The number of ways of forming a queue of 4 boys and 3 girls such...

Total number of boys $= 4$

Total number of girls $= 3$

Total number of children $= 7$

Total number of ways to arrange $7$ children in a queue $= 7! = 5040$

To find the number of ways where all $3$ girls are together, consider the $3$ girls as a single unit.

Total number of units to arrange $= 4$ boys $+ 1$ unit of girls $= 5$ units.

Number of ways to arrange these $5$ units $= 5! = 120$

Number of ways to arrange the $3$ girls among themselves $= 3! = 6$

Total number of ways where all girls are together $= 5! \times 3! = 120 \times 6 = 720$

Number of ways such that all the girls are not together $= \text{Total ways} - \text{Ways where all girls are together}$

$= 5040 - 720 = 4320$

Answer: $4320$