Three adults and three kids are to be seated at a circular table. In how many different ways can they be seated if each kid must be next to two adults? (Two seatings are considered the same if one can be rotated to form the other.)
P.S. This is slighty different from this question:
There are two cases to consider:
Case 1: Two children sit next to one adult, and one child sits between two adults.
To seat the three adults, there are 2! ways, since rotations of the same arrangement are considered the same.
To seat the three children, there are two possible arrangements:
The child sitting between the two adults can be chosen in 3 ways. The other two children can then be seated in 2! ways.
The two children sitting next to the same adult can be chosen in 3 ways. The other child can then be seated in 2 ways.
Therefore, the total number of ways to seat the adults and children in Case 1 is 2! * (3 * 2! + 3 * 2) = 24.
Case 2: One child sits next to each adult.
To seat the adults, there are 2! ways.
To seat the three children, there are 3! ways.
Therefore, the total number of ways to seat the adults and children in Case 2 is 2! * 3! = 12.
Therefore, the total number of ways to seat the three adults and three children at the circular table if each child must be next to two adults is 24 + 12 = 36.