Given: 6 boys and 6 girls.

$\therefore\:$ There are 12 persons.

All of them can be seated in $(12)!$ ways.

$i.e.,n(S)=(12)!$ where $S$ is sample space.

$E=$ Event of the girls and boys to be seated alternatively.

6 girls are seated in $6!$ ways. $G_1, G_2,G_3,G_4,G_5,G_6$

There are $7$ alternate places and $6$ Boys.

$\therefore$ $6$ boys can be seated in $7$ places in $^7P_6$ ways.

$\therefore\: n(E)=6!.^7P_6=\large\frac{6!.7!}{(7-6)!}$$=6!.7!$

$\therefore$ The required probability=$\large\frac{n(E)}{n(S)}=\frac{6!.7!}{(12)!}$