 20 questions correct by guessing is \( \frac{1}{12!} \).

The final answer is \( \boxed{\frac{1}{12!}} \)