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.

For $10$ girls we have $10!$ permutations. We have $3!$ for boys. We just put the boys in the right positions. Thus, the result is $10! \times 3!$.
