Three boys and three girls are to be seated around a circular table. Among them the boy X does not want  any girl as a neighbour and girl $Y$ does not want any boy as a neighbour. The number of possible arrangement is:

Clearly X should sit between the two remaining boys and Y should sit between the remaining two girls We have to arrange two groups $B_{1}X{B}_2$ and $G_{1}Y{G}_2$ along a circle. This can be done is $(2-1)!$ ways. But group of boys (girls) can be arranged in  $2!$ ways. Thus, the required number of ways is (2) (2) (1) = 4 ways.