How many teams of $5$ players out of $15$ girls and $10$ boys can be formed with at least $2$ boys and $2$ girls [with complement] 
How many teams of $5$ players out of 15 girls and 10 boys can be formed with at least 2 boys and 2 girls? The solution has to be with complement.

This is related to: How many ways to assemble a team of 5 out of 15 girls and 10 boys with limitations?
I think I'm doing something wrong with complement, is it treated like the negation of the given conditional? 
In this case it would be the general case without any limitations $\binom {25} 5$, minus a team with only boys: $\binom {10} 5$, or a team with only girls: $\binom {15} 5$ or a team with one boy and four girls: $\binom{15} 4 \binom{10}1$ or a team with with one girl and four boys: $\binom {15}1\binom {10} 4$.
The total is: $$\binom {25} 5-\left(\binom {10} 5+\binom {15} 5+\binom{15} 4 \binom{10}1+\binom {15}1\binom {10} 4\right) $$

That looks fine.  It's the exactly right way to count the complement.
By the Rule of Complements: the measure of an event ($A$) is the measure of the total space ($S$) minus the measure of the complement ($A^\complement$); and vice versa. $$\lvert A\rvert = \lvert S\rvert - \lvert A^\complement\rvert \\ \lvert A^\complement \rvert = \lvert S \rvert - \lvert A \rvert$$
The total space, or general case, is counted as you did: the ways to select any $5$ students from all $25$.
The complement is partitioned as you did: counting ways to select $5$ from $10$ boys, or $5$ from $15$ girls, or $4$ boys and $1$ girl, or $1$ boy and $4$ girls.
$$\dbinom{25}{5} -\left(\binom{10}{5}+\binom{15}{5} + \binom{10}{4}\binom{15}{1}+\binom{10}{1}\binom{15}{4} \right)$$
Which, if you wish to verify, will equal the ways to select $3$ boys and $2$ girls or $2$ boys and $3$ girls.
$$\binom{10}{3}\binom{15}{2}+\binom{10}{2}\binom{15}{3}$$
