There are 5 girls and 3 boys. In how many different ways you can
divide them into 2 teams (4 people each) if each team should have at
least 1 boy?
Here is my solution. If one boy should be in each team then just assign 2 boys to teams and then the question is: you have 5 girls and 1 boy. In how many different ways you can divide them into 2 teams (3 people each). If you select one team, another one is selected automatically. So in how many ways you can select 3 elements from 6. Which is $C_{6}^{3}=20$.
The only problem is that the solution in the book is 30. So where is an error in my solution and how can I achieve 30?
Best Answer
Number of pairs of teams where all boys are together: exactly $5$ (one for each girl).
Number of possible pairs of teams: $\frac{1}{2}\binom{8}{4}=35$.
The difference has to be the answer, i.e. $30$.