Choosing 2 color pairs from 6 balls I saw an example for choosing a full house from a deck of cards. The formula used was: $$\binom{13}{1}\binom{4}{2}\binom{12}{1}\binom{4}{1}\over\binom{52}{5}$$
In order to gain some intuition for why this works and understand how to apply this formular generally, I wanted to work through a smaller problem but I can't get a consistent answer to my smaller problem.
My smaller problem:
We have 6 balls, 2 Red 2 Yellow, 2 Green. We want to draw 4 balls without replacement. What are the odds of us getting 2 pairs of matched colors, disregarding order drawn?
Using this formula I'd expect this
$$\frac{\binom{3}{1}\binom{2}{2}\binom{2}{1}\binom{2}{2}}{
\binom{6}{4}} = \frac{6}{15}$$
But if I actually think about the options, there are only 3, {RRYY, RRGG, YYGG} so simply by counting I'd think it should be $\frac{3}{15}$. Can you help me understand which of these two answers is wrong and why?
 A: You are right, the probablity should be $\frac{3}{15} = \frac15$.
The problem with your formula is that you pick one color with $3 \choose 1$, then a second color with $2 \choose 1$. But the order of these two colors doesn't matter, so if you multiply $3 \cdot 2$ you have double counted. (There are only three ways to pick the two colors, not six.)
In the full house example, order does matter, because the first choice is for the pair and the second choice is for the three of a kind. AAKKK is a different hand from AAAKK.
A: +1 to the answer of 3rdMoment.  An alternative approach is
$$\frac{\binom{3}{2} \times \binom{2}{2} \times \binom{2}{2}}{\binom{6}{4}}. \tag1 $$
In (1) above, the denominator reflects that the order that  the balls are selected is to be regarded as irrelevant.  This must be consistently applied in the numerator.
In (1) above, in the numerator, the first factor represents the number of ways of selecting which two colors will be used.

Another alternative, which leads to the same answer is
$$\frac{\binom{3}{2} \times \binom{2}{2} \times \binom{2}{2} \times (4!)}{\binom{6}{4} \times (4!)}. \tag2 $$
In (2) above, the 2nd factor in the denominator reflects that the order that the balls are selected is to be regarded as relevant.  That is, the $4$ balls drawn can appear in any order.  Here, it is being assumed (for simplicity), that balls of the same color are distinguishable from each other.
Since this is a Probability problem, this artificial assumption is okay, only as long as the assumption is (also) strictly applied in the numerator.
In (2) above, in the numerator, the last factor represents the number of ways of permuting the order of selection of the $4$ selected balls.
A: Here’s another way to think about the question.  Ask yourself which balls are excluded.  Any ball can be picked first without affecting the probability.  The second ball to be excluded must have the same color as the first, which is true of exactly one of the remaining five balls.  So the chance that you excluded two balls of the same color, resulting in picking two pairs of matched colors, must be $\dfrac 15$.
