Combination Problem

If we forget that we need minimum one girl and one boy, then there are $\binom{7+8}{3}=455$ ways to choose three childs. We can choose three girls $\binom{8}{3}=56$ ways, and three boys $\binom{7}{3}=35$ ways. If we don't choose three boys, or three girls, then there is minimum one girl, and one boy. So the answer is: $455-56-35=\boxed{364}$ .

The number of ways to choose 1 boy and 1 girl is $8\times 7=56$ . Then, the third member of a group can be either boy or girl, so since $(8-1)+(7-1)=13$ boys and girls are left then, there are $56\times 13=728$ permutations. But, we have some overlaps here! Notice that it's the same if we chose first Alice and Bob and then Charlotte (I took random names to provide clearer explanation) and if we chose first Charlotte and Bob, and then Alice. It turns out that our 2 permutations are actually 1 combination. Hence, the number of combinations of selecting 3 students is $\frac{728}{2}=364$ .