Five students are to be arranged on a platform such that the boy $B_1$ occupies the second position and such…
- $4$
- $7$
- $8$
- $6$
Solution
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 $\mathrm{G}_1$ and $\mathrm{G}_2$ 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 $=2 ! \times 2 ! \times 2 !=8$Asked in: MHT CET 2023 (11 May Shift 1)