Pairs of Beads on a Ring I would like to count the number of inequivalent ways to arrange $2N$ colored beads on a ring if there are $N$ colors and $2$ beads of each color. 
By "inequivalent' I mean inequivalent under (1) rotating the ring (2) flipping the ring (3) exchanging any two beads of the same color.
 A: HINT Start by considering the ways to arrange your beads in a straight line.
Then it will become easier to know how to avoid counting repeated combinations.
A: First imagine the beads in a straight line. 
There are $(2N)!$ ways to arrange the beads if we don't care about duplicates, mirrors, etc.
But now we want to divide out the stuff we don't want. 
Every pair of beads that have the same color have been double-counted. In other words (Green1) (Green2) is the same as (Green2) (Green1). So for each pair of colors we divide by $2$. There are $N$ colors so we divide by $2^N$.
At this point we add $N!$ for a reason I am not sure about yet. Maybe someone else can offer an explanation in a separate answer. 
Now we divide by $2$ to account for the fact that any layout can be flipped over into a mirror layout.
Finally we divide by $N$ to account for the fact that we could start the chain in $N$ different spots, but we want to consider all of these as being the same ring.
$$\frac{\frac{(2N)!}{2^N} + N!}{2N}$$
Also see https://oeis.org/A137729
This is also a specific case ($n=2$) of "Number of distinct $k$-colored necklaces with $n$ beads per color" where
$$ A(n,k) = \sum_{d | n} \frac{\phi(n/d)(kd)!}{(d!)^kkn}$$
