The number of ways in which $5$ boys and $4$ girls can be arranged on a circular table such that no two girls sit together and two particular boys are always together is

Let two particular boys as one boy, we have only four boys which can be seated at a round table in $3!$ ways.
The two boys together can be arranged in $2$ ways.
So, boys can be seated in $2\times 3!$ ways.

$B_1{B}_2$ are together. Now $4$ girls can be seated at four places (marked $\times$) in $4!$ ways.
$\Rightarrow$ Required number of ways $=3!\times 2\times 4!$
$=6\times 2\times 24=288$ ways