What is the probability that at least one pottery project and at least one mixed media project will be​ shown? 
A Grade 11 Art class is offering students two choices for a​ project: a pottery project and a mixed media project.  Of the $48$ students in the​ class, $30$ have selected to do the pottery project and $28$ have selected to do the mixed media project​ (notice some students have decided to do​ both). If two students are selected at random from the class to show their finished​ project(s), what is the probability that at least one pottery project and at least one mixed media project will be​ shown?

So we have $20$ choose to do the media project and $18$ choose to do the mixed media project and $10$ choose to do both. I also know that the denominator is $48C2$. I think I have $$1-\frac{20C2 +18C2}{48C2}.$$
Does this make sense? If so that is $\dfrac{785}{1128}$?
 A: Yes, you are correct. If one project of each type is not shown then this can only be because both chosen students are doing just one project and their projects are the same type.
The probability that the two chosen students are both doing only the pottery project is $\frac{20 \times 19}{48 \times 47}$.
The probability that both are doing only the mixed media project is $\frac{18 \times 17}{48 \times 47}$.
So the probability that at least one project of each type is shown is
$1 - \frac{20 \times 19}{48 \times 47} - \frac{18 \times 17}{48 \times 47} = \frac{1570}{2256} = \frac{785}{1128}$
A: I am getting the answer 840/1128 but I could be wrong. Please do tell where I made a mistake before down-voting.

(Please excuse my handwriting)

CORRECTION #1: The no. of ways of selecting two students from group 3 would be 10*9 = 90 (and not 10*10 = 100) ... so, my new answer would be 830/1128.
CORRECTION #2: As @gandalf61 pointed out, it^ should be 45 (and not 90), giving a correct total of 785. Brain's not working today.
