Three white, two black balls: what are the chances that no black one remains after three draws? This is an easy combinatorics problem but I can never understand when order does or doesn't matter. There are a million examples but none that seem to give me any clue to how to solve exactly this problem.

We have $5$ balls in total: $3$ white, $2$ black. We draw them blindly one after the other until $3$ have been drawn. What is the probability that no black ball remains?

Working on this problem, I restated it as "What is the chance that we draw two black balls in three attempts?" I found $C(5,3) = 20$ ways to draw $3$ balls altogether from a set of $5$, but I can't for the life of me figure out how to do the same for the black balls.
An example I read had $9$ red balls in a set of $20$ or so balls, and $6$ draws. He said that it was $C(9,6)$ ways to draw the reds. That's just false, because that would mean there would be $C(2,3)$ ways for these black balls and what the heck is that? A negative number?
What am I missing?
 A: Think of it as a single draw of $3$ balls, which has the same "effect" as drawing one ball at a time:


*

*The total number of ways for drawing $3$ out of $5$ balls is $\binom53=10$

*The number of ways which include the $2$ black ones is $\binom31\cdot\binom22=3$

*Hence the probability of drawing the $2$ black ones is $\frac{3}{10}$

A: There is a difference between the "guy's red ball problem" and yours. In the "red ball problem" he has $9$ balls and he draws $6$ balls. Here $9>6$, so, the answer to the question:


*

*In how many ways can I draw only red balls
is indeed $C(9,6)$. In your case, you have $2$ black balls, but you draw $3$ balls, so here $2<3$, and therefore $C(2,3)$ does not make sense as an answer to the question


*In how many ways can I draw all black balls
(of course note that these questions are different in the first place!). So, how to proceed? 
The answer is the following: You are going to pick $3$ balls and you want $2$ of them to be black and necessarily $1$ of them to be white. So


*

*$C(2,2)$ ways to pick the two black balls

*$C(3,1)$ ways to pick one white ball


By the rule of product (or multiplication principle) multiply these two to get the total number of ways to perform this task. Lastly, the total ways to pick $3$ balls out of $5$ is as you correctly have equal to $C(5,3)$ (which by the way is equal to $10$ and not $20$ as you write), hence giving you a probability of $$\frac{C(2,2)C(3,1)}{C(5,3)}=\frac{3}{10}$$
