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?

The number of arrangements in which exactly one boy sits between the girls is 
$$12 \cdot 2! \cdot 12!$$
since there are twelve ways to choose the boy who sits between the girls, two ways of choosing the girl who sits to his left, one way of choosing the girl who sits to his right, and $12!$ ways of arranging the block of three people and the other eleven boys.
The number of arrangements in which exactly two boys sit between the girls is 
$$12 \cdot 11 \cdot 2! \cdot 11!$$
since there are twelve ways to choose the boy who sits in the first seat between the two girls, eleven ways to choose the boy who sits in the second seat between the two girls, two ways to choose the girl who sits to their left, one way of choosing the girl who sits to their right, and $11!$ ways to arrange the block of four people and the other ten boys.  
Notice that 
$$14! - 2!13! - 12 \cdot 2!12! - 12 \cdot 11 \cdot 2!11! = \binom{11}{2}2!12!$$
in agreement with the answers provided by drhab and Henning Makholm.
