How many ways can 6 pairs of 2 pegs be arranged in 12 holes… discarding duplicates read in opposite direction (re: music theory) I hope it’s OK for a non-mathematician (a musician) to ask a question here. My question is actually about music, but I’ve dressed it in plain clothes.
Suppose you have a line of 12 equally-spaced holes, and 6 sets of 2 pegs. The pegs are indistinguishable from each other, but each pair is connected by a stretchy (virtual) rubber band, stretchy enough so that any pair of pegs can sit in any pair of holes. The line of holes runs left to right. My question is: how many different ways can all 6 pairs of pegs be arranged in the 12 holes such that every pattern/arrangement of pegs/rubber bands is unique, also taking into account that where a pattern read from left to right is the same as a pattern read from right to left, then the two patterns are considered equivalent, and count as one pattern only, not two?
Ideally, I’m looking for the general case: h holes, s sets of p pegs, where h = s x p
Leaving aside left/right equivalence—I guess the number of ways in which the 6 pairs of 2 pegs can be arranged in 12 holes = 11x9x7x5x3 = 10,395 —equivalent to: h! / (p^s x s!)   Is that correct? 
If so, then the answer I’m looking for is 10,395 minus the number of times there exist two arrangements such that in one arrangement the pattern of pairs of pegs read left to right is the same as in another arrangement when read from right to left. Is there a way of finding this number without comparing 10,395 arrangements? 
I see this could be answered by writing a program that compares left and right patterns — but is there a maths solution for this? One a lay-person could usefully employ?
Apologies if the question is long-winded - when I try to precis it gets less clear, I think.  
Thanks so much! 
Padraic
Drawing of counting method
 A: The pairs of pegs with rubber bands can be thought of as labeling each pair, so in your example we could say we have 12 pegs labelled A,A,B,B,C,C,D,D,E,E,F,F. We want to know how many distinct words can be written in these letters, where two words are equivalent if they are reflections of each other. First without this equivalence the answer would be $\frac{12!}{2^{6}}$. There are $12!$ ways to order $12$ distinct objects, and then for each pair we've over-counted by a factor of 2, since for each word swapping any pair of pegs gives an equivalent word.
Now we can think about the equivalence. Think about reflecting a word through its center: ABACBF|EDDCFE $\rightarrow$ EFCDDE|FBCABA. Let $W$ be the set of all words before equivalence. Let $X$ be the set of words up to equivalence (so a pair of reflected words is one element). Let $P$ be the set of palindromes (words that are the same read forwards and backwards). Palindromes are exactly the words that are unchanged by reflection through the center. Now using Burnside's Lemma we have:
$$|X| = \frac{|W| + |P|}{2}$$
where putting the set in bars means the size of the set. We know $|W| = \frac{12!}{2^{6}}$ and we want to know $|X|$, so we just need to count $|P|$. To count palindromes we need to place $6$ distinct letters down in any order, and then the last $6$ letters are forced. For example if we pick ACBFED we get the palindrome ACBFED|DEFBCA. There are $6!$ ways to do this, so $|P| = 6!$. We then have $$|X| = \frac{1}{2}\left(\frac{12!}{2^{6}} + 6!\right)$$. 
In general with $s$ sets of $p$ pegs we can still count $|W|$ as $|W| = \frac{(s\cdot p)!}{p!^s}$ since now we over count by a factor of $p!$ rearrangements for each set $s$ of pegs. To count |P| we can do something similar to before as well. If there is only 1 set of pegs then $|X| = 1$, and from now on we'll assume there's more than 1 set of pegs. If $p$ is odd and there is more than 1 set of pegs we can't have any palindromes, so $|X| = \frac{|W|}{2} =\frac{(s\cdot p)!}{2(p!^s)}$. If $p$ is even, then from each set of pegs half of the pegs will be in the left half of the word, and the other half reflected on the right half. We need to count distinct left halves, which are words of length $\frac{s\cdot p}{2}$ using $s$ sets of $\frac{p}{2}$ letters. There are $\frac{\left(\frac{s\cdot p}{2}\right)!}{\left(\frac{p}{2}\right)!^s}$ of these. We get: $$|X| =\frac{1}{2}\left(\frac{(s\cdot p)!}{p!^s} + \frac{\left(\frac{s\cdot p}{2}\right)!}{\left(\frac{p}{2}\right)!^s}\right)$$
