# How many chords exist in total? [closed]

Is there any count of chords that exist? I mean for instance we have only 12 major triad 1-3-5 chords and 12 minor triad 1-3-5 chords, we also have 12 sus2 and 12 sus4 chords, we also have augs and different +6,+7,+9,+11,+13 chords.

How many would there be in total for two hands (piano) if not take very complex large chords that would require more than 5 fingers on each hand? Okay, and probably not more than 10 keys.

Maybe there is a diagram of all those chords?

Edit: A chord is more than 1 note. So G5 counts. Inversions don't count.

## closed as too broad by Richard, Doktor Mayhem♦Feb 12 '18 at 9:52

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• Can't see this being very useful to future readers. There probably is a finite number, but what good would it be? – Tim Feb 9 '18 at 15:23
• Why the downvotes? – xerotolerant Feb 9 '18 at 16:56
• IMHO, this question is "not useful" (one of the three reasons seen when hovering over the downvote button). Basically, there are countless possible chords, unless you put arbitrary restrictions on how you count them, and then what's the point? Perhaps it's about creating some algorithm or catalog, but with so many chords, even with the restrictions, it's hard to see how such a thing would be useful. It's like asking how many words there are in the English language between three and eight letters long and then wanting to write a novel with only those words or something. – Todd Wilcox Feb 9 '18 at 21:17
• @xerotolerant The end of yo''s answer says it all: "this is meaningless combinatorics". – Todd Wilcox Feb 9 '18 at 22:11
• Looking at Ted Greene's Chord Chemistry, I count that he has catalogued ~2450 different voicings and inversions which are playable on the guitar. This may or may not be a more useful way to approach this, depending on your perspective. I don't know if there is a similar catalogue for piano, but I suspect someone has done this. – David Bowling Feb 10 '18 at 7:09

It's difficult to say. A particular set of notes can conceivably be called by many different names. By your rules, and following the generally accepted conventions of chord symbol rules, what would we call D♯ and F played together? If we respelled F as E♯, it could be called D♯(sus2omit5) or D♯(add9omit3,5). Do those count separately?

Then, are we talking about the note D♯ or the piano key sometimes called D♯? If it's the latter and we kept F as F, we could also include E♭(sus2omit5), E♭(add9omit3,5), and F7(omit3,5). And how about G♭♭7(omit3,5)?

I could go on, but what I can't do is think of a properly named chord that has just D♯ and F in it. Yet, how can anyone say those two notes can't be sounded together? You see the problem?

So if a chord is defined as 2 to 10 unique pitch classes and inversions don't count, there would still be additional things to take into consideration, and counting the combinations is way more than simple math. You need to consider the compatibility of each interval. You can't, for example, have a diminished 5th and an augmented fourth in the same chord, can you? That's the same note. But let's say we define a chord as a root plus one or more of the following intervals:

• A second, either minor or major.
• A third, either minor or major.
• A fourth, either perfect or augmented.
• A fifth, either diminished, perfect, or augmented.
• A sixth, either minor or major.
• A seventh, either diminished, minor, or major.
• A ninth, either minor, major, or augmented.
• An eleventh, either perfect or augmented.
• A thirteenth, either minor or major.

...you would come up with 46655 chords for each root note. But before you go multiplying that by every root, you'd have to eliminate chords which don't make sense. A minor third with an augmented fifth is really just a different major chord, so let's reject those. You wouldn't want something like a Csus2add9 chord, because that's duplicating a pitch class that's already in the chord. You wouldn't want something like Cadd♯9omit3 because that ♯9 is better identified as a minor third. So, if you take away everything that doesn't make sense, you're down to 7637 possibilities, by my count. Many of these are enharmonic, but should still count separately. For example, C+7add♯4 and C7♭5add♭6 are different by interval, but the same by piano keys depressed.

Whatever the count, you'd have to multiply that by each possible root, of which there are 12. If you count typical enharmonics, there are 17. If you count B♯, C♭, E♯, and F♭, then there are 21. If you count double sharps and flats, that's 35 different root notes. 35 times 7637 (if you trust me on that) is 267,295 different chords. Again, you'd have tons of enharmonic chords.

This is an interesting theoretical question. I wrote an algorithm to explore the possibilities and I asked a few questions here to help me do it. (See this, this, and this if you're interested.) You can check out the fruits of my labor here:

http://tomweissmusic.com/chords/

• I like your answer. I didn't take into account that D#(sus2omit5)=EB(sus2omit5). Thus makes it a lot harder since as a chord they are equal but as a name they are different and thus occurs a lot in music theory. – SovereignSun Feb 10 '18 at 9:21

Tuplets: 12 base notes and above each you have 11 choices: 12*11 = 132.

Triplets: 12 base notes, 11 choices for 2nd, 10 choices for third: 12*11*10 = 1320.

Etc.

If we stop at 5, we get 108372 chords.

If we stop at 10, then first, some of our chords will need keys too high to ever exist on a piano. Ignoring this, we get 344058132 chords.

The theoretical maximum is 12, because then you choose all notes and there's nothing to add. The number you get there is 1302061332 chords.

However, this is meaningless combinatorics, so to say.

• Are such chords like C,C#,E among these numbers? – SovereignSun Feb 9 '18 at 16:07
• @SovereignSun As I say, meaningless combinatorics. In practice, I would say it's no more that couple hundreds on each base note, meaning couple thousands in total. – yo' Feb 9 '18 at 16:21
• And then the BIG problem is being able to name a lot of them, meaningfully... – Tim Feb 9 '18 at 16:34
• @Tim Yep, I've see a chord denoted H- ☉ (H meaning B, then a dash and then a circled dot). Crazy enough. Anyway, I'm out. – yo' Feb 9 '18 at 16:38
• This is wrong. The OP says inversions don't count, so, e.g., the number of 12-note chords is 1 (ignoring voicing). There is only one such chord, not 1302061332. You're counting sets of notes as if the order were significant, but it's not. CEG is the same chord as GCE (ignoring voicing). See en.wikipedia.org/wiki/Combination – Ben Crowell Feb 10 '18 at 1:22

You say:

How many would there be in total for two hands (piano) if not take very complex large chords that would require more than 5 fingers on each hand? Okay, and probably not more than 10 keys.

But:

Inversions don't count.

It seems to me that these two statements contradict each other. If you ignore voicings, doublings, and inversions, then all we're doing is labeling any subset of the chromatic scale as a chord, and I doubt that there are any such subsets that can't be pretty easily fingered on the piano with two hands in a close voicing. For example, there is only one 12-note chord, which consists of all the notes of the chromatic scale. Hitting all those notes simultaneously on a piano keyboard is actually pretty easy.

So assuming we ignore the piano fingering issue, the answer is that there are 4083 chords. C is either in the chord or not, so that's two choices. C# is either in it or not, so two more choices. Continue on in this way, and you get 2x2x2...x2 with 12 factors of 2, or 2 to the twelfth power, which is 4096. However, of these one is an empty set (no notes at all), which doesn't count, and 12 are single notes, which you also don't want to count. So we have 4096-1-12=4083.

The number gets quite a bit smaller if you want to count only chord types. For example, you could say you don't want to distinguish a C triad from a C# triad. The count gets roughly 12 times smaller if you do this, but not exactly 12 times because some chords are symmetrical, e.g., Cdim7 is the same as F#dim7.

• +1 for a well-explained way to look at it. – trw Feb 10 '18 at 15:06

I actually answered a question similar to this awhile back here. The question was concerning scales, but since I calculated the amount of scales with all length of scales that includes any combination of notes that could also be considered a chord. Note that these were all calculated based on 12 tone, equal temperament and were only in a one octave range.

For instance, with Tritonic (3 note) scales there are 220 unique versions with 55 of those being unique modes. If you don't want to count inversions there are even less than that. Personally I think inversions sound distinct enough to be their own chords, but things like semitone clusters will definitely sound the same.

Here's a Google Spreadsheet I did on it. Any other style of chord you could possibly want is in there.