There are 5 positions. Given that $\mathrm{B}_1$ occupies $2^{\text {nd }}$ position
$\therefore \quad \mathrm{B}_1$ can be arranged in 1 way. As $G_1$ and $G_2$ are always together, none of them can take $1^{\text {st }}$ position.
$\therefore \quad \mathrm{G}_1, \mathrm{G}_2$ and one of the remaining students can be arranged on $3^{\text {rd }}, 4^{\text {th }}$ and $5^{\text {th }}$ position when $\mathrm{G}_1$ and $\mathrm{G}_2$ are always together in $2 ! \times 2$ ! Ways.
And remaining 2 students can be arranged in 2 ! Ways.
$\therefore \quad$ The required number of arrangements
$\begin{aligned}
& =2 ! \times 2 ! \times 2 ! \\
& =8
\end{aligned}$