# Probability Puzzle: 9 People and 9 Seats (Seeking a Better Method)

Q: There are 3 universities- University A, B & C. Each university will send 3 students, where the 3 students from each university represent the faculty of Health, Engineering and Law to participate in a seminar. The seating plan is as follow.

If no two students from the same university nor two students of the same faculty will sit in the seat with the same letter or number, what is the probability that the Engineering student from University A will sit immediately to the right of the Law student from University B?

I've found out that the total possibility of arrangements fulfilling the constraints will be 72 (2 x 3P3 x 3P3, as there are 2 general patterns, 6 arrangement possibility for each pattern). And the possibility for the Engineering student from University A will sit immediately to the right of the Law student from University B is 12 (6 x 2). So the answer to the puzzle should be 1/6.

However, is there a better and less complicated method to solve this? As in to solve the entire puzzle just with one step or using a certain formula?

$$P($$Law student from B has at least one seat to her right$$) = {2 \over 3}$$
$$P($$Engineering student from A is in the seat immediately to the right$$) = {1 \over 4}$$
$$P($$Law student from B has exactly zero seats to her right$$) = {1 \over 3}$$
$$P($$Engineering student from A is in that non-existent seat$$) = 0$$ (just not possible!)
$$P($$Engineering student from A is immediately to the right of Law student from B$$) = {2 \over 3} \times {1 \over 4} +{ 1 \over 3} \times 0 = {2 \over 12} = {1 \over 6}$$