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VoiceLeading Possibilities:
PitchClass Assignment Combinations, Permutations, and Directional Considerations
Dr. Jody Nagel
June 12  July 3, 2005

List of Appendices
Appendix 0
Appendix 1
Appendix 2
Appendix 3
Appendix 4
Appendix 5
Appendix 6
Appendix 8

Given two chords, how many unique ways are there for these chords to be voiced? (For now, we will assume that any one voice moves no more than an octave.)
Let us take a particular situation. A flute, oboe, and clarinet are to play a complete Cmajor chord (C, E, G) followed by a complete Dmajor chord (D, F#, A). How many ways can this progression be voiced? Starting with the flute in the first chord, one of three pitchclasses might be assigned, and in the second chord, one of three pitchclasses might again be assigned. This creates 3 x 3 = 9 possible twopitchclass melodic fragments that the flute might play. These are: roottoroot, roottothird, roottofifth; thirdtoroot, thirdtothird, thirdtofifth; fifthtoroot, fifthtothird, and fifthtofifth. These will be abbreviated here as: 1.1, 1.3, 1.5; 3.1, 3.3, 3.5; 5.1, 5.3, 5.5.
After one of these fragments is assigned to the flute, we will turn our attention to the oboe. The oboe can only be assigned one of two remaining pitchclasses for the first chord, and again only one of two remaining pitchclasses for the second chord. This yields 2 x 2 = 4 possible twopitchclass melodic fragments that the oboe might play. Thus, if the flute is assigned 1.1, then the oboe may be assigned 3.3, 3.5, 5.3, or 5.5. Or, if the flute is assigned 5.1, then the oboe may be assigned 1.3, 1.5, 3.3, or 3.5. (Etc.)
Once the flute and oboe have been assigned their pitchclasses, this will leave only 1 x 1 = 1 possible twopitchclass melodic fragment that the clarinet can play.
Since any of the original nine flute choices may be coupled with any of four oboe choices (and then only one clarinet choice), this gives a total of 3^{2} x 2^{2} x 1^{2} = 36 possible ways that the pitchclasses might be assigned to the three instruments. These are listed below.

Flute 
Oboe 
Clar. 
1.1 
3.3 
5.5 
1.1 
3.5 
5.3 
1.1 
5.3 
3.5 
1.1 
5.5 
3.3 


Flute 
Oboe 
Clar. 
3.1 
1.3 
5.5 
3.1 
1.5 
5.3 
3.1 
5.3 
1.5 
3.1 
5.5 
1.3 


Flute 
Oboe 
Clar. 
5.1 
1.3 
3.5 
5.1 
1.5 
3.3 
5.1 
3.3 
1.5 
5.1 
3.5 
1.3 



1.3 
3.1 
5.5 
1.3 
3.5 
5.1 
1.3 
5.1 
3.5 
1.3 
5.5 
3.1 


3.3 
1.1 
5.5 
3.3 
1.5 
5.1 
3.3 
5.1 
1.5 
3.3 
5.5 
1.1 


5.3 
1.1 
3.5 
5.3 
1.5 
3.1 
5.3 
3.1 
1.5 
5.3 
3.5 
1.1 



1.5 
3.1 
5.3 
1.5 
3.3 
5.1 
1.5 
5.1 
3.3 
1.5 
5.3 
3.1 


3.5 
1.1 
5.3 
3.5 
1.3 
5.1 
3.5 
5.1 
1.3 
3.5 
5.3 
1.1 


5.5 
1.1 
3.3 
5.5 
1.3 
3.1 
5.5 
3.1 
1.3 
5.5 
3.3 
1.1 




To generalize, if "N" monophonic instruments each play one of "N" pitchclasses in some first chord of "N" pitchclasses (which may or may not contain duplications) and then each play one of "N" pitchclasses in some second chord of "N" pitchclasses (which may or may not contain duplications), and where the total pitchclass content of both chords must be fully utilized, then the number of possible ways (i.e., permutations) that the pitchclasses might be assigned to the "N" instruments is:
N^{2} x (N1)^{2} x . . . x 2^{2} x 1^{2} = (N!)^{2}
However, in order to fully determine the number of ways that two chords can be voiced, it must be remembered that, in twelvepitchclassesperoctave "mod12" pitch space, a voice may move "up" or "down" when moving from the first pitchclass to the second pitchclass. Each instrument independently has this "upordown" option. Returning to our original example of a flute, oboe, and clarinet playing a Cmajor chord followed by a Dmajor chord, we will recall that there are 36 possible ways that the pitchclasses might be assigned. The "upordown" motion of the flute, oboe, and clarinet, respectively, yield the following possibilities:
upupup 
upupdown 
updownup 
updowndown 
downupup 
downupdown 
downdownup 
downdowndown 
There are 2^{3} = 8 possible "upordown" motioncombinations possible for each of the 36 pitchclass assignments. This gives a grand total of 36 x 8 = 288 possible voicings for which three instruments can play these two 3note chords.
(See Appendix 0.)
(Also, see Appendix 3.)
If "N" instruments successively play two Nnote chords, and each instrument is capable of moving up or down to its second note, then the grand total number of permutations in which these chords might be voiced is:
(N!)^{2} x 2^{N}
COMMONTONES
There is one "small" detail remaining. If the two chords have one or more "commontones," then, if one instrument is actually assigned a commontone, we must decide if (1) the commontone is required to remain at the same octave, or if (2) the commontone is either allowed to remain at the same octave or to move up one octave or to move down one octave. Clearly in the first case, our "grand total" will be somewhat reduced; and just as clearly in the second case, our "grand total" will be somewhat increased. We will call the first case "Rule 1," for determining V_{1} possible voiceleadings of "N" instruments successively play two Nnote chords. And we will call the second case "Rule 2," for determining V_{2} possible voiceleadings of "N" instruments successively play two Nnote chords. Note that if the two chords have no commontones, then V_{1} = V_{2}.
ONE COMMONTONE
Let's consider the case of our wind trio playing a complete Cmajor chord (C, E, G) followed by a complete Eflatmajor chord (Eb, G, Bb). There is one potential commontone between the two chords, but this commontone may, or may not, be held common to one instrument. If the flute happens to be assigned the single commontone 5.3 (GG), then there are four pitchclass assignments that will be effected.
However, the same could be said if we started the problem from the oboe's pointofview or the clarinet's pointofview; each time there will be four cases that are effected. So, (four cases) for (each of three instruments) means that 4 x 3 = 12 of our 36 pitchclass assignments will need to be modified. For N = 3, this is 1/N of the (N!)^{2} cases (1/3 of the 36 cases). The other 24 pitchclass assignments do not actually involve holding the commontone in any one of the instruments.
If we insist on holding the commontone at the same octave (i.e., use "Rule 1"), then there are 2 x 2 x 1 = 4 "upordown" options for each of the 12 "oneheldcommontone" cases. There remain 8 "upordown" options for the 24 "noheldcommontones" cases. Our grand total is (8 x 24) + (4 x 12) = 192 + 48 = 240 voiceleading possibilities. The general case (with the singleoctave restriction) for counting voiceleading possibilities for two Npitchclass chords containing precisely one commontone is:
[
(N!)^{2} x
(n1)/n x
(2^{N}) ]
+
("noheldcommontones" cases)
[ (N!)^{2} x
1/n x
(2^{N1}) x
1 ]
("oneheldcommontone" cases)
If, following "Rule 2," the commontone is either allowed to remain at the same octave or to move up one octave or to move down one octave, then there are 2 x 2 x 3 = 12 "upordownorsame" options for each of the 12 "onecommontone" cases. Now our grand total is (8 x 24) + (12 x 12) = 192 + 144 = 336 voiceleading possibilities. The general case (without the singleoctave restriction) for counting voiceleading possibilities for two Npitchclass chords containing precisely one commontone is:
[
(N!)^{2} x
(n1)/n x
(2^{N}) ]
+
("noheldcommontones" cases)
[
(N!)^{2} x
1/n x
(2^{N1}) x
3 ]
("oneheldcommontone" cases)
TWO COMMONTONES
When chords contain more than one commontone, the situation becomes more complex since sometimes just one or just some subset of the potential commontones are, indeed, actually held as commontones.
Let's consider the case of our wind trio playing a complete Cmajor chord (C, E, G) followed by a complete Aminor chord (A, C, E). There are two potential commontones between the two chords, but either one of the commontones or both of the commontones may, or may not, be held common to one instrument.
a) In the case of flute 1.1 (CA); oboe 3.3 (EC); clarinet 5.5 (GE), no commontone is held, and therefore all 2^{3} = 8 "upordown" options apply.
b) In the case of flute 1.3 (CC); oboe 3.1 (EA); clarinet 5.5 (GE), one commontone is held by the flute. If the flute is required to hold the commontone at the same octave (Rule 1), then there are only 2^{2} x (2  1) = 4 "upordown" options. If the flute is either allowed to remain at the same octave or to move up one octave or to move down one octave (Rule 2), then there are 2^{2} x (2 + 1) = 12 "upordownorsame" options.
c) In the case of flute 1.3 (CC); oboe 3.5 (EE); clarinet 5.1 (GA), two commontones are held, by the flute and the oboe. If both the flute and oboe are required to hold the commontone at the same octave (Rule 1), then there are only 2^{1} x (2  1) x (2  1) = 2 "upordown" options. If both the flute and oboe are either allowed to remain at the same octave or to move up one octave or to move down one octave (Rule 2), then there are 2^{1} x (2 + 1) x (2 + 1) = 18 "upordownorsame" options.
The general question is: How do these commontone cases effect our "grand total" of 288, which, you will recall, had been true for two 3note chords that lack commontones? The number 288 was derived by the following multiplications:
(3^{2} x 2^{2} x 1^{2}) x 2^{3} = (3!)^{2} x 2^{3} = 36 x 8 = 288.
In the case of our wind trio playing a Cmajor chord progressing to an Aminor chord, of the 36 possible pitchclass assignments, 6 employ both of the two commontones, 12 employ just one of the two commontones, and 18 make no use of commontones. (See Appendix 1.)
 The flute might play one or the other of the two commontones for its pitchclass assignment, 1.3 (CC) or 1.5 (EE). This will effect eight of the 36 possible pitchclass assignments.
However, this is true from each of the three instrumental pointsofview. Thus each instrument contributes to 8 possible pitchclass assignments involving at least one held commontone. As can be seen from the above chart, four of the eight pitchclass assignments actually involve two held commontones and this is also true from each of the three instrumental pointsofview.
 If two commontones are indeed held, then they could be found in any of the three pairs of instruments, and each instrument could have either of the two possible commontones.
Thus, 2 x 3 = 6 possible pitchclass assignments involve two held commontones. Each instrument would be involved in 2 x 2 = 4 of the 6 total pitchclass assignments involving two held commontones.
 Now, each of the three instruments has 8 possible pitchclass assignments involving at least one held commontone, but from these 8 assignment possibilities we must subtract the four twoheldcommontone cases involved with the other two instruments. There are 8  (2 x 2) = 4 possible pitchclass assignments for each of the three instruments that will turn out to be a oneheldcommontone case. Therefore, there are 3 x 4 = 12 possible pitchclass assignments involving just one held commontone.
 The final case, where no commontones are held, can be found by subtracting the 6 twoheldcommontone cases and the 12 oneheldcommontone cases from the total (3!)^{2} = 36 cases. There are 36  6  12 = 18 noheldcommontone cases.
a) When we insist on "Rule 1" (held commontones being held in the same octave), then there are (6 x 2 x 1 x 1) + (12 x 2 x 2 x 1) + (18 x 2 x 2 x 2) = 204 voiceleading possibilities.
b) If we choose "Rule 2" (held commontones are free either to be held at the same octave or to move up an octave or to move down an octave), then there are (6 x 2 x 3 x 3) + (12 x 2 x 2 x 3) + (18 x 2 x 2 x 2) = 396 voiceleading possibilities. (See Appendix 3.)
THREE COMMONTONES
Next, let us consider the case of our wind trio playing a single Cmajor chord and then repeating that chord. An instrument can either "arpeggiate" up or down from one pitchclass to another, or can simply repeat the initial pitchclass. This is, then, the situation of having "two" 3note chords with three "commontones." Of the 36 possible pitchclass assignment permutations, 6 employ all three "commontones," 18 employ just one of the three "commontones," and 12 make no use of "commontones." Incidentally, if three commontones are employed, there are 3 x 3 x 3 = 3^{3} = 27 "upordownorsame" options. (See Appendix 2.) Let's derive these numbers: 6, 18, 12.
 Each instrument might play 1 of the 3 commontone pitchclass assignments. Begin by imagining the Flute playing 1.1 (CC). Again, if the first instrument's pitchclass assignment is made using a commontone pitchclass assignment, then there remain 4 total pitchclass assignment permutations possible for the other two instruments, of which 2 of the 4 involve two additional commontones.
In the above chart, the Flute could just as well have been assigned 3.3 or 5.5. Thus, each of the 3 instruments contributes 3 x 4 = 12 pitchclass assignment permutations involving at least one commontone.
 If three commontones are indeed employed, then the first instrument could have any of the three possible commontones. This would leave two possibilities for the second instrument and, after making that choice, only one possibility for the third instrument. Thus, 3 x 2 x 1 = 3! = 6 possible pitchclass assignments involve three commontones.
 There cannot be a case where only two of the three commontones are used; after two commontones are assigned, there is only an additional commontone left to choose for the third instrument. There are 0 twocommontone cases.
 Each of the three instruments now has 12 possible pitchclass assignments involving at least one commontone, but from these 12 assignment possibilities we must subtract the six threecommontone cases involved with the other two instruments. There are 12  (3!) = 6 possible onecommontone pitchclass assignments for each of the three instruments. Thus, there are 3 x 6 = 18 possible pitchclass assignments involving just one commontone.
 The final case, where no commontones are held, can be found by subtracting the 6 threecommontone cases and the 18 onecommontone cases from the total (3!)^{2} = 36 cases. There are 36  6  18 = 12 nocommontone cases.
a) When we insist on following "Rule 1" (holding commontones at the same octave), then there is only one voiceleading option for each of the 6 threeheldcommontone cases. There are four options for each of the 18 oneheldcommontone cases, and all eight options for each of the 12 noheldcommontone cases. There is, then, a total of V_{1} = (6 x 1x1x1) + (0 x 2x1x1) + (18 x 2x2x1) + (12 x 2x2x2) = 174 voiceleading possibilities.
b) If following "Rule 2" (held commontones are free either to be held at the same octave or to move up an octave or to move down an octave), then there are 27 voiceleading options for each of the 6 threeheldcommontone cases. There are 12 options for each of the 18 oneheldcommontone cases, and the usual eight options for each of the 12 noheldcommontone cases. There is, then, a total of V_{2} = (6 x 3x3x3) + (0 x 2x3x3) + (18 x 2x2x3) + (12 x 2x2x2) = 474 voiceleading possibilities.
The ultimate question here becomes: How many voicings are possible for a group of monophonic instruments playing a pair of chords that contain any given number of pitchclasses and an anysized subset of commontones between the two chords? From this point forward, we will label our variables as follows: N = the number of instruments, and the number of pitchclasses in each of the two chords. K = the number of commontones that can be found between the two chords. D = the number of held commontones, tones that are, indeed, actually held as a commontone by one single instrument. U = (K  D) = the number of unheld commontones. Y = the number of combinations of pitchclass assignments possible for a given (N, K, D). X = ( Y x N! ) = the number of permutations of pitchclass assignments possible for a given (N, K, D). Note the distinction of how Y is defined. There will always be N! combinations and (N!)^{2} permutations for a given N. However, we are defining Y to mean the number of combinations of pitchclass assignments possible for a given (N, K, D). This is necessary because, for different values of K and D, there will be different numbers of "upordown" possibilities. For example, in the case of the Cmajor and Aminor triads, which we just examined, we stated for (3, 2, 0) that Y = 3, X = (Y x N!) = 18; for (3, 2, 1) that Y = 2, X = 12; and for (3, 2, 2) that Y = 1, X = 6. The 18 permutations possible for (3, 2, 0) need to be multiplied by 8 up/down choices to obtain 144 total voicing possibilities. On the other hand, the 6 permutations possible for (3, 2, 2) need to be multiplied by 2 up/down choices to obtain 12 total voicing possibilities. Where do these values for Y (and therefore, for X) come from, in the general case?
To discover the answer to this question, we will break the problem into discreet parts. Ironically, it will be a bit easier to see the underlying pattern that governs this problem if we use a slightly larger value for N. We will consider the voicing possibilities for a pair of 5note chords, with 3 commontones, 2 of which are held commontones, and 1 of which is unheld. This is the (N, K, D) of (5, 3, 2), which would apply to a pair of chords that we will symbolize by [A, B, C, D, E] and [A, B, C, F, G]
Step1: We will at first consider only the combinations involving only the noncommontones plus the unheld commontones. Let's decide to let A and B represent the held commontones and C, for the time being, can be the unheld commontone. Thus, we consider only the combinations of ND = 52 = 3 pitchclasses [C, D, E] moving to [C, F, G], where C is unheld. These 4 combinations are:

C F
D C
E G

C F
D G
E C

C G
D C
E F

C G
D F
E C


This number of combinations, 4, we will label "B" and will discuss in detail momentarily. For now, realize that "B" is equal to the lowermagnitude number of combinations, Y_{2}, that are possible when [ N_{2} = ND; K_{2} = KD; D_{2} = DD = 0 ], or, in this case, when Y_{2} is based on [3, 1, 0], and therefore B = 4 combinations.
(See Appendix 6.)
Step2: There are P(3, 3) = (N_{2})! = (ND)! = 3! = 6 relative permutations for each of the B=4 combinations involving only the noncommontones plus the unheld commontones. For example, for the combination [ C F
D C
E G ], the first instrument could play the first fragment, while the other two instruments play either of the remaining two fragments. Or the second instrument could play the first fragment, while the first and third instruments play either of the remaining two fragments, and etc. The six permutations of this single combination are:

C F
D C
E G

C F
E G
D C

D C
C F
E G

D C
E G
C F

E G
C F
D C

E G
D C
C F


Each of the 4 combinations would have 6 permutations, so the total number of permutations so far is 4 x 6 = 24. (See Appendix 6.)
Step3: The possible permutations of the two chosen held commontones [A and B] over five positions is given by: P(5, 2) = 5! / (52)! = [ ( 5 x 4 x 3! ) / 3! ] = 20. The general case is: P(N, D) = [ N! / (ND)! ]. The 20 permutations for [A and B] are shown below. The other pitchclasses, for now, are simply shown as dashes.

A A 
B B 
  
  
  
A A 
  
B B 
  
  
A A 
  
  
B B 
  
A A 
  
  
  
B B 
B B 
A A 
  
  
  
  
A A 
B B 
  
  
  
A A 
  
B B 
  
  
A A 
  
  
B B 
B B 
  
A A 
  
  
  
B B 
A A 
  
  
  
  
A A 
B B 
  
  
  
A A 
  
B B 
B B 
  
  
A A 
  
  
B B 
  
A A 
  
  
  
B B 
A A 
  
  
  
  
A A 
B B 
B B 
  
  
  
A A 
  
B B 
  
  
A A 
  
  
B B 
  
A A 
  
  
  
B B 
A A 

Each of the 20 permutations of the selected 2 held commontones could be coupled with each of the 24 relative permutations of the other ND pitchclasses. So we now have 20 x 24 = 480 permutations. (See Appendix 6.)
Step4: The commontones that could have been chosen for the two held commontones are any two of the three: AA, BB, CC. There are C(3, 2) = 3! / ( (2!)(32)! ) = 3 "sub"combinations. The general case is: K! / [ (D!) x (KD)!) ]. If we allow U = KD to equal the number of unheld commontones, then we have K! / ( D! x U! ) subcombinations of the D held commontones. In our current consideration, the possible "sub"combinations of "2outof3" commontones are:


Finally, each of these 3 subcombinations could have governed the preceding 480 permutations. Thus, there are 480 x 3 = 1,440 total permutations for the case of [5, 3, 2]. (See Appendix 6.)
The total permutations of pitchclass assignments for (N, K, D), where U = KD, can be arrived at by multiplying together the four preceding terms:
K!
x N!
x
(ND)! x
B
D! x U! (ND)!
After canceling the (ND)! term from the numerator and denominator we end up with the very convenient formula:
For [N, K, D], X permutations =
N! x K! x B
D! x U!
And also,
For [N, K, D], Y combinations =
K! x B
D! x U!
At least, it would be VERY convenient if we were a bit clearer as to how to determine the value of B. First, we will complete the story for our case of [5, 3, 2].
Step5:
Since there are D = 2 held commontones, and ND = 3 noncommontones plus unheld commontones, then
(1) When we insist that commontones be held in the same octave, there are 2 x 2 x 2 x 1 x 1 = 8 "upordown" options. The general case is that there are 2^{(ND)} x 1^{D} = 2^{(ND)} "upordown" options. So, for [5, 3, 2], the 1,440 permutations of pitchclass assignments each have 8 "upordown" options for a semigrand total of 11,520 possible voicings among the 5 instruments.
(2) If commontones are free either to be held at the same octave or to move up an octave or to move down an octave, then there are 2 x 2 x 2 x 3 x 3 = 72 "upordownorsame" options. The general case is that there are 2^{(ND)} x 3^{D} "upordownorsame" options. So, for [5, 3, 2], the 1,440 permutations of pitchclass assignments each have 72 "upordownorsame" options for a semigrand total of 103,680 possible voicings among the 5 instruments.
Step6:
If we really want to know how many total voicings are available for N=5, and K=3, then we would have to add together the K+1 semigrand totals. In this case, the K+1 = 4 values for D that are possible are D=0; D=1; D=2; or D=3. The "Grand Total" for N=5 and K=3 is as follows:

D = 
Permutations X = 
Case (1) "upordown" options 
Case (1) semi grand total of voicings 
Case (2) "up ordownor same" options

Case (2) semi grand total of voicings 
0 
7,680 
2^{5} = 32 
245,760 
2^{5}x3^{0} = 32 
245,760 
1 
5,040 
2^{4} = 16 
80,640 
2^{4}x3^{1} = 48 
241,920 
2 
1,440 
2^{3} = 8 
11,520 
2^{3}x3^{2} = 72 
103,680 
3 
240 
2^{2} = 4 
960 
2^{2}x3^{3} = 108 
25,920 
Grand Total 
14,400 

338,880 

617,280 

In comparison, the Grand Total number of voicings for [5, 0, 0] is 460,800.
We must consider how to find the value for "B." Remember that "B" is equal to the lowermagnitude number of combinations, Y_{2}, that are possible when [ N_{2} = ND; K_{2} = KD; D_{2} = DD = 0 ]. What if D already equals zero? We still have no way to determine directly what the value of "B" is. Our attention, then, will now focus on cases of [N, K, 0].
Let's examine the following numerical chart. It shows how many combinations of pitchclass assignments can be made for twonote chords and for threenote chords. For now, the following chart shows only those cases when D = 0, or, in other words, when potential commontones are not actually held by any one instrument.


N 
K 
D 
Y 
C_{0} 
2 
0 
0 
2 
C_{1} 
2 
1 
0 
1 
C_{2} 
2 
2 
0 
1 





D_{0} 
3 
0 
0 
6 
D_{1} 
3 
1 
0 
4 
D_{2} 
3 
2 
0 
3 
D_{3} 
3 
3 
0 
2 

Now, notice that the Ydifference between D_{0} and D_{1} equals C_{0}. The Ydifference between D_{1} and D_{2} equals C_{1}. The Ydifference between D_{2} and D_{3} equals C_{2}. Let's make one more test. We will examine how many combinations of pitchclass assignments can be made for threenote chords and for fournote chords.


N 
K 
D 
Y 
D_{0} 
3 
0 
0 
6 
D_{1} 
3 
1 
0 
4 
D_{2} 
3 
2 
0 
3 
D_{3} 
3 
3 
0 
2 





E_{0} 
4 
0 
0 
24 
E_{1} 
4 
1 
0 
18 
E_{2} 
4 
2 
0 
14 
E_{3} 
4 
3 
0 
11 
E_{4} 
4 
4 
0 
9 

Again, notice the Ydifference between E_{0} and E_{1} equals D_{0}. The Ydifference between E_{1} and E_{2} equals D_{1}. The Ydifference between E_{2} and E_{3} equals D_{2}, and the Ydifference between E_{3} and E_{4} equals D_{3}. The value of Y for a given (N, K, 0), is based on the preceding values of Y.
(See Appendix 5.)
We will examine a particular case of (N, K, D), say (4, 2, 0), which is given as E_{2} (Y = 14) in the above chart. E_{2} = E_{1}  D_{1}. Also, E_{1} = E_{0}  D_{0}; and D_{1} = D_{0}  C_{0}. Furthermore, E_{0} = 4! = 24, D_{0} = 3! = 6, C_{0} = 2! = 2. So, to substitute terms, E_{2} = (E_{0}  D_{0})  (D_{0}  C_{0}). Thus, E_{2} = (1)E_{0}  (2)D_{0} + (1)C_{0} = (1)4!  (2)3! + (1)2! = 14. The coefficients of the combined terms are drawn from Pascal's famous "Triangle." By examining Appendix 5, one can quickly see that, for each value of N (with D=0), the number of combinations possible for (N, K, 0) is based on the difference between Y_{N} when (N, K1, 0) and Y_{(N1)} when (N1, K1, 0).
We can conclude from this that, when D=0, the value of B is: [ (C_{1})(N)! 
(C_{2})(N1)! + (C_{3})(N2)! ± . . .], where the coefficients, C_{x}, are determined by the (KD)th line of "Pascal's Triangle," and the entire expression will contain KD+1 terms. Notice that the successive terms alternate between subtraction and addition. As mentioned earlier, I refer to this value as "B." (The reason I chose this letter is because, at first, I did not see the pattern to these numbers, so I kept asking myself what could it "be?")

When for [N, K, D] D=0, 
Y 
= 
K! x B 
= 
K! x B 
= 
K! x B 
= 
B 
then Y = B. 


D! x U! 

D! x (KD)! 

0! x K! 



If for [N, K, D] D>0, then B = Y_{2} when [N_{2}=(ND); K_{2}=(KD); D_{2}=(DD)=0 ] We can, therefore, for all cases, restate the series of products of decreasing factorials and Pascal coefficient as: B = [ (C_{1})(ND)!  (C_{2})(ND1)! + (C_{3})(ND2)! ± . . .]
The formula for determining X permutations of pitchclass assignments for ALL cases of [N, K, D], and exclusively in terms of N, K, and D, is:
X = [ ( N! x K! ) / ( D! x (KD)! ) ] x { (C_{1})(ND)!  (C_{2})(ND1)! + (C_{3})(ND2)! ± . . .}
The numerator's bracketed {} portion will contain KD+1 terms. The Coefficients, Cx, are determined by the (KD)th line of "Pascal's Triangle." The operations of subtraction and addition alternate between successive terms.
To find the number of total voicings for [N, K, D], we multiply X by all the "upordown" options. V_{1} will equal the total voicings in the case where we insist that commontones be held in the same octave. V_{2} will equal the total voicings in the case when commontones are free either to be held at the same octave or to move up an octave or to move down an octave. Our final formulas to use, then, are:
V_{1} = X x 2^{(ND)}
and
V_{2} = X x 2^{(ND)} x 3^{D}
And finally (the big drum roll!), to find the Grand Total number of voicings possible for a given N and K, we must simply add up all K+1 semigrandtotals (all the V_{1}'s, or all the V_{2}'s) for the various possible values of D.
A small but simple detail remains, though. Since music uses nowadays (mostly) a 12pitchclassperoctave equaltempered system, we must realize that, if the chords to be played have more than 6 pitchclasses (N > 6), then the minimal number of commontones, K_{min}, becomes greater than zero. We have seen that, if N < 7, then K may range from 0 to N, and D may range from 0 to K. The number of GrandTotals for N, when N < 7, is N+1, and the number of semigrandtotals for (N, K) is K+1. However, if N > 6, then K may only range from K_{min} = 2N12 to N, though D is still free to range from 0 to K. In particular:

For N = 
the minimum value of K K_{min} = (2N  12) = 
7 
2 
8 
4 
9 
6 
10 
8 
11 
10 
12 
12 

This will only effect the number of "GrandTotals" that must be found for a given N. When N > 6, the number of GrandTotals will be (N+1)  (2N  12) = (13  N), though the number of semigrandtotals for (N, K) is still K+1. The process itself, for finding the values of the "semigrandtotals" and summing them into a Grand Total, is not effected. For a summary of N=0 through N=12, see
Appendix 4. (430 K.)
If the problem is enlarged to include voice motion greater than an octave, then the "grand total" skyrockets to ever larger numbers of voiceleading possibilities. When voices are constrained to move no more than an octave, but the number of voices increases, again, as we have seen, the "grand total" grows very quickly. If our wind trio is expanded to a quartet and the ensemble plays a viiº^{7}  I progression (where, say, the Ichord is specified to have a doubled root), then there are 4^{2} x 3^{2} x 2^{2} x 1^{2} = 576 pitchclass assignments. Since there are no commontones in this progression, the 4 instruments have 2^{4} = 16 "upordown" options. Thus, when playing these two chords, our brave quartet has 576 x 16 = 9,216 different voiceleading possibilities.
If an instrument is allowed to rest rather than play a note, or if two instruments optionally may select the same choice from the specified pitchclass sets (creating "incomplete" chords with arbitrary numbers of doublings), then the number of possibilities again increases. Consider our wind trio playing a "Cmajor chord" followed by a "Dmajor chord," but where the three instruments may each independently choose a first note (or a rest) and a second note (or a rest). There are now four objects (3 pitchclasses and a rest) from which to choose for the flute's first note and for the flute's 2nd note, giving the flute 16 possible pitchclassorrest assignments (including the possibility of two rests!) The oboe and the clarinet each have the same 16 possibilities. Thus, with this degree of freedom, there are 163 = 4096 total possible pitchclassorrest assignments. This does not take into account yet the 8 "upordown" options, and before you try calculating that, remember that you have to "deduct" the "upordown" option from instruments that have been assigned one or two rests!
In all of these previous considerations, we've not mentioned register. Are chords "required" to be in "close" position, or can they be in "open" position, or in any odd registral relationship whatsoever. Do individual instruments need to be kept in specified limited registers? All these factors make for an even deeper excursion into mathematical analysis.
After you've had fun with all that, then take the plunge into equaltempered scales that divide the octave into less or more than 12 pitchclasses! (Just kidding: the only thing that is effected then is the minimal number of commontones required for a given N that is greater than half the number of pitches per octave.)
After seeing the final entry of Appendix 4 (430 K.), we are overwhelmed at how huge the number of possible voicings becomes. When 12 instruments play a succession of two twelvetone chords, the number of possible voicings (for case 2) is: 1,549,462,668,226,094,812,800. This is one and a half sextillion voicings! The cosmologists tell us that the time cycle from the "Big Bang" to the "Big Crunch" is roughly 40 billion years. If all possible 12tone voicings were attempted during cosmological history, over 38 billion per year would need to be examined for all 40 billion years! (That's about 104 million voicings per Earth day.)
Besides the intrinsic interest of applying mathematical combinatorics to the pitchclass possibilities of music, it should also be quite a relief to composers to realize that, if three instruments playing a single pair of 3note chords can have that many hundreds of voicing possibilities, then, when a third chord and a fourth chord are added and an entire composition is worked out, there is very little likelihood of running out of compositional possibilities anytime in the near future!


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