G2.2 Three boys $B_{1}, B_{2}, B_{3}$ and three girls $G_{1}, G_{2}, G_{3}$ are to be seated in a row according to the following rules:
1) A boy will not sit next to another boy and a girl will not sit next to another girl,
2) Boy $B_{1} `$ must sit next to girl $G_{1}$
If $s$ is the number of different such seating arrangements, find the value of $s$.
First, arrange the three boys in a line, there are 3 ! permutations. $\mathrm{B}_{1} \mathrm{~B}_{2} \mathrm{~B}_{3}, \mathrm{~B}_{1} \mathrm{~B}_{3} \mathrm{~B}_{2}, \mathrm{~B}_{2} \mathrm{~B}_{1} \mathrm{~B}_{3}, \mathrm{~B}_{2} \mathrm{~B}_{3} \mathrm{~B}_{1}, \mathrm{~B}_{3} \mathrm{~B}_{1} \mathrm{~B}_{2}, \mathrm{~B}_{3} \mathrm{~B}_{2} \mathrm{~B}_{1}$.
If $\mathrm{B}_{1}$ sits in the middle, there are two different cases. For instance, $\mathrm{B}_{2} \mathrm{~B}_{1} \mathrm{~B}_{3}$. Then the possible seating arrangements are:
$B_{2} G_{2} B_{1} G_{1} B_{3} G_{3}, B_{2} G_{3} B_{1} G_{1} B_{3} G_{2}, G_{2} B_{2} G_{1} B_{1} G_{3} B_{3}$ or $G_{3} B_{2} G_{1} B_{1} G_{2} B_{3}$,
$\mathrm{G}_{3} \mathrm{~B}_{2} \mathrm{G}_{2} \mathrm{~B}_{1} \mathrm{G}_{1} \mathrm{~B}_{3}, \mathrm{G}_{2} \mathrm{~B}_{2} \mathrm{G}_{3} \mathrm{~B}_{1} \mathrm{G}_{1} \mathrm{~B}_{3}, \mathrm{~B}_{2} \mathrm{G}_{1} \mathrm{~B}_{1} \mathrm{G}_{3} \mathrm{~B}_{3} \mathrm{G}_{2}$ or $\mathrm{B}_{2} \mathrm{G}_{1} \mathrm{~B}_{1} \mathrm{G}_{2} \mathrm{~B}_{3} \mathrm{G}_{3}$.
If $\mathrm{B}_{1}$ sits in the left end or right end, there are four different cases. For instance, $\mathrm{B}_{2} \mathrm{~B}_{3} \mathrm{~B}_{1}$, then the possible seating arrangements are:
$\mathrm{B}_{2} \mathrm{G}_{2} \mathrm{~B}_{3} \mathrm{G}_{3} \mathrm{~B}_{1} \mathrm{G}_{1}, \mathrm{~B}_{2} \mathrm{G}_{3} \mathrm{~B}_{3} \mathrm{G}_{2} \mathrm{~B}_{1} \mathrm{G}_{1}, \mathrm{G}_{2} \mathrm{~B}_{2} \mathrm{G}_{3} \mathrm{~B}_{3} \mathrm{G}_{1} \mathrm{~B}_{1}, \mathrm{G}_{3} \mathrm{~B}_{2} \mathrm{G}_{2} \mathrm{~B}_{3} \mathrm{G}_{1} \mathrm{~B}_{1}$,
$\mathrm{B}_{2} \mathrm{G}_{2} \mathrm{~B}_{3} \mathrm{G}_{1} \mathrm{~B}_{1} \mathrm{G}_{3}, \mathrm{~B}_{2} \mathrm{G}_{3} \mathrm{~B}_{3} \mathrm{G}_{1} \mathrm{~B}_{1} \mathrm{G}_{2}$.
$\therefore$ Total number of seating arrangements $=2 \times 8+4 \times 6=40$
Method 2 Label the 6 positions as
\begin{tabular}{|l|l|l|l|l|l|}
\hline 1 & 2 & 3 & 4 & 5 & 6 \\
\hline
\end{tabular}
$\mathrm{B}_{1}$ and $\mathrm{G}_{1}$ sit next to each other, their positions can be $12,23,34,45$ or 56 , altogether 5 ways.
$\mathrm{B}_{1}$ and $\mathrm{G}_{1}$ can interchange positions to $\mathrm{G}_{1} \mathrm{~B}_{1}, 2$ different ways.
For the other positions, the two other boys and the two other girls can sit in $2 \times 2=4$ ways.
For instance, if $\mathrm{B}_{1} \mathrm{G}_{1}$ sit in the $2-3$ positions,
1 1 then $\mathrm{B}_{2}, \mathrm{~B}_{3}, \mathrm{G}_{2}, \mathrm{G}_{3}$ can sit in the following 4 ways:
$\mathrm{G}_{2} \mathrm{~B}_{1} \mathrm{G}_{1} \mathrm{~B}_{2} \mathrm{G}_{3} \mathrm{~B}_{3}, \mathrm{G}_{3} \mathrm{~B}_{1} \mathrm{G}_{1} \mathrm{~B}_{2} \mathrm{G}_{1} \mathrm{~B}_{3}, \mathrm{G}_{2} \mathrm{~B}_{1} \mathrm{G}_{1} \mathrm{~B}_{3} \mathrm{G}_{3} \mathrm{~B}_{2}, \mathrm{G}_{3} \mathrm{~B}_{1} \mathrm{G}_{1} \mathrm{~B}_{3} \mathrm{G}_{2} \mathrm{~B}_{2}$.
The total number of sitting arrangements $=5 \times 2 \times 4=40$ ways