Hey guys, I have been wondering about a somewhat useless question for a very long time now. How many three-fingered chords can a human press on the piano with only one hand? Take the maximum distance of a chord to be an octave long. (My hand is short haha. :heh:)

Well, I'm not math wiz, however it seems to be a pretty easy answer: One.

You have 5-fingers on one of your hands (let's hope!) and if you're playing a 3-fingered chord, you only have 2-fingers left, and you'd need 6-fingers to play "two, 3-fingered chords."

Might I ask where this question popped up from? xD

I tried with Cm + Ab sus4. But then, I get also an AbM7 which is 4 fingers chord if I remove the C#.

Sorry guys, that's not what I meant... :unsure: My apologies for making the question hard to interpret. What I meant was, how many different types of 3-note combinations can a human play with one hand, on a piano with 88 keys? This involves only combinations that don't exceed an octave's length. For example, taking the key of C,

C, D, E
C, D, F
C, D, G
C, D, A
C, D, B
C, D, C

D, E, F
D, E, G
D, E, A
D, E, B
D, E, C
D, E, D

See where I'm going with this? I know the number will reach hundreds, so I'm just wondering if there's any easy way to calculate the exact total. :)

P.S Mazeppa - I just thought of this while studying probability in my high/secondary school. That was a few years back.

Well, let's see.

We have 12 possible notes per key, which makes the number of possible combinations 12 x 12 x 12, or 1728 possible combinations (if I did my math correctly). 1728 is representative of the possible combination per octave, so if we approximate 88 keys to be about 7.5 octaves, we have 1725 x 7.5 = 12,960 possible combinations. (I wonder, though, if my math is right for the last one.)

Combinatorics to the rescue! You can use the nPk (Permutation (http://mathworld.wolfram.com/Permutation.html)) and nCk (Combination (http://mathworld.wolfram.com/Combination.html)) equations to determine repeatable and unique patterns. These both read as

nPk <=> n!/(n-k)!
nCk <=> n!/(k!*(n-k)!)

where n is the total number of possibilities, and k is the number of n values to form a desired pattern, and the postfix function factorial (!) corresponds to the function x! <=> f(x) = [ x * f(x-1) | x>0 ].

We can simplify nCk by canceling equations such that we can reuse the nPk operation's result.


nCk <=> n!
-------------
(k!*(n-k)!)

= n! 1
----------- * ----
(n-k)! k!

Notice the equation on the left hand side matches the nPk equation. Substituting that in we have,



nCk = nPk 1
* ----
k!

= nPk
-----
k!

So, since you asked about chords of three within an octave, we have twelve unique notes which must be in patterns of three to form the chord. So n = 12, and k = 3. This gives us 12P3, counting duplicate chords, and 12C3 not counting duplicates.

so that means


12P3 <=> 12!/(12-3)!

= 12*11*10*9*8*7*6*5*4*3*2*1
----------------------------
(9*8*7*6*5*4*3*2*1)

= 12*11*10

= 1320 repeatable chords.

And



12C3 <=> 12P3 / 3!

= 12*11*10
----------
3*2*1

= 220 unique chords.

That's within one octave, and counting non-standard chords (so chords like C, B, Ab are in this count). Thus, we know that a piano host approximately 8 octaves. Thus, we can take the nPk equation results and multiply that by 8, resulting in 10,560 repeatable chords. Similarly with nCk being 8 octaves, this gives us 1760 unique chords.

Ah, and my prophecy of M swinging by was true. ;)

This turned out to be quite an enlightening topic!

Oh yes that's it, Permutation and Combination. Thanks M (and Clarinetist too). ^_^ Haha, next step to becoming a master pianist, memorize all 1760 chords! XD

Awww... Sorry, but I think you've made a mistake M...(But maybe i'm wrong). For me, the answer is 5830 chords. I'll try to explain how i've come to this result, but since I'm not anglophone, I apologize for my english^^

My reasoning is the follows :

We want a chords of three notes, with at most one octave between the first and the third note.

First we choose the lowest note (i think it's C on a grand-piano). Then we count how many chords of 3 notes are possible between this note and its octave, what is 13C3 (12 notes for A to G plus A 8va).

Now, we'll add all the chords which end with a note upper than the octave of the lowest note. The number of these chords is 12C2*12*7. Indeed, fix a single note, then there are 12C2 chords possible who end with her. We have to multiply by 12 to cover the 12 notes from C# 8va to C, and by 7 for the 7 octaves after (C-C 8va).

Thus, we obtain finaly 13C3 + 7*12*12C2, which is equal to 5830.

The problem I have with your reasoning is that I see not how you can obtain the chords G-B-D..

Awww... Sorry, but I think you've made a mistake M...(But maybe i'm wrong). For me, the answer is 5830 chords. I'll try to explain how i've come to this result, but since I'm not anglophone, I apologize for my english^^

My reasoning is the follows :

We want a chords of three notes, with at most one octave between the first and the third note.

First we choose the lowest note (i think it's C on a grand-piano). Then we count how many chords of 3 notes are possible between this note and its octave, what is 13C3 (12 notes for A to G plus A 8va).

Now, we'll add all the chords which end with a note upper than the octave of the lowest note. The number of these chords is 12C2*12*7. Indeed, fix a single note, then there are 12C2 chords possible who end with her. We have to multiply by 12 to cover the 12 notes from C# 8va to C, and by 7 for the 7 octaves after (C-C 8va).

Thus, we obtain finaly 13C3 + 7*12*12C2, which is equal to 5830.

The problem I have with your reasoning is that I see not how you can obtain the chords G-B-D..

I apologize if I misunderstood something but I can't follow your math.

Recounting the octave note would be to recount it on every next octave multiple. Thus in the C8 round of combanations, you would count the C8 chords down to C7. In the C7 round of combinations you would recount the C7 chords counted in the C8 round and permute down to C6. On the C6 round of chords you would recount C6 from the C7 round and then permute down to C5, ad fundum. A completely unique set must include all unique elements. What you are doing is taking the super set of possibilities and adding in a subset from the superset, resulting in an overlapping union with many duplicate chords.

Hmmm, I think you misunderstood (Sorry if I can't get through to you, and if I haven't understood what you said...). The unions I take are all disjoints (I consider that a note and its octave are different). Indeed, since everytime I take a different note for upper note of the chord, and this for every set of combination, any set of combinations cannot have a shared chord with another.
(I can index every set of combinations (which haven't a duplicate chord) with the upper note of the chords, different for each set)


EDIT :

I though of a way to explain this "more visually" : Fix a note, then there are 12C2 chords possible which BEGIN on this note. multiply by 12 for the notes composing the scale, and by 7 for the octaves. Now there is still the last octave. Apply the same reasoning, there are 12C2 chords beginning on C8, 11C2 on C#8, 10C2 on D8, etc.

Finaly we obtain 7*12*12C2 + (12C2+11C2+10C2+ ... ) = 7*12*12C2 + 13C3

Since all the chords are different for one set of combination, and that the chords for one set begin on a different note of another set, at least, it's a lower bound (I think)