There are 4 adults and 6 children sitting around a round table. If there must be at least one child between any two adults, then how many ways are there for them to sit around the table? Rotations are considered the same, while reflections are distinct.

There are 4 adults and 6 children. Adults can be arranged in (4-1)!=3!=6ways.

Out of the 6 children, at least 1 can sit between 2 adults. For this case, the maximum number of children who can sit between 2 adults is 2. Therefore, to determine the total number of ways for at least 1 child to sit between 2 adults we shall add as shown below,

Number of ways to arrange the children is $\binom{6}{1}+\binom{6}{2}=6+15=21ways$.

Now the total number of ways is 6*21=126ways

Therefore, there are 126 ways for them to sit around the table.