Equivalent probability solutions We have 3 white, 3 red and 5 black balls and we want the probability of getting 3 black balls or 1 of each. 
The solution to this is 5C3/12C3 + (3C1)(4C1)(5C1)/12C3
however, I simply did (5/12)(5/11)(5/10) + (3/12)(4/11)(5/10).
now, i am trying to understand why it is wrong and i've been trying to wrap around the idea that those 2 aren't equivalent, but i really can't. Can someone explain to me how I should proceed when solving these types of problems? What's the best way? I always get confused in these type of questions.
 A: There is a small typo in the problem statement. You probably mean $4$ red.
Both your expressions are not right, but for different reasons. Doing it your way, the probability of $3$ black should have been
$$\frac{5}{12}\cdot \frac{4}{11}\cdot \frac{3}{10}.$$
This is because we are sampling without replacement. Once we have taken out a black, there are only $4$ left out of a total of $11$. So the probability the second is black, given that the first was, is $\frac{4}{11}$. And given the first two were black, the probability that the third is black is $\frac{3}{10}$. (Your $5$-$5$-$5$ may be essentially a typo.)
For the one of each colour part, you got $\frac{3}{12}\cdot \frac{4}{11}\cdot \frac{5}{10}$. That's the probability of a white, then a red, then a black. But we can get one of each in several other ways, like black, then white, then red.
It turns out there are $3!$ different orders in which one can get one of each colour. It also turns out that they all have the same probability: you can write one or two of them to check this does happen. 
So you would get the right answer for the all different part if you multiplied the number you got for that part by $6$.
Remark: The "choose" procedure of the official solution can be more versatile, so it is good to learn to use it.
