The number of ways in which 5 boys and 3 girls can be seated on a round table, if a particular boy $B_1$ and…
The number of ways in which 5 boys and 3 girls can be seated on a round table, if a particular boy $B_1$ and a particular girl $G_1$ never sit adjacent to each other, is
7 !
$5 \times 6$ !
$6 \times 6$ !
$5 \times 7$ !
Solution
$\begin{aligned} & \text { First, we arrange } 4 \text { boys and } 2 \text { girls (excluding } \\ & \mathrm{B}_1 \text { and } \mathrm{G}_1 \text { ) around the table, which can be done } \\ & \text { in } 5!\text { ways. } \\ & \text { In any such arrangement, } \mathrm{B}_1 \text { and } \mathrm{G}_1 \text { can be } \\ & \text { arranged in } 6 \text { available gaps in }{ }^6 \mathrm{P}_2=6 \times 5 \text { ways. } \\ & \therefore \quad \text { Total number of arrangements }=5!\times 6 \times 5 \\ & =6!\times 5\end{aligned}$