Picking a Pair from 2 Colored Groups A box contains three pairs of blue gloves and two pairs of green gloves. Each pair consists of a left-hand glove and a right-hand glove. Each of the gloves is separate from its mate and thoroughly mixed together with the others in the box. If three gloves are randomly selected from the box, what is the probability that a matched set (i.e., a left- and right-hand glove of the same color) will be among the three gloves selected? 
Approach:
Since I have 5 pairs of gloves, I pick one ${5 \choose 1}$. Now I have to pick a color from 2 available, Blue or Green ${2 \choose 1}$. From the paired, colored gloves I just picked, I have to 3 on the Blue side and 2 on the Green side but since the 3 on the Blue side are identical, same for the 2 on the Green side, I pick one of the pairs and pick 2 of them ${2 \choose 2}$. For the last glove I can pick any glove. So I can pick out of the 4 remaining  ${4 \choose 1}$, and since it doesn't matter what color, I pick one from that pair ${2 \choose 1}$. Hence,
$${5 \choose 1}{2 \choose 1}{2 \choose 2}{4 \choose 1}{2 \choose 1}$$
divide this by ${10 \choose 3}$. This is however not the answer. What am I doing wrong?
 A: We probably need to divide into cases. The following "grinding things out" way is not elegant, but it works. There are $4$ cases, a lot, but they come in very similar pairs. 
We count the number of "good" choices of $3$ gloves.
(i) Three green. There are $\binom{4}{3}$ ways to do this, all good, since all such choices lead to a matching pair.
(ii) Three red. There are $\binom{6}{3}$ ways to choose $3$ red. Of them, all but $2$ (all left and all right) are good. So the number of good of this type is $\binom{6}{3}-2$.
(iii) Two green, one red. We must pick one left green and one right green, which can be done in $\binom{2}{1}\binom{2}{1}$ ways. For each such way, we can pick any red. That gives $\binom{2}{1}\binom{2}{1}\binom{6}{1}$ good.
(iv) Two red, one green. The same reasoning as the one above gives $\binom{3}{1}\binom{3}{1}\binom{4}{1}$. 
A: The problem is your counting procedure does not produce an outcome that could be produced by the original pick three procedure. So your counting scheme is to lay the five pairs out in front of you and pick one pair, but then you say to pick a color; however when you pick one of the five gloves, that fixes the color already, so from there you cannot recover.
It looks like your strategy is to number the 10 gloves and chose three of them (that is how i interpret your denominator) so when you want to count the ways to get a matching pair, you want the number of ways to choose 3 out of 10 numbers where the numbers correspond to a matching pair. 
If you number the left blue gloves 1 to 3, right blue gloves 4 to 6, and the left green cloves 7 to 8 and right green gloves 9 to 10, you want the number of ways to grab one from each of the first two groups  plus the ways to grab one from each of the last two groups 
These two events are disjoint since you are picking three gloves out (cannot satisfy both grabbing green and grabbing blue full set simultaneously, so that makes life easier.
So, what is the number of ways to have a full set of green gloves?
What is the number of ways to have a full set of blue gloves?
Count these separately first. But be careful, one procedure here is pick one from first group, one from second group and pick one from the rest, but you may run into a double count situation when the first and third glove come from the same group (cannot distinguish picking 1 then 3 vs picking 3 then 1)
So perhaps you can separate the count further to a case by case scenario where you organize by how many of each type you pick.
I.e. the number of ways to choose two left blues and a right blue, etc then one left blue and two right blues, etc, then one left blue one right blue and a green, etc.
 Seems a tad messy, but will get you what you want. Then maybe after you can come up with a more clever organization.

Comment elaborating: 
The main point is that you want to avoid over counting . If you lay out the five pairs in front of you, you could choose the first blue pair, or the second blue pair, but also the left of the first and right of the second... it quickly gets ugly. A double count would occur if you picked the second pair, and the third glove from the first pair,  vs if you first picked the left of the first , the right of the second, and for the third glove picked the other in the second pair.
The organization comes with experience and you need quick ways to identify when overcounts happen... but sometimes it is hidden too well...I like to number everything so I can explicitly find two situations that end up with a double count.
