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?

First arrange nine Bs and two Gs in any order you want, which can be done in $\binom{11}2$ ways. Then add three more Bs between the two Gs.
Finally multiply by $2!\cdot 12!$ if you want to count lists of names rather than just sequences of genders.
