The number of ways in which 5 boys and 4 girls can be arranged on a circular table such that no two girls sit together and two particular boys are always together is

Step-by-step solution

Let two particular boys as one boy, we have only four boys which can be seated at a round table in 3 ! ways. The two boys together can be arranged in 2 ways. So, boys can be seated in 2 × 3 ! ways. B 1 B 2 are together. Now 4 girls can be seated at four places (marked × ) in 4 ! ways. ⇒ Required number of ways = 3 ! × 2 × 4 ! = 6 × 2 × 24 = 288 ways