In a group of 3 girls and 4 boys, there are two boys $B_1$ and $B_2$. The number of ways, in which these girls and boys can stand in a queue such that all the girls stand together, all the boys stand together, but $B_1$ and $B_2$ are not adjacent to each other, is:
- (1) 144
- (2) 72
- (3) 96
- (4) 120
Solution:
Let the group of girls be $G$ and the group of boys be $B$.
Since all girls stand together and all boys stand together, we can arrange the two groups ($G$ and $B$) in $2!$ ways.
Number of ways to arrange 3 girls among themselves = $3!$.
Number of ways to arrange 4 boys such that $B_1$ and $B_2$ are NOT adjacent:
Total arrangements of 4 boys = $4! = 24$.
Arrangements where $B_1$ and $B_2$ are together: Consider $(B_1B_2)$ as one unit. We arrange 3 units in $3!$ ways, and $B_1, B_2$ can switch places in $2!$ ways.
$$ \text{Together} = 3! \times 2! = 6 \times 2 = 12 $$
$$ \text{Not Adjacent} = 24 – 12 = 12 $$
Total required ways:
$$ = (\text{Arrangement of Groups}) \times (\text{Arrangement of Girls}) \times (\text{Arrangement of Boys}) $$
$$ = 2! \times 3! \times 12 $$
$$ = 2 \times 6 \times 12 = 144 $$
Ans. (1)