The number of ways in which six boys and six girls can be seated at a round table so that no two girls sit together and two particular girls do not sit next to a particular boy is

The boys can be seated in  $\overline{)5}$ ways.
If the girls $g_1$  and $g_2$  not want to sit by the side of  $A_1$ (says). The two gaps.
$A_6-A_{1}and{A}_1-A_2$  must be filled by two of the remaining and ${ }^4{P}_2$  ways
The other four gaps can be filled in  $\overline{)4}$ ways 
$$\begin{gathered} R.N.W=5!\times {}^4{P}_2\times 4! \\ =5!\times \frac{4!}{2!}\times 4!=2.6!4! \end{gathered}$$