Suppose we uniformly and randomly select a permutation from the 20! permutations of 1, 2, 3 …., 20. What is the probability that 2 appears at an earlier position that any other even number in the selected permutation? A. $$\frac{1}{2}$$ B. $$\frac{1}{{10}}$$ C. $$\frac{{9!}}{{20!}}$$ D. None of these

The correct answer is $\boxed{\frac{1}{2}}$.

There are $20!$ possible permutations of the numbers 1, 2, 3, …, 20. In half of these permutations, 2 will appear before any other even number. This is because there are $10$ even numbers, and 2 can appear in any of the $10$ positions before any other even number. Therefore, the probability that 2 appears at an earlier position than any other even number is $\frac{10}{20} = \boxed{\frac{1}{2}}$.

Option A is incorrect because it is the probability that 2 appears at an earlier position than any number. Option B is incorrect because it is the probability that 2 appears at an earlier position than any odd number. Option C is incorrect because it is the probability that 2 appears in the first position.