Number of ways to pick a team of 4 with at least 1 girl and 1 boy from 4 girls and 4 boys. To find the number of ways to pick a team of $4$ with at least $1$ girl and $1$ boy from $4$ girls and $4$ boys, I thought to manually assign the team with $1$ girl and $1$ boy to begin with.
$$G, \quad B, \quad G \textrm{ or } B, \quad G \textrm{ or } B$$ 
The number of ways to pick $1$ girl and $1$ boy from $4$ girls and $4$ boys is
$${4 \choose 1} \times {4 \choose 1}$$
The number of ways to pick the $2$ remaining team members from $6$ people is
$${6 \choose 2}$$
Using this logic the number of teams with at least $1$ girl and $1$ boy is $4 \times 4 \times 15 = 240$, but the total number of teams with no restrictions is ${8 \choose 4} = 70$, so there is definitely something wrong with my thinking.
How do I arrive to the correct answer with this line of reasoning. I already know you can use the fact that it is $70 - (\textrm{number of teams with all boys or all girls}) = 70 - 2 = 68$, but I would like to understand how to do it the way I thought of.
 A: A team of four with at least one boy and at least one girl will have either three boys and one girl, two boys and two girls, or one boy and three girls.  Hence, the number of admissible teams is
$$\binom{4}{3}\binom{4}{1} + \binom{4}{2}\binom{4}{2} + \binom{4}{1}\binom{4}{3} = 68$$
which agrees with the answer you obtained by subtracting those teams composed only of boys or only of girls from the total number of teams that could be formed.
Why was your attempt wrong?
In designating a particular boy and a particular girl as the boy and girl who must be on the team, you counted each team with more than one boy or more than one girl multiple times.  In particular, you counted each team with three boys and one girl three times, once for each way of designating one of the boys as the boy on the team.
\begin{array}{c c}
\text{designated boy} & \text{designated girl} & \text{additional boys}\\ \hline
\text{Alexander} & \text{Barbara} & \text{Clifford, David}\\
\text{Clifford} & \text{Barbara} & \text{Alexander, David}\\
\text{David} & \text{Barbara} & \text{Alexander, Clifford}
\end{array}
By symmetry, you also counted each team with one boy and three girls three times, once for each way you could have designated one of the girls as the girl on the team.
You counted each team with two boys and two girls four times, once for each of the two ways you could have designated one of the boys as the boy on the team and once for each of the two ways you could have designated one of the girls as the girl on the team.
\begin{array}{c c}
\text{designated boy} & \text{designated girl} & \text{additional children}\\ \hline
\text{Alexander} & \text{Barbara} & \text{Claire, David}\\
\text{David} & \text{Barbara} & \text{Alexander, Claire}\\
\text{Alexander} & \text{Claire} & \text{Barbara, David}\\
\text{David} & \text{Claire} & \text{Alexander, Barbara}
\end{array}
Notice that 
$$\color{red}{\binom{3}{1}}\binom{4}{3}\binom{4}{1} + \color{red}{\binom{2}{1}\binom{2}{1}}\binom{4}{2}\binom{4}{2} + \color{red}{\binom{3}{1}}{\binom{4}{1}\binom{4}{3}} = \color{red}{240}$$
A: The counting
$${4 \choose 1} \cdot {4 \choose 1}\cdot {6 \choose 2}=240$$
is the number of teams with 1 "labelled" girl and 1 "labelled" boy. In order to "unlabel" them split ${6 \choose 2}$ into ${3 \choose 2}{3 \choose 0}+{3 \choose 0}{3 \choose 2}+{3 \choose 1}{3 \choose 1}$ and divide each term by the total numbers of boys and girls:
$$\underbrace{{4 \choose 1}}_{\text{1 lab. girl}} \cdot \underbrace{{4 \choose 1}}_{\text{1 lab. boy}} \cdot\left( \frac{1}{3\cdot 1}\underbrace{{3 \choose 2}{3 \choose 0}}_{\text{2 girls and 0 boys}}+\frac{1}{1\cdot 3}\underbrace{{3 \choose 0}{3 \choose 2}}_{\text{0 girls and 2 boys}}+\frac{1}{2\cdot 2}\underbrace{{3 \choose 1}{3 \choose 1}}_{\text{1 girl and 1 boy}}\right)=68.$$
