jmc

There are 30 ways to draw the first white ball, 29 ways to draw the second, and 28 ways to draw the third. However, since order doesn't matter, we must divide by $3!$ to get $\frac{30\times 29\times 28}{3!}=4060$ ways to draw three white balls. There are 20 ways to draw the first red ball and 19 ways to draw the second, however, since order doesn't matter, we must divide by $2!$ to get $\frac{20\times 19}{2!}=190$ ways to draw two red balls. So the total number of outcomes for both the red and the white balls is $4060\times 190=\boxed{771,400}$.