How many ways can we draw three balls such that at least two are red? We have $7$ balls, three red, two white and two blue. How many ways can we select three of them such that at least two are red?
So, my answer was:
If the balls are identical, then there are $3$ ways: $RRR$, $RRB$ and $RRW$.
If the balls are distinguishable, then there are ${7}\choose{3} $=$35$ total possibilities, but there are ${4}\choose{3}$=$4$ ways that have no red balls, and $2 \cdot$${4}\choose{2}$=$12$ with only one ball so there are 19 with 2 red balls or more.
But the answer given is 13.
Could anyone explain this to me?

You are correct that there are $\binom{7}{3}$ ways to select three of the seven balls and that there are $\binom{4}{3}$ ways to select none of the red balls.  However, there are $\binom{3}{1}\binom{4}{2}$ ways to select exactly one of the three red balls and two of the remaining four balls, which yields
$$\binom{7}{3} - \binom{4}{3} - \binom{3}{1}\binom{4}{2} = 13$$
ways to select at least two red balls.
Alternatively, we can select exactly two red balls in $\binom{3}{2}\binom{4}{1}$ ways since we must select two of the three red balls and one of the other four balls, and we can select all three red balls in $\binom{3}{3}$ ways.  Since these cases are mutually exclusive, we obtain
$$\binom{3}{2}\binom{4}{1} + \binom{3}{3} = 13$$
selections with at least two red balls.
