\(10\) persons are seated at a round table. The probability that two particular person sit together is

\(\frac{2}{9}\)

Here, 10 people are seated at the round table.

Total number of ways in which \(10\) persons can sit at a round table is \((10-1)!=9!\)

Now, if we regard two particular persons as one person, then we will be left with only 9 persons. These 9 persons can be seated along with a round table in (9 - 1)! = 8!

Also, those two particular persons can be arranged among themselves in 2! ways.

∴ Number of favorable events = 8! × 2!

⇒ Required probability = (8! × 2!)/9! = 2/9