+686 Three adults and three children are to be seated at a circular table. In how many different ways can they be seated if each child must be next to at least one adult? (Two seatings are considered the same if one can be rotated to form the other.)
0 Lets do complementary counting for this.
The total number of ways to arrange all 3 adults and 3 children is:
\(\frac{6!}{6}=120\)
However we overcounted the cases of aaaccc in circle.
Total number of aaaccc cases are:
Fix the children block on top of the circle and the adults block on the bottom so we dont have to divide for rotation.
There are \(3!\) ways to arragne the children and \(3!\) to arrange the adults so:
\(3!\times3!=36\)
So: \(120-36=\boxed{84}\)
