Pub quizz: So How many chords are there?

[james0tucson]

Not including the various voicings inversions & notes.
Trying to collect all of them google is not helping.

Read "The Foundations of Atonal Music" by Allen Forte and learn how to express harmony in terms of "pitch class sets" instead of "chords".

[duncanparsons]

I'm with the math idea.. given that a chord has at least three notes (two notes is just an interval, not a chord), and there are 12 notes to use: its a mulitple combinations question..

NB - this is looking compositionally, rather than single instrument, so it can be that all notes for the chord are spread across multiple sources, so a piano may play a rootless chord, but that root can be supplied by another voice.

12 available, 3+ must be used, so work out combinations for 3note, 4note, 5note..12note

Code: Select all

nCr =   n!
     -------
     r!(n-r!)

For us, this translates to

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12Cr =   12!
      --------
      r!(12-r!)

where r is 3,4,5,6,7,8,9,10,11 and add them, and an additional 1 for when all 12 notes are used (which would fail in that formula)

So,

Code: Select all

12! = 12x11x10x9x8x7x6x5x4x3x2x1 = 479001600
11! =    11x10x9x8x7x6x5x4x3x2x1 =  39916800
10! =       10x9x8x7x6x5x4x3x2x1 =   3628800
 9! =          9x8x7x6x5x4x3x2x1 =    362880
 8! =            8x7x6x5x4x3x2x1 =     40320
 7! =              7x6x5x4x3x2x1 =      5040
 6! =                6x5x4x3x2x1 =       720
 5! =                  5x4x3x2x1 =       120
 4! =                    4x3x2x1 =        24
 3! =                      3x2x1 =         6
 2! =                        2x1 =         2
 1! =                          1 =         1

And with that you can do all the calculations..


12C3 + 12C4 + 12C5 + 12C6 + 12C7 + 12C8 + 12C9 + 12C10 + 12C11 + 12C12 =
12!/3!9! + 12!/4!8! + 12!/5!7! + 12!/6!6! + 12!/7!5! + 12!/8!4! + 12!/9!3! + 12!/10!2! + 12!/11!1! + 1 =
220 + 495 + 792 + 924 + 792 + 495 + 220 + 66 + 12 +1 = 4017

Now what you choose to call each of these is up to you, for instance is FACE FM7 or Am6 in 3rd inversion?

Does that help?

[duncanparsons]

NB, if you want to take into account the ordering of notes, it's actually quite a bit easier, just 12!+11!+10!+9!+8!+7!+6!+5!+4!+3!.. I'll let you do the math on that

[frederik D]

I do agree with the math-guys on this one. You can very easily calculate all the possibilities with quite common math formula's (as duncanparsons pointed out).

That's the only logical and coherent way to do this, because if you start with chord names and symbols you'll end up with quite a mess... (once you move to chords using more then three notes there are quite a bit of chords using the same notes, albeit in different inversions - but I presume we're not taking those in account as different chords)

Of course you can argue about the "musical value" of this method (are all those combinations of notes "musical chords"?), but as long as you define a chord as composite of more than two (different) notes this is the way to do it.

[jancivil]

just to be a noodge: F A C E isn't really an A minor anything in music.

[vurt]

if you need more than 3 then youre doin it wrong.

[jancivil]

three's two many.

cf. John Lee Hooker.