Five persons A, B, C, D and E occupy seats in a row at random. The probability that A and B sit next to each other is:
Step 1 — Total arrangements: 5 persons in a row give \(5! = 120\) outcomes.
Step 2 — Favourable arrangements: Treat A and B as a single block. Arrange 4 units in \(4! = 24\) ways; A and B internally arrange in \(2! = 2\) ways.
\[\text{Favourable} = 4!\times 2! = 48\]
Step 3 — Probability:
\[P = \frac{48}{120} = \frac{2}{5}\]
Therefore the required probability is \(\dfrac{2}{5}\).
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