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:

Step 1: Arrange the groups of girls and boys

First, consider the group of 3 girls as a single unit (G) and the group of 4 boys as another single unit (B). These two units can be arranged in a queue in $2!$ ways. This means either the girls' block comes first, or the boys' block comes first.

Step 2: Arrange the girls within their group

Within the group of 3 girls, the girls can arrange themselves in $3!$ ways. This is because there are 3 distinct girls, and their internal arrangement matters.

Step 3: Calculate total arrangements for boys without restrictions

Within the group of 4 boys, the boys can arrange themselves in $4!$ ways if there were no restrictions. This is the total number of permutations for the 4 boys.

Step 4: Calculate arrangements where $B_1$ and $B_2$ are adjacent

To find the number of arrangements where $B_1$ and $B_2$ are adjacent, treat them as a single unit. Now we have 3 units to arrange (the $B_1B_2$ pair, $B_3$, and $B_4$), which can be done in $3!$ ways. Additionally, $B_1$ and $B_2$ can swap positions within their unit in $2!$ ways.

Step 5: Calculate arrangements where $B_1$ and $B_2$ are NOT adjacent

The number of ways the boys can stand together such that $B_1$ and $B_2$ are not adjacent is the total number of ways to arrange the 4 boys minus the number of ways where $B_1$ and $B_2$ are adjacent.

Step 6: Calculate the total number of ways

Multiply the number of ways to arrange the girl and boy blocks, the number of ways to arrange girls within their block, and the number of ways to arrange boys within their block such that $B_1$ and $B_2$ are not adjacent.

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