Five students are to be arranged on a platform such that the boy \(B_1\) occupies the second position and such that the girl \(\mathrm{G}_1\) is always adjacent to the girl \(G_2\). Then, the number of such possible arrangements is

There are 5 positions. Given that \(\mathrm{B}_1\) occupies \(2^{\text {nd }}\) position
\(\therefore \mathrm{B}_1\) can be arranged in 1 way. As \(\mathrm{G}_1\) and \(\mathrm{G}_2\) are always together, none of them can take \(1^{\text {st }}\) position.
\(\therefore \mathrm{G}_1, \mathrm{G}_2\) 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.
\(\therefore\) The required number of arrangements \(=2 ! \times 2 ! \times 2 !=8\)