Apparently, the answer to this is:
$^{8}C_2\cdot^{7}C_4$ + $^{8}C_3\cdot^{7}C_3$ + $^{8}C_4\cdot^{7}C_2$ = 4410
But why is $^{8}C_2\cdot^{7}C_2\cdot^{11}C_2$ incorrect? by this I mean choosing 2 boys from 8 boys, then 2 girls from 7 girls, and finally 2 random students from the remaining 11 students. This is somehow equal to 32340 which is absurdly large as $^{15}C_6$ is only 5005.
Let girls be $G_1, G_2,...,G_7$ and boys be $B_1, B_2,...,B_8$. When you do $\binom{7}{2}\binom{8}{2}\binom{11}{2}$, you are overcounting some cases.
For example, let $G_1$ and $G_2$ be the girls, $B_1$ and $B_2$ be the boys and $B_3$ and $B_4$ be the randomly chosen students. But, this is as same as choosing $B_3$ and $B_4$ as the boys and $B_1$ and $B_2$ chosen randomly. And there are too mant overcounted cases so we end up with much larger result (For example, when we choose them as in the example, we count this case $\binom{4}{2} = 6$ times).