Find 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 2 particular boys are always together?Options276296306288

To solve this problem, we need to use the principles of permutations and combinations. Here are the steps:

1. Since it's a circular arrangement, we fix one position. Let's consider the two particular boys as a single entity since they always have to be together. So, we now have 4 boys (including the entity of 2 boys) and 4 girls to arrange.

2. The girls cannot sit together, so we place the boys first. In a circular arrangement, (n-1)! is used, so we arrange the 4 boys in (4-1)!= 3! = 6 ways.

3. Now, we have 5 places (between the boys) to arrange the 4 girls. This can be done in 5P4 = 5!/(5-4)! = 120 ways.

4. The two boys considered as a single entity can be arranged among themselves in 2! = 2 ways.

5. Therefore, the total number of ways is 6 (ways to arrange boys) * 120 (ways to arrange girls) * 2 (ways to arrange the two particular boys) = 1440 ways.

So, none of the options given (276, 296, 306, 288) is correct. The answer should be 1440.