In how many ways can 2 adults, 2 girls, and 2 boys be seated around a circular table? In how many ways can 2 adults, 2 girls, and 2 boys be seated around a circular table if the adults are to sit together and the boys and girls are to alternate?
My answer was that there's 8 ways - the adults can swap, while the boys and girls each have 2! ways of sitting. Is this correct?
 A: The parents can swap, the boys can swap and the girls can swap. That's $8$. However, in addition, you can also have the boys swap with the girls, which makes the answer $16$.
A: Let the two adults be a mom and a dad. Let's view the problem from the perspective of the mom through four independent observations:


*

*Her husband can be either to her left or the right, giving a factor of $2$.

*Wherever her husband does not sit, either a daughter or a son can sit, giving a factor of $2$.

*The child next to her is either the first or the second child of that sex, giving a factor of $2$.

*The child of the opposite sex that sits one space away from her can be either the first or the second, giving a factor of $2$.
These four independent observations fully describe the state of the table for $2^4 = 16$ possibilities.
A: We can have 2 patterns: $AAbgbg$ or $AAgbgb$.
In each of these we can permutate inside groups of adults, boys, and girls, each in $2!$ ways. Therefore we have
$$2\cdot 2!\cdot 2! \cdot 2! = 16$$
ways of arranging the seating.
Anyway, if there are adults with different sexes and we don't want Males sit next to Females, we still have 2 patterns: $MFbgbg$ and $FMgbgb$, but the adults can't permutate, therefore in this situation we have
$$2\cdot 2!\cdot 2! = 8$$
ways.
A: My answer is 2! x 2 x 2 x 2 = 16.

Reasoning:

1. consider swapping parents

2. then choose boys or girls to group with parents

3. with that group, permutate with table (i.e (3-1)!)

4. then there will be 2 slots to fit between BGB (I assume that boy is group with parents in step 2 & 3)
A: In case of round table we have two tasks to counter excess to linear problem:
1) Fixing the end.
2)Countering the direction.
In this case we fix the  position  by two 2 parents who can permute in two ways. Then going in clockwise sense we can fix any one of the 4 kids in 4 ways then we shall have only one option for rest of positions keeping in mind the restriction.
We can repete same process in counterclockwise position.
Hence we shall have $2(1×2×4×1×1×1)$ arrangements
