If 2 boys and 2 girls are to be arranged in a row so that the girls are not next to each other, how many possible arrangements are there?

The correct option is 12.

To find the total number of arrangements where the girls are not next to each other, we can use the gap method.

Step 1: Arrange the boys first.
There are 2 boys. The number of ways to arrange 2 boys in a row is given by 2! (2 factorial):
$2!=2\times 1=2$

Step 2: Identify the available gaps for the girls.
When 2 boys are seated, they create gaps around and between them where the girls can sit without being adjacent to each other.
Let the positions of the boys be denoted as B₁ and B₂:
_ B₁ _ B₂ _
As shown above, there are 3 possible positions (gaps) for the girls.

Step 3: Arrange the girls in the available gaps.
We have 2 girls and 3 gaps available. The number of ways to choose and arrange 2 girls in 3 available gaps is given by the permutation formula ³P₂:
${}_{3}P_2=\frac{3!}{(3-2)!}=\frac{6}{1}=6$

Step 4: Calculate total arrangements.
By using the fundamental counting principle, we multiply the number of ways to arrange the boys by the number of ways to place the girls in the gaps:
$\text{Total arrangements}=2\times 6=12$

Thus, there are 12 possible arrangements where the 2 girls are not next to each other.