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 adjacent to each other, is:

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:

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

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 $$\boxed{144}.$$

This corresponds to Option D.

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