The number of ways, of forming a queue of 4 boys and 3 girls such that all the girls are not together, is :

Total number of persons = 4 boys + 3 girls = 7.

Step 1 : Total possible queues without any restriction
Arranging 7 distinct persons in a line gives $$7! = 5040$$ ways.

Step 2 : Arrangements where all three girls stand together
Treat the 3 girls as one single block G.
Then we have: G (one block) + 4 individual boys ⇒ total units = 5.
These 5 units can be permuted in $$5!$$ ways.
Within the girls’ block, the 3 girls can be ordered among themselves in $$3!$$ ways.
Hence, arrangements with all girls together = $$5! \times 3! = 120 \times 6 = 720$$.

Step 3 : Required arrangements (girls not all together)
Required count = (all possible arrangements) − (arrangements with girls together)
$$\Rightarrow 7! - 5!\,3! = 5040 - 720 = 4320.$$

Hence, the number of ways to form the queue so that the three girls are not all together is $$4320$$.

Option D which is: $$4320$$