In how many ways can the 10 kids in my class be seated in a circle if John and Sam insist on being seated diametrically opposite each other and if Stacy and Roger also insist on being diametrically opposite each other?

40320

never mind, it's 5760 counting Stacy and Rogers

To find the number of ways the 10 kids can be seated in a circle with certain conditions, we can use the concept of permutations.

Step 1: Find the total number of ways to arrange the 10 kids in a circle without any restrictions.

The number of ways to arrange 'n' objects in a circle is (n-1)!, as the first object can be placed in any position, and the remaining (n-1) objects can be arranged in (n-1)! ways.

So, the total number of ways to arrange the 10 kids in a circle without any restrictions is (10-1)! = 9!.

Step 2: Consider John and Sam as a single unit and place them diametrically opposite each other.

Treating John and Sam as a single unit, we have 8 units to arrange in a circle. The number of ways to arrange these 8 units is (8-1)! = 7!.

Step 3: Consider Stacy and Roger as a single unit and place them diametrically opposite each other.

Treating Stacy and Roger as a single unit, we now have 7 units to arrange in a circle. The number of ways to arrange these 7 units is (7-1)! = 6!.

Step 4: Multiply the results from Step 2 and Step 3 to get the total number of ways to arrange the 10 kids with the specified conditions.

By the multiplication principle, the total number of ways to arrange the 10 kids with John and Sam diametrically opposite and Stacy and Roger diametrically opposite is 7! * 6! = 5040 * 720 = 362,880.

Therefore, there are 362,880 possible ways to seat the 10 kids in a circle with the given conditions.