I need help here plz
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.)
This is a difficult problem, but we can solve it using Burnside's Lemma. Burnside's Lemma is a counting theorem that allows us to count the number of distinct ways of arranging items in a group, while taking into account the symmetry of the group.
First, we need to find the number of fixed points of each possible rotation. A fixed point is a seat that remains unchanged after a rotation.
For a single adult, there are 2 fixed points: the seat directly opposite the adult, and the seat directly to the right of the adult.
For two adults, there are 4 fixed points: the two seats directly opposite the adults, and the two seats directly to the right of the adults.
For three adults, there are 6 fixed points: each of the seats.
So, the number of seatings with no fixed points is 0. The number of seatings with exactly one fixed point is 6 * 2, the number of seatings with exactly two fixed points is 4 * 3, and the number of seatings with exactly three fixed points is 2 * 4.
Finally, we divide the total number of seatings by the number of rotations, which is 6 in this case (since a circular table has 6 rotational symmetries), to get the number of distinct seatings:
(6 * 2 + 4 * 3 + 2 * 4) / 6 = 12 + 12 + 8 / 6 = 32 / 6 = 5.33...
So, there are approximately 5 distinct ways to seat the three adults and three children at the circular table such that each child is next to at least one adult.