Say we have an urn filled with 28 white balls, 1 red and 1 black ball. The probability of getting at least one of each red and black ball (even if in this case the distinction of -at least- does not matter...), can be computed like so:
$$\sum_{r=1}^1 \left(\sum_{b=1}^1 \frac{{1 \choose r}{1 \choose b}{28 \choose 3-r-b}}{{30\choose 3}}\right)=0.00689655...$$
As I have learned by now, another (MUCH more efficient way, once more copies of colored balls are in the urn, ie. 4 red and 4 black...) of computing this, is by calculating the probability of NOT getting a red ball + NOT getting a black ball + getting only white balls and subtracting the result of this addition from 1, like so:
$$1-2(\frac{{29 \choose 3}}{{30 \choose 3}}) + \frac{{28 \choose 3}}{{30 \choose 3}}=0.00689655...$$
Now, here is where I am pulling my hair for hours, and I really hope someone can shed some light!
Let's simply replace one white ball with a yellow one in the urn (so instead of 28 we now have 27 white, 1 yellow, 1 red and 1 black ball).
$$\sum_{y=1}^1\left(\sum_{r=1}^1 \left(\sum_{b=1}^1 \frac{{1 \choose y}{1 \choose r}{1 \choose b}{27 \choose 3-y-r-b}}{{30 \choose 3}}\right)\right)=0.000246305...$$
Alright, so far, so good. Now let's try the more efficient computation method:
$$1-3(\frac{{29 \choose 3}}{{30 \choose 3}}) + \frac{{27 \choose 3}}{{30 \choose 3}}=-0.9795566$$
What? Why? Why are those results not matching? How can there be any difference between the two?
They're not just not matching; one of them is negative, so you already know which of them is wrong.
The problem may already start where you say "the probability of not getting a red ball + not getting a black ball + getting only white balls". This is not a description of the correct inclusion-exclusion calculation that follows. You need to add the probabilities of not getting a red ball and not getting a white ball and then subtract the probability of getting only white balls, i.e. getting neither a red, nor a black ball.
It seems that because you didn't understand that what you were doing in the first case was inclusion-exclusion, you failed to apply it properly in the second case. My advice would be to read up on inclusion-exclusion, understand why what you did in the first case was correct even though you described it incorrectly, then apply it to the second case (and then reconsider whether you think that this is a "MUCH more efficient way" to solve the present problem).