Help how are you supposed to do this
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.)
+195 First, we can seat the three adults in any order around the table. This can be done in (3-1)! = 2! = 2 ways, since rotations of the same arrangement are considered the same.
Next, consider the three children. There are two possible arrangements:
1. Two children sit next to one adult, and one child sits between two adults.
2. One child sits next to each adult.
Case 1: Two children sit next to one adult, and one child sits between two adults.
To count the number of ways to seat the children in this case, we can choose one of the adults to have two children next to them, and then choose which two children sit next to that adult. The remaining child must sit between the other two adults. There are three ways to choose which adult has two children next to them, and then 2! ways to arrange the two children next to that adult. Therefore, there are 3 x 2! = 6 ways to seat the children in this case.
Case 2: One child sits next to each adult.
To count the number of ways to seat the children in this case, we can choose which child sits next to which adult. There are 3! ways to do this.
Therefore, the total number of ways to seat the adults and children is:
2! (6 + 3!)
= 2 x (6 + 6)
= 24
Therefore, there are 24 different ways to seat the three adults and three children at the circular table if each child must be next to at least one adult.
+118680 Thanks for answering Justin. I just want to have a go at it too 
If all the children are seperated by adults
Place an adult down anywhere there are 2! possibilities for the other 2 adults and 3! for the children
2! * 3! = 2 * 6 = 12 ways to seat them
If 2 of the chilren sit together and 2 adults sit together.
3 ways to chose the two children then 2 ways to seat them. they start the circle so that is 6 ways for them
there are 2 places the 3rd child can go
6*2 so far
then there are 3! =6 ways to seat the adults.
6*2*6 = 72 ways
Total = 12+72 = 84 ways [I think]
+195 Your solution and answer are correct! There are 84 ways to seat 3 children and 3 adults in a circle so that no child is next to another child and two adults are sitting together. Good job!
