In a group of 3 girls and 4 boys, there are two boys B_1 and B_2.

Question

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 ad...

Accepted Answer

Let's break the problem down step by step:

There are two blocks because all girls must stand together and all boys must stand together. The two blocks can be arranged in:

$2\,$

The girls can be arranged among themselves in:

$3!=6$

For the boys (4 in total), they must be arranged such that the specific boys ${B}_{1}$ and ${B}_{2}$ are not adjacent.

First, calculate the total number of arrangements of 4 boys:

$4!=24.$

Next, count the arrangements where ${B}_{1}$ and ${B}_{2}$ are adjacent. Think of ${B}_{1}$ and ${B}_{2}$ as a single unit. This unit can be arranged in:

$2!=2{B}_{1}{B}_{2}$

Now, with this new unit, we have 3 units in total (the ${B}_{1}{B}_{2}$ unit and the other 2 boys), which can be arranged in:

$3!=6$

So, the number of arrangements where ${B}_{1}$ and ${B}_{2}$ are adjacent is:

$2!	imes 3!=2	imes 6=12.$

Therefore, the number of valid arrangements for the boys where ${B}_{1}$ and ${B}_{2}$ are not adjacent is:

$24-12=12.$

Finally, multiply all the factors together:

$=2	imes 6	imes 12=144.$

Thus, the number of ways in which the girls and boys can stand in the queue under the given conditions is $144.$

This corresponds to Option D.

Solutions · 1