(This is copy and pasted from the math part of this website)

I have looked into trying to figure what are all the possible types of note set combinations there are and how I would go by listing them if possible. Turns out this is harder than I thought. Here are the 12 notes

1, b2, 2, b3, 3, 4, #4/b5, 5, b6, 6, b7, 7

A type of set would be something like this


The thing is that set has inversion possibilities such as 3,1,5,7 and I do not want to count any of them as part of the list. No repeating notes either such as 1,1,3,5 or 1,b3,5,b3.

I've thought about the inversion case of 3,1,5,7 and convert that same set into 1,b6,b3,5 which is actually the same notes (which I'm trying to avoid as well). But what might be useful gleaming from this is using '1' as a base reference and instead try using each interval in order (from lowest to highest) for every single set on the list for example.. (These are just random examples)

1,b2,2,3 / 1,b2,3,4 / 1,2,3,4 / 1,4,5,b6,b7

I could use 1,b3,5,b6 but then I would have to not list 1,3,5,7 in the list because they are the same 'set type' So in the end its kind of difficult to find which note sets are already there and which are not, which leaves me to believe that I am missing a certain condition. Or maybe rather theres actually an easier way to look at all this that I haven't thought of yet?

So the question is how do I list all the different set types of notes without unintentionally repeating one?

(Edit added the last question to clear things up hopefully)

closed as unclear what you're asking by David Bowling, MattPutnam, Shevliaskovic, Tim, Richard Nov 1 '18 at 13:28

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  • I can't tell what your question is, but how is it that { 1, b3, b6, 7 } and { 1, 3, 5, 7 } are the same 'set type'? You might be interested in the mathematical notion of equivalence classes. You seem also to be assuming 12TET, but there are other temperaments. – David Bowling Oct 24 '18 at 3:48
  • Ooops I think I made a huge mistake give me a second lol. – pizzaking Oct 24 '18 at 4:06
  • There we go! Its 1, b3, 5, b6. So basically I convert 3, 1, 5, 7 into 1,b3,5,b6 by changing the '3' from 3,1,5,7 into '1' and basing the rest of the intervals into that new reference point so now its 1,b3,5,b6. And yes I'm doing 12 equal temperant – pizzaking Oct 24 '18 at 4:12
  • The question is how do I go about listing all the different set types without repeating one unintentionally? – pizzaking Oct 24 '18 at 4:21
  • See my answer in the math stackexchange here – weee Oct 24 '18 at 12:29

From what I can tell, you don't care about order since inversions don't count as a different combinations so this is indeed a combinatoric question. Assuming you know how combinations work (if not, then check this):

12C1 + 12C2 + 12C3 + ... + 12C10 + 12C11 + 12C12 = 12 + 66 + 220 + 495 + 792 + 924 + 792 + 495 + 220 + 66 + 12 + 1 = 4095 different combinations.

What this does is find the number of combinations we can choose from a group of twelve using one to twelve notes to build a set.

As for your last thought, 1,b3,5,b6 and 1,3,5,7 being a same set, I am not sure what you mean by this so I am unable to proceed any further. Could you provide an example?

  • Sure. So here we have 1 3 5 7. Lets say its C E G B = C major 7. If I had arranged it to 3 1 5 7 it would be E C G B = still C Major 7. Now if I turned that '3' from 3 1 5 7 into '1'' and based the rest of the intervals from that it would be 1 b6 b3 5 or in other words 1, b3, 5, b6 which would be E G B C = C Major 7. All I'm doing is shifting the number reference around so I would always start on a '1'. Likewise if for example if you have a C major 6 chord and a Amin7 chord I consider both of them being the 'same set' since they both have the same notes – pizzaking Oct 24 '18 at 5:00
  • That combinatorics thing makes a lot of sense actually! For me now its actually just a matter of figuring how do I go by listing these different note sets without repeating one by accident – pizzaking Oct 24 '18 at 5:13

You may be interested in an article by Julian Hook titled "Why are There Twenty-Nine Tetrachords? A Tutorial on Combinatorics and Enumeration in Music Theory."

In general we don't like to give link-only answers, but I'm ending my answer here for two reasons:

  1. Hook gives the information much more succinctly than I ever could.
  2. And since this is housed at Music Theory Online, there's almost zero chance this link will ever die. If it does, you can certainly find it online elsewhere by title/author/journal.
  • That is a fantastic link; thanks for sharing. – David Bowling Oct 25 '18 at 3:33

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