Combinations of 6 students among 20, at least one male

I have the following problem: A class has 20 students, 16 females, 4 males. Find all possibilities of choosing 6 students so that at least one is male.

I did the following, pick one male from the 4, you have 4 possibilities, then you are left with 19 students, pick 5, so I get 4*${19}\choose{5}$ possibilities.

However, one can notice that if you want no one to be male, then the possibilities are ${16}\choose{6}$, so, one could argue that the solution is instead ${20}\choose{6}$ - ${16}\choose{6}$.

So, which is solution is correct, and why?

$$4* {{19}\choose{5}}$$ or $${{20}\choose{6}} - {{16}\choose{6}}$$

It's funny, I made the exact same mistake recently. The problem with the first approach is that it contains double counting.

Say you start out by picking male $A$ among the four males $A, B, C, D$, and then you choose $5$ people among the remaining $19$. Maybe one of these five people is male $B$, and the rest are $x, y, z, w$.

But you could also originally have picked male $B$, and then male $A$ could have been among the $5$ people that you then choose out of $19$, where the rest are also $x, y, z, w$. It's the same choice as before.

The second solution is correct.