QJEEM25Mab2c1: JEE Main 2025 Mathematics - Class 11 JEEM25Mab2c1 Solution: In a group of 3 girls and 4 boys, there are two boys and . The…

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:

$$2\text{ ways}\text{(i.e., girls first then boys, or boys first then girls).}$$

The girls can be arranged among themselves in:

$$3!=6\text{ ways.}$$

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:

$$2!=2\text{ ways (since }{B}_1\text{ and }{B}_2\text{ can swap positions).}$$

Now, with this new unit, we have 3 units in total (the $B_1B_2$ unit and the other 2 boys), which can be arranged in:

$$3!=6\text{ ways.}$$

So, the number of arrangements where $B_1$ and $B_2$ are adjacent is:

$$2!\times 3!=2\times 6=12.$$

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:

$$\text{Total ways}=2\times 6\times 12=144.$$

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

This corresponds to Option D.