For those who don't know, 12 Tone Equal Temperament (12TET) simply means that, for a given frequency $f$, each step between $f$ and $2f$ can be written as $2^{k/12}f$, where $k$ denotes the $k$th step from $f$. In music, these notes are usually denoted as $$\lbrace C,C\#,D,D\#,E,F,F\#,G,G\#,A,A\#,B\rbrace$$Let us call this set $S$. I will define a "scale" to be a nonempty subset of $S$.

My First Thought Process

Taking $S$ as the above set, how many scales can be formed?

My logic is given the set

$\lbrace 1,2,3,4 \rbrace$

Then possible subsets are

$\lbrace 1,2,3,4\rbrace$

$\lbrace 1,2,3\rbrace$

$\lbrace 1,2,4\rbrace$

$\lbrace 1,3,4\rbrace$

$\lbrace 2,3,4\rbrace$

$\lbrace 1,2\rbrace$

$\lbrace 1,3\rbrace$

$\lbrace 1,4\rbrace$

$\lbrace 2,3\rbrace$

$\lbrace 2,4\rbrace$

$\lbrace 3,4\rbrace$

$\lbrace 1\rbrace$

$\lbrace 2\rbrace$

$\lbrace 3\rbrace$

$\lbrace 4\rbrace$

$\lbrace \emptyset \rbrace$

$1 + 4 + 6 + 4 + 1$

So for 12 members, I would get $2^{12}=4096$.

My Question:

What other mathematical ways are there to come to this number $4096$ besides the binomial theorem? Specifically, why might your particular method be interesting from a musical standpoint?


The musical scale A minor (A, B, C, D, E, F, G, A) is generally considered different from the scale C major (C, D, E, F, G, A, B, C). In fact the same notes make seven different scales: major (Ionian), minor (Aeolian), Dorian, Phrygian, Lydian, Mixolydian, and Locrian, depending on which note is considered the first note in the scale.

Your method counts these as only one scale.

If you want to consider only scales that start at C, however, in order to avoid the problem of counting different scales on the same notes, then it does not seem right to count sets that do not contain C; for how can a scale start at a note that is not in the scale? In that case you have exactly $2^{11} = 2048$ possible scales, depending on which of the $11$ other notes between C and the next higher C are in the scale.

And that is not considering scales (such as the melodic minor) that are played differently when the frequencies are rising than when they are falling.

  • $\begingroup$ @David Good observation. The whole point of equal temperament was to enable transposition, so starting somewhere other than the first note but using the same pattern of intervals doesn’t really give you a different scale (claims about how different keys feel different notwithstanding). However, I disagree with you in one regard: the OQ’s method does consider A minor as different from C major because the pattern of intervals is different, i.e., they select different sets of 7 notes from the available 12. $\endgroup$ – amd Sep 27 '15 at 17:48
  • 1
    $\begingroup$ @amd One question is whether C (in OP's set notation) represents the tonic of the scale or a specific frequency. I interpreted it as frequency. Whether A minor is really different from C minor is then a matter of whether it's "the same scale, transposed" or whether you are practicing playing it on an instrument. (We agree that OP's method already counted C minor.) I mentioned A minor only to point out that this is an ambiguity to be resolved; I then resolved it by counting only scales that start at C. $\endgroup$ – David K Sep 27 '15 at 19:10

The 2048 can be reduced to 351 distinct scale shapes if you're prepared to accept David K's reduction to scales having at least one 'start' note (e.g. his C) which takes it to 2048, and then also consider the 'modes' of any given k-sized scale as basically the same shape but starting on a different one of its k notes.

You then have 1 set with one note (and 11 notes) because there's really only one way to have a scale with only one note (or 11 notes, 12 with 1 left out) in it. It doesn't matter which 'key' they're in - they're the same scale 'type' (=shape). Similarly there are only 6 2-note (and, by the symmetry of omitting two notes, 10-note) scales because they are just a pair of notes with - successively - 1, 2, 3, 4, 5 and 6 semitones between them. Once you get to 7 semitones between them you're effectively the same 'shape' as a scale with 5 semitones between them (and so forth). The remaining scales may be counted as 19 (3-set/9-set), 43 (4-set/8-set), 66 (5-set/7-set) and 80 6-set. Thus 351 = 80 + 2*66 + 2*43 + 2*19 + 2*6 + 2*1 + 1 (that last 1 is the full chromatic 12-note set).

Oh, and Miles Davis's 'space between the notes' isn't the silence but the interval (in semitones).


You don't really need the Binomial Theorem to count subsets. When building a subset, you either include or exclude each element. For $n$ elements, that yields $2^n$ possibilities, less one for the empty set.

As for musical validity, that's outside of my brief and a subject for a different forum, I suspect. I would, however disagree with you about silence. As Miles Davis said, "Music is the space between the notes."

  • $\begingroup$ Binomial approach had a distinct advantage because each coefficient corresponds to the number of scales possible for a given number. So for example, there is 1 12note scale and 12 11 note scales, etc. So I think that reveals useful musical information. $\endgroup$ – Stan Shunpike Sep 27 '15 at 1:34
  • $\begingroup$ Perhaps so, but you asked for other ways to count the number of scales. $\endgroup$ – amd Sep 27 '15 at 17:41

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