A delegation of 5 students is to be selected. There are 10 boys and 13 girls to choose from. The team is to include at least one boy and one girl. 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.
 A: 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.
A: I would do this: Since there are 10 boys there are 10 ways to choose the first boy.  There are then 9 boys left to choose from.  Since there are 13 girls there are 13 ways to choose the first girl.  There are then 12 girls left to choose from so there are 9+ 12= 21 students left to choose from.  There are 21 ways to choose the third student, 20 ways to choose the fourth student, and 19 ways to choose the fifth student.  There are 10(13)(21)(20)(19) ways to choose the five students in that order.
But order is not important so we have to divide by 5! to discount the 5! ways to order five students.
There are $\frac{10(13)(21)(20)(19)}{5(4)(3)(2)}= 5(13)(7)(19)= 8645$ different ways to choose the five students.
