Thread: Number of possible musical chords and scales

I hope I put this thread in the correct spot!

This is my first post and let me start by saying I am definitely not a mathematician. I am in fact a musician and I started thinking last night about just how many possible combinations of 3, 4, 5, 6, 7, 8, 9, 10, 11, and 12 notes there are above a given note and within one octave. I'm not sure if not knowing about music will affect anyone's ability to understand what I'm talking about so I'll try to fill in the gaps now.

There are twelve notes in the octave in western (European) music - that is, if you start on C and play every note up to the next C you will have played 12 notes (the next C being the 13th note). Up until the 20th century most chords consisted of 3-5 notes, and scales 7 notes (the notes of the scales being what chords are built up from). In a chord or a scale there is never two of the same note, that is C+E+G is not considered any different to C+E+G+C. The same goes for scales, you can't have two D flats in one scale. At the turn of the 20th century, composers began experimenting with bigger chords (chords with more notes) and new scales. So we now have music using scales and chords that have anywhere from 1 to 12 notes. Essentially, anything is possible, but what I want to know is exactly what is possible.

(If you do know a lot about music please ignore the fact that in tonal music chords are constructed directly in relation the the scales upon which the music is based and the effect a chord has in music is entirely dependent on context- I know that already. For this exercise I want to consider chords a scale as independent sonorities.)

I figured out (I think) how many combinations there where of 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, and 12 notes within an octave above a given note. Here is what I got. There is a definite pattern here and I would love to know if there was a simpler way of expressing it;
For a single note there is possibility - the root note,
2 note combinations there are 11 - the root note plus each of the other 11 notes,
3 notes = 10+9+8+7+6+5+4+3+2+1 = 55 - the root note plus all combinations of 2 notes within the remaining 11 notes,
4 notes = 9+(8x2)+(7x3)+(6x4)+(5x5)+(4x6)+(3x7)+(2x8)+(1x9) = 165 - the root note plus all combinations of 3 notes within the remaining 11 notes,
5 notes = 8+(7x(1+2))+(6x(1+2+3))+(5x(1+2+3+4))+(4x(1+2+3+4+ 5))+(3x(1+2+3+4+5+6))+(2x(1+2+3+4+5+6+7))+(1x(1+2+ 3+4+5+6+7+8)) = 330
6 notes = 7+(6x(1+(1+2)))+(5x(1+(1+2)+(1+2+3)))+(4x(1+(1+2)+ (1+2+3)+(1+2+3+4)))+(3x(....))+(2x(....))+(1x(.... )) = 462
7 notes = 6+(5x(....))+(4x(....)).... = 462
8 notes = 5+(4x(....))+(3x(....)).... = 330
9 notes = 4+(3x(....))+(2x(....)).... = 165
10 notes = 3+(2x(....))+(1x(....)) = 55
11 notes = 2+(1x9) = 11
12 notes = 1

Grand total = 2048

I really hope I have written this in a way that makes sense. Like I said, I am no mathematician.

I would like to know two things. First; Is there a more elegant way of expressing this? Second. If I wanted the extend it into two octaves, keeping in mind that if the notes in the second octave are identical the those in the first it wouldn't count as different (as long as one note is different then it is okay), how would I do that?

Thanks.

I'm interested in both mathematics and music (I used to teach music to children as well) so I find your post really interesting.

Originally Posted by Shunarjuna

There are twelve notes in the octave in western (European) music - that is, if you start on C and play every note up to the next C you will have played 12 notes (the next C being the 13th note). Up until the 20th century most chords consisted of 3-5 notes, and scales 7 notes (the notes of the scales being what chords are built up from). In a chord or a scale there is never two of the same note, that is C+E+G is not considered any different to C+E+G+C. The same goes for scales, you can't have two D flats in one scale. At the turn of the 20th century, composers began experimenting with bigger chords (chords with more notes) and new scales. So we now have music using scales and chords that have anywhere from 1 to 12 notes. Essentially, anything is possible, but what I want to know is exactly what is possible.

You might like to use group theory to help you out. In traditional 12-tone Western music, chords are based on two-note intervals. An interval can be "added" to another interval to form another interval. For example, the interval from C to E♭ is a minor third, and the interval from E♭ to G is a major third; when you add these two intervals, you get the interval from C to G, which is a perfect fifth. Hence, we can say that "minor-third plus major-second equals perfect-fifth". (I know musicians don't really say that but just bear with me; I'm illustrating what I mean by "adding" musical intervals.) If you don't count notes more than an octave apart as different from those within the same octave, then such additions will not produce intervals larger than the major seventh (C–B); for example, major-sixth (C–A) plus perfect-fourth (A–D) just yields a whole-tone interval (C–D). The octave interval, in particular, is identified with the unison interval (C–C).

The beauty of all this is that set of the unison interval and all the eleven two-note intervals forms a cyclic group of order 12 under addition of intervals!

Group theory is a fascinating branch of mathematics with applications in a wide variety of non-mathematical areas – including, as you can see, music. Try learning it.

Originally Posted by Shunarjuna

I figured out (I think) how many combinations there where of 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, and 12 notes within an octave above a given note. Here is what I got. There is a definite pattern here and I would love to know if there was a simpler way of expressing it;
For a single note there is possibility - the root note,
2 note combinations there are 11 - the root note plus each of the other 11 notes,
3 notes = 10+9+8+7+6+5+4+3+2+1 = 55 - the root note plus all combinations of 2 notes within the remaining 11 notes,
4 notes = 9+(8x2)+(7x3)+(6x4)+(5x5)+(4x6)+(3x7)+(2x8)+(1x9) = 165 - the root note plus all combinations of 3 notes within the remaining 11 notes,
5 notes = 8+(7x(1+2))+(6x(1+2+3))+(5x(1+2+3+4))+(4x(1+2+3+4+ 5))+(3x(1+2+3+4+5+6))+(2x(1+2+3+4+5+6+7))+(1x(1+2+ 3+4+5+6+7+8)) = 330
6 notes = 7+(6x(1+(1+2)))+(5x(1+(1+2)+(1+2+3)))+(4x(1+(1+2)+ (1+2+3)+(1+2+3+4)))+(3x(....))+(2x(....))+(1x(.... )) = 462
7 notes = 6+(5x(....))+(4x(....)).... = 462
8 notes = 5+(4x(....))+(3x(....)).... = 330
9 notes = 4+(3x(....))+(2x(....)).... = 165
10 notes = 3+(2x(....))+(1x(....)) = 55
11 notes = 2+(1x9) = 11
12 notes = 1

Grand total = 2048

I really hope I have written this in a way that makes sense. Like I said, I am no mathematician.

I would like to know two things. First; Is there a more elegant way of expressing this? Second. If I wanted the extend it into two octaves, keeping in mind that if the notes in the second octave are identical the those in the first it wouldn't count as different (as long as one note is different then it is okay), how would I do that?

Thanks.

Yes, what you've done is computing for to Did you notice that That is not a coincidence.

For your second question, try exploring the properties of the cyclic group of order 24. Good luck!

Wow!! Thanks for that reply. It was very helpful. I'll look into group theory. I did figure out, however, how to extend the possibilities into two octaves. It is actually very simple. Assuming that only one note needs to be different between the two octaves, you can just multiply the 2048 possibilities by 2048 and then minus 2048 so that you don't double up any chord or scale. 2048 x 2048 - 2048 = 4,192,256. Of course that is assuming each group of notes begins on the same note, which a composer may not want. So the real number of possibilities is...? I have not idea, but I'll try and figure it out once I've looked into group theory.
Thanks again,
Shunarjuna.

You're welcome.

I'd like to comment on another fascinating thing about the mathematical aspect of musical intervals. If you base your music on the 12-note scale of the octave, then the intervals form a cyclic group of order 12. Now, if you start with any note (say C), go up by a certain interval, then up again by the same interval, and again, and again, you will eventually end up with the note you started with. But do you pass through all 12 notes in the process? Well, you do with the interval of the perfect fifth (C–G). So do you with the interval of the perfect fourth (C–F). In fact, going up in fourths is the same as going down in fifths, so that is no surprise. This is called the circle of fifths (or fourths). Any more such circles, those that take you through all 12 notes? The circle of major thirds (C–E) does not because going up by this interval only takes you through three notes (C, E, G♯). Well, there are just two such 12-notes circles besides the circles of fourths and fifths, namely that of the semitone (C–C♯) and of the major seventh (C–G) (noting that going up in sevenths is the same as going down in semitones).

In mathematical parlance, the four intervals semitone, perfect fourth, perfect fifth, and major seventh are said to be the generators of the cyclic group of order 12.

Now if you try the same idea with the 24-note scale of two octaves, based on the cyclic group of order 24, you will find that there are eight intervals whose circles go through all 24 notes of the scale (namely, the four intervals referred to above and the four corresponding ones an octave higher). In other words, there are eight generators for the cyclic group of order 24.

I hope I put this thread in the correct spot!

This is my first post and let me start by saying I am definitely not a mathematician. I am in fact a musician and I started thinking last night about just how many possible combinations of 3, 4, 5, 6, 7, 8, 9, 10, 11, and 12 notes there are above a given note and within one octave. I'm not sure if not knowing about music will affect anyone's ability to understand what I'm talking about so I'll try to fill in the gaps now.

There are twelve notes in the octave in western (European) music - that is, if you start on C and play every note up to the next C you will have played 12 notes (the next C being the 13th note). Up until the 20th century most chords consisted of 3-5 notes, and scales 7 notes (the notes of the scales being what chords are built up from). In a chord or a scale there is never two of the same note, that is C+E+G is not considered any different to C+E+G+C. The same goes for scales, you can't have two D flats in one scale. At the turn of the 20th century, composers began experimenting with bigger chords (chords with more notes) and new scales. So we now have music using scales and chords that have anywhere from 1 to 12 notes. Essentially, anything is possible, but what I want to know is exactly what is possible.

(If you do know a lot about music please ignore the fact that in tonal music chords are constructed directly in relation the the scales upon which the music is based and the effect a chord has in music is entirely dependent on context- I know that already. For this exercise I want to consider chords a scale as independent sonorities.)

I figured out (I think) how many combinations there where of 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, and 12 notes within an octave above a given note. Here is what I got. There is a definite pattern here and I would love to know if there was a simpler way of expressing it;
For a single note there is possibility - the root note,
2 note combinations there are 11 - the root note plus each of the other 11 notes,
3 notes = 10+9+8+7+6+5+4+3+2+1 = 55 - the root note plus all combinations of 2 notes within the remaining 11 notes,
4 notes = 9+(8x2)+(7x3)+(6x4)+(5x5)+(4x6)+(3x7)+(2x8)+(1x9) = 165 - the root note plus all combinations of 3 notes within the remaining 11 notes,
5 notes = 8+(7x(1+2))+(6x(1+2+3))+(5x(1+2+3+4))+(4x(1+2+3+4+ 5))+(3x(1+2+3+4+5+6))+(2x(1+2+3+4+5+6+7))+(1x(1+2+ 3+4+5+6+7+8)) = 330
6 notes = 7+(6x(1+(1+2)))+(5x(1+(1+2)+(1+2+3)))+(4x(1+(1+2)+ (1+2+3)+(1+2+3+4)))+(3x(....))+(2x(....))+(1x(.... )) = 462
7 notes = 6+(5x(....))+(4x(....)).... = 462
8 notes = 5+(4x(....))+(3x(....)).... = 330
9 notes = 4+(3x(....))+(2x(....)).... = 165
10 notes = 3+(2x(....))+(1x(....)) = 55
11 notes = 2+(1x9) = 11
12 notes = 1

Grand total = 2048

I really hope I have written this in a way that makes sense. Like I said, I am no mathematician.

I would like to know two things. First; Is there a more elegant way of expressing this? Second. If I wanted the extend it into two octaves, keeping in mind that if the notes in the second octave are identical the those in the first it wouldn't count as different (as long as one note is different then it is okay), how would I do that?

Thanks.

Wow terrifying answer awesome. you got it exactly are a math teacher Slyvia?

You might like to use group theory to help you out. In traditional 12-tone Western music, chords are based on two-note intervals. An interval can be "added" to another interval to form another interval. For example, the interval from C to E♭ is a minor third, and the interval from E♭ to G is a major third; when you add these two intervals, you get the interval from C to G, which is a perfect fifth. Hence, we can say that "minor-third plus major-second equals perfect-fifth". (I know musicians don't really say that but just bear with me; I'm illustrating what I mean by "adding" musical intervals.) If you don't count notes more than an octave apart as different from those within the same octave, then such additions will not produce intervals larger than the major seventh (C–B); for example, major-sixth (C–A) plus perfect-fourth (A–D) just yields a whole-tone interval (C–D). The octave interval, in particular, is identified with the unison interval (C–C).