**SOLUTION:**

Number of ways they can be seated

=9!=504(6!)

2 Americans, 2 Chinese, 2 British, 4 others

Let, A=>2 Americans together

B=>2 British together

C=>2 Chinese together

∴ Number of ways in which at least 2people of same nationality are together

**=A+B+C−[A∩B+B∩C+C∩A]−A∩B∩C**

**Number of ways in which 2 Americans sit together across a round table**

=8(2!) and the same is for 2 British and 2 Chinese

Thus total ways = 3(8!2!)

**Number of ways in which 2 Americans and 2 Britishers sit together across a round table**

= (7!2!2!) and the same is for 2 British and 2 Chinese and for 2 Americans and 2 Chinese and sitting together

Thus total double intersections = 3(7!2!2!)

**Number of ways when 2 Americans,2 British and 2 Chinese sit together (triple intersection) across a round table =**(6!2!2!2!)

Thus, A+B+C−[A∩B+B∩C+C∩A]−A∩B∩C

=3(8!2!)−3(7!2!2!)−(6!2!2!2!)

=6!(260)

Thus, Required number of ways = 6![504−260]=244(6!)

**OPTION (C)**