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?

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 now we have 4 entities (3 boys and 1 entity of 2 boys) and 4 girls to arrange.

2. The boys can be arranged around the table in (4-1)!=3! ways (since in circular permutations, we subtract one from the total number of items).

3. The two boys considered as a single entity can arrange themselves in 2! ways.

4. Now, we have to arrange the girls. Since no two girls can sit together, we can think of arranging the girls in the spaces between the boys. There are 4 spaces between the boys (including the space before the first boy and after the last boy). The girls can be arranged in these spaces in 4P4 ways.

5. Therefore, the total number of arrangements is (3! * 2! * 4P4) = (6 * 2 * 24) = 288 ways.

Sure, here is the step by step solution:

Step 1: Since it is a circular arrangement, we fix one person first. Let's fix one of the boys. Now we have 4 boys and 4 girls left.

Step 2: Two particular boys are always together. We can consider these two boys as one entity. So, now we have 3 boys and 1 entity (2 boys) and 4 girls.

Step 3: As no two girls can sit together, we have to place the girls in between the boys. We have 4 places (between the boys) and 4 girls. The number of ways to arrange the 4 girls in 4 places is 4!.

Step 4: Now, we have 4 entities (3 boys and 1 entity of 2 boys) left. The number of ways to arrange these 4 entities in a circle is (4-1)! = 3!.

Step 5: But the 2 boys (who are considered as one entity) can be arranged among themselves in 2! ways.

Step 6: Therefore, the total number of ways is 4! * 3! * 2! = 576 ways.