Answer is (C) 7/60
The probability of the event T1 ∩ T2 ∩ T3 ∩ T4 is
Now, the probability that at least one pair of students sit adjacent to each other is
The probability that at least two pairs of students sit adjacent to each other is
The probability that at least three pairs of students sit adjacent to each other is
The total arrangements is
5! = 5 × 4 × 3 × 2 × 1 = 120
Therefore, the required probability is
(120 - 192 + 108 - 24 + 2)/120 = 14/120 = 7/60