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.
Best Answer
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: $$\underbrace{{4 \choose 1}}_{\text{1 lab. girl}} \cdot \underbrace{{4 \choose 1}}_{\text{1 lab. boy}} \cdot\left( \frac{1}{3\cdot 1}\underbrace{{3 \choose 2}{3 \choose 0}}_{\text{2 girls and 0 boys}}+\frac{1}{1\cdot 3}\underbrace{{3 \choose 0}{3 \choose 2}}_{\text{0 girls and 2 boys}}+\frac{1}{2\cdot 2}\underbrace{{3 \choose 1}{3 \choose 1}}_{\text{1 girl and 1 boy}}\right)=68.$$