counting double 
lets say there are 42 balls in an urn, 3 green balls, 6 orange balls, 33 red ones. 
a) What is the probability of getting all 3 colours? given that you are taking 4 balls

my first intuition was:
$$
\frac{\binom{3}{1}\binom{6}{1}\binom{33}{1}\binom{39}{1}}{\binom{42}{4}}
$$
as in: pick one from the red ones, the green ones and the orange ones, and the last one can be whatever.
BUT the answer states that this way is double counting all ways of getting the balls.
Can anybody explain why this is the case?
 A: Suppose that the balls are labelled $g_1,g_2,g_3,o_1,\ldots,o_6,r_1,\ldots,r_{33}$. Consider the set $\{g_1,g_2,o_1,r_1\}$. Your calculation counts it once with $g_1$ as the chosen green ball, $o_1$ as the chosen orange ball, $r_1$ as the chosen red ball, and $g_2$ as the extra ball. Unfortunately, it counts it again with $g_2$ as the chosen green ball, $o_1$ as the chosen orange ball, $r_1$ as the chosen red ball, and $g_1$ as the extra ball, so it counts that set of four balls twice. A little thought shows that this happens with every set of $4$ balls that contains at least one ball of each color: exactly one color appears twice, so your numerator counts that set of four balls twice. Thus, your probability is exactly twice what it should be.
You can also argue that there are three cases: 


*

*$\dbinom32\cdot6\cdot33$ sets with two green balls and one of each of the other colors;  

*$3\cdot\dbinom62\cdot33$ sets with two orange balls and one of each of the other colors; and  

*$3\cdot6\cdot\dbinom{33}2$ sets with two red balls and one of each of the other colors,


for a probability of
$$\frac{3\cdot6\cdot33+3\cdot15\cdot33+3\cdot6\cdot528}{\binom{42}4}=\frac{11,583}{111,930}\;.$$
A: This is the way I would do it, but I'm sure there are simpler/easier ways. This is very intuitive too.
You have to account for the different patterns you can get all 3 colors at, which are:
GOR, GRO, ROG, RGO, ORG, OGR
Think of the situation as: (G && O && R) || (G && R && O) || ... For "&&" [AND's], you multiply. For "||" [OR's], you add.
You find the probability of each of these and add them up. Keep in mind that the denominator gets smaller with each one (since you are taking 1 ball out).
$GOR = \frac{3}{42}*\frac{6}{41}*\frac{33}{40}; GRO = \frac{3}{42}*\frac{33}{41}*\frac{6}{40}; ROG = \frac{33}{42}*\frac{6}{41}*\frac{3}{40} ... $
Note that the fraction will always be $\frac{3*6*33}{42*41*40}$, so you don't actually need to calculate each one over and over.
There are 6 different valid patterns. So all you have to do is multiply the fraction by 6. The answer you get is $\frac{297}{5740}$.
EDIT: Your denominator was wrong because you forgot that it's not always 42.
