Number of ways of arranging ten girls and three boys if the boys separate the girls into groups of sizes $3, 3, 2, 2$ Ten girls are to be divided into groups of sizes $3,3,2,2$. Also, there are $3$ boys. Number of ways of linear sitting arrangement such that between any two groups of girls, there is exactly one boy (no boy sits at either extreme end)?
MY SOLUTION:
$10$ girls can be divided into groups of sizes $3,3,2,2$ in 
$$\frac{10!}{3!3!2!2!2!2!}$$ 
ways which gives me unique combination of groups.
I can then arrange these groups in $4!$ ways, and people within them in $3!3!2!2!$ ways. 
Finally the $3$ boys in $3!$ ways. 
Seems correct? It gives $$\frac{10! 4! 3!}{2! 2!}$$ ways.

Your solution is correct.
Here is another way to think about the problem:
We will seat the girls in two blocks of three seats and two blocks of two seats.  There are $\binom{4}{2}$ ways to choose the positions of the blocks of three seats.  Once the blocks of seats have been arranged, there are $\binom{10}{3}$ ways to select the girls who will sit in the leftmost block of three seats and $3!$ ways to arrange them in those seats, $\binom{7}{3}$ ways to select which of the remaining seven girls who will sit in the other block of three seats and $3!$ ways to arrange them in those seats, $\binom{4}{2}$ to select which of the four remaining girls will sit in the leftmost block of two seats and $2!$ ways to arrange them in those seats, and $2!$ ways to arrange the girls who will sit in the other block of two seats.  The three boys can be arranged in the three seats that separate the blocks of girls in $3!$ ways.
$$\binom{4}{2}\binom{10}{3} \binom{7}{3}\binom{4}{2}\binom{2}{2}3!3!2!2!3! = \frac{4!}{2!2!} \cdot \frac{10!}{3!7!} \cdot \frac{7!}{3!4!} \cdot \frac{4!}{2!2!} \cdot \frac{2!}{2!0!} \cdot 3!3!2!2!3! = \frac{4!}{2!2!} \cdot 10! \cdot 3!$$
as you found.
