X and Y stand in a line at random with 10 other people. What is the probability that there are3 people between X and Y?

To solve this problem, we need to consider two things: the total number of ways X, Y, and the other 10 people can stand in line, and the number of ways X and Y can stand with exactly 3 people between them.

Step 1: Total number of arrangements There are 12 people in total (X, Y, and 10 others). They can stand in line in 12! (12 factorial) ways, which is 12111098765432*1 = 479,001,600 ways.

Step 2: Favorable arrangements We can think of X and the 3 people between X and Y as a single entity. This entity and Y and the remaining 7 people can be arranged in 9! = 362,880 ways. But within the entity, X can be at either end, so we multiply by 2. Also, the 3 people between X and Y can arrange themselves in 3! = 6 ways. So the total number of favorable arrangements is 9!23! = 4,354,560.

Step 3: Probability The probability is the number of favorable arrangements divided by the total number of arrangements, which is 4,354,560 / 479,001,600 = 0.0091 or about 0.91%.