I am asked the following question:
A delegation of 5 students is to be selected for a MUN conference. There are 10 boys and 13 girls to choose from. If the team is to include at least one boy and one girl, in how many ways can the delegation be selected?
I had two approaches for this, but they unfortunately resulted in different answers. The second approach is wrong but I am not sure why.
Approach #1: Calculating all possibilities and subtracting the amount of teams that only contains girls or boys
$$\binom{23}{5} – \binom{10}{5} – \binom{13}{5} = 32,110$$
Approach #2: Establishing a team with one boy and one girl, and for the other 3 positions, selecting any student. After that, dividing by $3!$, which are the ways we can sort these three "groups of students" (the girl, the boy and the other three students), and then by $3!$ again, the possible ways to distribute the three students in their little group.
$$\frac{13 \cdot 10 \cdot 21 \cdot 20 \cdot 19}{3! \cdot 3!} = \text{(not an integer)}$$
Any guidance is highly appreciated.
Best Answer
The problem with the second approach is that it introduces duplicates and does not give you the right answer.
You are first choosing one boy and one girl in $~13 \cdot 10~$ ways. You are then selecting $3$ from the remaining $21$ as $\displaystyle {21 \choose 3}$. Say the first selection is
$\{G1 ~ B1\}, \{G2 ~G3~ B2\}$
Say the second selection is,
$\{G1 ~ B2 \}, \{G2~G3~B1 \}$
But both the selections comprise of the same girls and same boys!
Using complementary method (your first approach) is the best. The other approach will be to add four cases which is more tedious.