A circular dartboard is labeled with each of the integers from 1 to 20

Karishma's comment aside (which is valid), here is how I approached this. I much prefer to use combinatorics on these kinds of questions; in fact, I'm not aware of any other method to solve for circular arrangements.

That said, recall that the probability here is always \(\frac{# of favorable outcomes}{# of total outcomes}\). Let's tackle the first—the total number of outcomes—to get it out of the way.

Total Number of Outcomes
With circular arrangements, the formula is very simple: the total number of arrangements is \((n - 1)!\), where n = the number of entities. With 20 distinct numbers, the total number of arrangements is \(19!\).

Number of Favorable Outcomes
Now, the slightly trickier bit. There are two things to keep in mind here: first, that we are completely agnostic as to whether 2 comes before 1, or vice-versa. Second, that this is still a circular arrangement.

Recall the formula for calculating the number of arrangements when two items must be together: \((n)!(y - n + 1)!\); where n is the number of elements we are keeping together and y is the total number of elements. In this case, it would be \(2!(20-2+1)!\), or \(2!19!\). However, this is a circular arrangement, meaning we must subtract 1 from the total number of arrangements, for 2!18!.

Combine into a Fraction
Glancing at the answer choices, we're clearly looking for a simplified fraction here, so the name of the game is expressing the unsimplified fraction and paring that down. Given that probability = \(\frac{# of favorable outcomes}{# of total outcomes}\), our fraction is \(\frac{2!18!}{19!}\). Our answer choices are denominated either in terms of a 19 or a 20, so let's try and get there (this means AC A is completely out since there's no way our fraction will ever get to be 1/19!).

If we rewrite our fraction as \(\frac{2!18!}{19 * 18!}\), something becomes clear immediately. The 18! will cancel out to become \(\frac{2!}{19}\), or simply \(\frac{2}{19}\), leaving AC E as the correct answer.­­­