Arrangement of 12 boys and 2 girls in a row. 12 boys and 2 girls in a row are to be seated in such a way that at least 3 boys are present between the 2 girls.
My result:
Total number of arrangements = 14!
P1 = number of ways girls can sit together = $2!×13!$

Now I want to find P2 the number of ways in one boy sits between the two girls and then P3 the number of ways in which two boys sit between the two girls. How to find these two?

With stars and bars we find that there are $\binom{9+2}{2}=\binom{11}{2}=55$
different summations $a+b+c=9$ where $a,b,c$ are nonnegative integers.
So there are also $55$ different summations $a+b+c=12$ under the extra
condition that $b\geq3$.
Here $b$ corresponds with the number of boys sitting in between the 
girls.
There are $12!$ arrangements for the boys and $2!$ for the girls,
so we end up with:
$$55\times12!\times2!$$ posssibilities.
