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 :

<p>Let's break the problem down step by step:</p> <p><p>There are two blocks because all girls must stand together and all boys must stand together. The two blocks can be arranged in:</p> <p>$$2 \text{ ways} \quad \text{(i.e., girls first then boys, or boys first then girls).}$$</p></p> <p><p>The girls can be arranged among themselves in:</p> <p>$3! = 6 \text{ ways.}$</p></p> <p><p>For the boys (4 in total), they must be arranged such that the specific boys $B_1$ and $B_2$ are not adjacent.</p></p> <p><p>First, calculate the total number of arrangements of 4 boys:</p> <p>$4! = 24.$</p></p> <p><p>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:</p> <p>$2! = 2 \text{ ways (since }B_1\text{ and }B_2\text{ can swap positions).}$</p> <p>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:</p> <p>$3! = 6 \text{ ways.}$</p> <p>So, the number of arrangements where $B_1$ and $B_2$ are adjacent is:</p> <p>$2! \times 3! = 2 \times 6 = 12.$</p></p> <p><p>Therefore, the number of valid arrangements for the boys where $B_1$ and $B_2$ are not adjacent is:</p> <p>$24 - 12 = 12.$</p></p> <p><p>Finally, multiply all the factors together:</p> <p>$\text{Total ways} = 2 \times 6 \times 12 = 144.$</p></p> <p>Thus, the number of ways in which the girls and boys can stand in the queue under the given conditions is $\boxed{144}.$ </p> <p>This corresponds to Option D.</p>