Number of ways to appoint a class leaderships with 36 students A class has 16 girls and 20 boys. The class leadership should be made by 4 students that should has at least one girl. In how many ways the choices can be made.
I think this can be solve with combinations, and every group of class leadership is divided in four possible cases.

*

*GIRL GIRL GIRL GIRL

*BOY GIRL GIRL GIRL

*BOY BOY GIRL GIRL

*BOY BOY BOY GIRL

What to do next?
 A: Method N°1
As you said, there are 4 possible ways to have a class leadership

*

*4 Girls

*3 Girls 1 Boy

*2 Girls 2 Boys

*1 Girl 3 Boys

From there, we simply need to compute each of these cases and to sum these up:

*

*Case 1 gives $\binom{16}{4}$ because we have $\binom{16}{4}$ ways to pick 4 girls from a group of 16.


*Case 2 gives $\binom{16}{3} \times \binom{20}{1}$ because we have $\binom{16}{3}$ ways to pick 3 people in a group of 16 and $\binom{20}{1}$ ways to pick 1 people in a group of twenty.


*Case 3 gives $\binom{16}{2} \times \binom{20}{2}$ (We repeat the same process here, and for the 4th case)


*Case 4 gives $\binom{16}{1} \times \binom{20}{3}$
So the total number of possible class leaderships are
$$\binom{16}{4} + \binom{16}{3} \times \binom{20}{1} + \binom{16}{2} \times \binom{20}{2} + \binom{16}{1} \times \binom{20}{3} = 54 \ 060$$
Method N°2
We can also simply notice that the question asks for the number of groups of four people, except the groups that are only made of boys.
This lead to a simple computation: the number of ways to choose 4 people among 36, minus the number of ways to choose 4 boys among 20.
Thus, we have:
$$\binom{36}{4} - \binom{20}{4} = 54 \ 060$$
Which gives the same result as with the first method
