Jody Nagel
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Voice-Leading Possibilities
Pitch-Class Assignment Combinations, Permutations, and Directional Considerations
 
 
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Voice-Leading Possibilities:
Pitch-Class 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 C-major chord (C, E, G) followed by a complete D-major chord (D, F#, A). How many ways can this progression be voiced? Starting with the flute in the first chord, one of three pitch-classes might be assigned, and in the second chord, one of three pitch-classes might again be assigned. This creates 3 x 3 = 9 possible two-pitch-class melodic fragments that the flute might play. These are: root-to-root, root-to-third, root-to-fifth; third-to-root, third-to-third, third-to-fifth; fifth-to-root, fifth-to-third, and fifth-to-fifth. 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 pitch-classes for the first chord, and again only one of two remaining pitch-classes for the second chord. This yields 2 x 2 = 4 possible two-pitch-class 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 pitch-classes, this will leave only 1 x 1 = 1 possible two-pitch-class 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 32 x 22 x 12 = 36 possible ways that the pitch-classes 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" pitch-classes in some first chord of "N" pitch-classes (which may or may not contain duplications) and then each play one of "N" pitch-classes in some second chord of "N" pitch-classes (which may or may not contain duplications), and where the total pitch-class content of both chords must be fully utilized, then the number of possible ways (i.e., permutations) that the pitch-classes might be assigned to the "N" instruments is:
 
N2 x (N-1)2 x . . . x 22 x 12  =  (N!)2
 


However, in order to fully determine the number of ways that two chords can be voiced, it must be remembered that, in twelve-pitch-classes-per-octave "mod-12" pitch space, a voice may move "up" or "down" when moving from the first pitch-class to the second pitch-class. Each instrument independently has this "up-or-down" option. Returning to our original example of a flute, oboe, and clarinet playing a C-major chord followed by a D-major chord, we will recall that there are 36 possible ways that the pitch-classes might be assigned. The "up-or-down" motion of the flute, oboe, and clarinet, respectively, yield the following possibilities:
 

up-up-up up-up-down up-down-up up-down-down
down-up-up down-up-down down-down-up down-down-down

There are 23 = 8 possible "up-or-down" motion-combinations possible for each of the 36 pitch-class assignments. This gives a grand total of 36 x 8 = 288 possible voicings for which three instruments can play these two 3-note chords. (See Appendix 0.)   (Also, see Appendix 3.)
 
If "N" instruments successively play two N-note 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 2N

COMMON-TONES
 
There is one "small" detail remaining. If the two chords have one or more "common-tones," then, if one instrument is actually assigned a common-tone, we must decide if (1) the common-tone is required to remain at the same octave, or if (2) the common-tone 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 V1 possible voice-leadings of "N" instruments successively play two N-note chords. And we will call the second case "Rule 2," for determining V2 possible voice-leadings of "N" instruments successively play two N-note chords. Note that if the two chords have no common-tones, then V1 = V2.
 
ONE COMMON-TONE
 
Let's consider the case of our wind trio playing a complete C-major chord (C, E, G) followed by a complete E-flat-major chord (Eb, G, Bb). There is one potential common-tone between the two chords, but this common-tone may, or may not, be held common to one instrument. If the flute happens to be assigned the single common-tone 5.3 (G-G), then there are four pitch-class assignments that will be effected.
 
            VoiceLeadingPict1
 
However, the same could be said if we started the problem from the oboe's point-of-view or the clarinet's point-of-view; 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 pitch-class 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 pitch-class assignments do not actually involve holding the common-tone in any one of the instruments.
 
If we insist on holding the common-tone at the same octave (i.e., use "Rule 1"), then there are 2 x 2 x 1 = 4 "up-or-down" options for each of the 12 "one-held-common-tone" cases. There remain 8 "up-or-down" options for the 24 "no-held-common-tones" cases. Our grand total is (8 x 24) + (4 x 12) = 192 + 48 = 240 voice-leading possibilities. The general case (with the single-octave restriction) for counting voice-leading possibilities for two N-pitch-class chords containing precisely one common-tone is:
 
    [ (N!)2   x    (n-1)/n   x    (2N) ]     +         ("no-held-common-tones" cases)
 
    [ (N!)2   x    1/n   x    (2N-1)   x    1 ]          ("one-held-common-tone" cases)
 
If, following "Rule 2," the common-tone 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 "up-or-down-or-same" options for each of the 12 "one-common-tone" cases. Now our grand total is (8 x 24) + (12 x 12) = 192 + 144 = 336 voice-leading possibilities. The general case (without the single-octave restriction) for counting voice-leading possibilities for two N-pitch-class chords containing precisely one common-tone is:
 
    [ (N!)2   x    (n-1)/n   x    (2N) ]     +         ("no-held-common-tones" cases)
 
    [ (N!)2   x    1/n   x    (2N-1)   x    3 ]          ("one-held-common-tone" cases)
 
 
TWO COMMON-TONES
 
When chords contain more than one common-tone, the situation becomes more complex since sometimes just one or just some subset of the potential common-tones are, indeed, actually held as common-tones.
 
Let's consider the case of our wind trio playing a complete C-major chord (C, E, G) followed by a complete A-minor chord (A, C, E). There are two potential common-tones between the two chords, but either one of the common-tones or both of the common-tones may, or may not, be held common to one instrument.
 
a) In the case of flute 1.1 (C-A); oboe 3.3 (E-C); clarinet 5.5 (G-E), no common-tone is held, and therefore all 23 = 8 "up-or-down" options apply.
 
b) In the case of flute 1.3 (C-C); oboe 3.1 (E-A); clarinet 5.5 (G-E), one common-tone is held by the flute. If the flute is required to hold the common-tone at the same octave (Rule 1), then there are only 22 x (2 - 1) = 4 "up-or-down" 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 22 x (2 + 1) = 12 "up-or-down-or-same" options.
 
c) In the case of flute 1.3 (C-C); oboe 3.5 (E-E); clarinet 5.1 (G-A), two common-tones are held, by the flute and the oboe. If both the flute and oboe are required to hold the common-tone at the same octave (Rule 1), then there are only 21 x (2 - 1) x (2 - 1) = 2 "up-or-down" 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 21 x (2 + 1) x (2 + 1) = 18 "up-or-down-or-same" options.
 
The general question is: How do these common-tone cases effect our "grand total" of 288, which, you will recall, had been true for two 3-note chords that lack common-tones? The number 288 was derived by the following multiplications:
 
           (32 x 22 x 12) x 23 = (3!)2 x 23 = 36 x 8 = 288.
 
In the case of our wind trio playing a C-major chord progressing to an A-minor chord, of the 36 possible pitch-class assignments, 6 employ both of the two common-tones, 12 employ just one of the two common-tones, and 18 make no use of common-tones. (See Appendix 1.)
  • The flute might play one or the other of the two common-tones for its pitch-class assignment, 1.3 (C-C) or 1.5 (E-E). This will effect eight of the 36 possible pitch-class assignments.
                VoiceLeadingPict2
    However, this is true from each of the three instrumental points-of-view. Thus each instrument contributes to 8 possible pitch-class assignments involving at least one held common-tone. As can be seen from the above chart, four of the eight pitch-class assignments actually involve two held common-tones and this is also true from each of the three instrumental points-of-view.
     
  • If two common-tones 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 common-tones.
                VoiceLeadingPict3
    Thus, 2 x 3 = 6 possible pitch-class assignments involve two held common-tones. Each instrument would be involved in 2 x 2 = 4 of the 6 total pitch-class assignments involving two held common-tones.
     
  • Now, each of the three instruments has 8 possible pitch-class assignments involving at least one held common-tone, but from these 8 assignment possibilities we must subtract the four two-held-common-tone cases involved with the other two instruments. There are 8 - (2 x 2) = 4 possible pitch-class assignments for each of the three instruments that will turn out to be a one-held-common-tone case. Therefore, there are 3 x 4 = 12 possible pitch-class assignments involving just one held common-tone.
     
  • The final case, where no common-tones are held, can be found by subtracting the 6 two-held-common-tone cases and the 12 one-held-common-tone cases from the total (3!)2 = 36 cases. There are 36 - 6 - 12 = 18 no-held-common-tone cases.
a) When we insist on "Rule 1" (held common-tones 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 voice-leading possibilities.
 
b) If we choose "Rule 2" (held common-tones 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 voice-leading possibilities. (See Appendix 3.)
 
THREE COMMON-TONES
 
Next, let us consider the case of our wind trio playing a single C-major chord and then repeating that chord. An instrument can either "arpeggiate" up or down from one pitch-class to another, or can simply repeat the initial pitch-class. This is, then, the situation of having "two" 3-note chords with three "common-tones." Of the 36 possible pitch-class assignment permutations, 6 employ all three "common-tones," 18 employ just one of the three "common-tones," and 12 make no use of "common-tones." Incidentally, if three common-tones are employed, there are 3 x 3 x 3 = 33 = 27 "up-or-down-or-same" options. (See Appendix 2.) Let's derive these numbers: 6, 18, 12.
  • Each instrument might play 1 of the 3 common-tone pitch-class assignments. Begin by imagining the Flute playing 1.1 (C-C). Again, if the first instrument's pitch-class assignment is made using a common-tone pitch-class assignment, then there remain 4 total pitch-class assignment permutations possible for the other two instruments, of which 2 of the 4 involve two additional common-tones.
                VoiceLeadingPict4
    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 pitch-class assignment permutations involving at least one common-tone.
     
  • If three common-tones are indeed employed, then the first instrument could have any of the three possible common-tones. 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 pitch-class assignments involve three common-tones.
                VoiceLeadingPict5
  • There cannot be a case where only two of the three common-tones are used; after two common-tones are assigned, there is only an additional common-tone left to choose for the third instrument. There are 0 two-common-tone cases.
     
  • Each of the three instruments now has 12 possible pitch-class assignments involving at least one common-tone, but from these 12 assignment possibilities we must subtract the six three-common-tone cases involved with the other two instruments. There are 12 - (3!) = 6 possible one-common-tone pitch-class assignments for each of the three instruments. Thus, there are 3 x 6 = 18 possible pitch-class assignments involving just one common-tone.
     
  • The final case, where no common-tones are held, can be found by subtracting the 6 three-common-tone cases and the 18 one-common-tone cases from the total (3!)2 = 36 cases. There are 36 - 6 - 18 = 12 no-common-tone cases.
a) When we insist on following "Rule 1" (holding common-tones at the same octave), then there is only one voice-leading option for each of the 6 three-held-common-tone cases. There are four options for each of the 18 one-held-common-tone cases, and all eight options for each of the 12 no-held-common-tone cases. There is, then, a total of V1 = (6 x 1x1x1) + (0 x 2x1x1) + (18 x 2x2x1) + (12 x 2x2x2) = 174 voice-leading possibilities.
 
b) If following "Rule 2" (held common-tones 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 voice-leading options for each of the 6 three-held-common-tone cases. There are 12 options for each of the 18 one-held-common-tone cases, and the usual eight options for each of the 12 no-held-common-tone cases. There is, then, a total of V2 = (6 x 3x3x3) + (0 x 2x3x3) + (18 x 2x2x3) + (12 x 2x2x2) = 474 voice-leading 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 pitch-classes and an any-sized subset of common-tones between the two chords? From this point forward, we will label our variables as follows: N = the number of instruments, and the number of pitch-classes in each of the two chords. K = the number of common-tones that can be found between the two chords. D = the number of held common-tones, tones that are, indeed, actually held as a common-tone by one single instrument. U = (K - D) = the number of unheld common-tones. Y = the number of combinations of pitch-class assignments possible for a given (N, K, D). X = ( Y x N! ) = the number of permutations of pitch-class 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 pitch-class 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 "up-or-down" possibilities. For example, in the case of the C-major and A-minor 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 5-note chords, with 3 common-tones, 2 of which are held common-tones, 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]
 
Step-1: We will at first consider only the combinations involving only the non-common-tones plus the unheld common-tones. Let's decide to let A and B represent the held common-tones and C, for the time being, can be the unheld common-tone. Thus, we consider only the combinations of N-D = 5-2 = 3 pitch-classes [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 lower-magnitude number of combinations, Y2, that are possible when [ N2 = N-D; K2 = K-D; D2 = D-D = 0 ], or, in this case, when Y2 is based on [3, 1, 0], and therefore B = 4 combinations. (See Appendix 6.)
 
Step-2: There are P(3, 3) = (N2)! = (N-D)! = 3! = 6 relative permutations for each of the B=4 combinations involving only the non-common-tones plus the unheld common-tones. 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.)
 
Step-3: The possible permutations of the two chosen held common-tones [A and B] over five positions is given by: P(5, 2) = 5! / (5-2)! = [ ( 5 x 4 x 3! ) / 3! ] = 20. The general case is: P(N, D) = [ N! / (N-D)! ]. The 20 permutations for [A and B] are shown below. The other pitch-classes, 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 common-tones could be coupled with each of the 24 relative permutations of the other N-D pitch-classes. So we now have 20 x 24 = 480 permutations. (See Appendix 6.)
 
Step-4: The common-tones that could have been chosen for the two held common-tones are any two of the three: AA, BB, CC. There are C(3, 2) = 3! / ( (2!)(3-2)! ) = 3 "sub"-combinations. The general case is: K! / [ (D!) x (K-D)!) ]. If we allow U = K-D to equal the number of unheld common-tones, then we have K! / ( D! x U! ) sub-combinations of the D held common-tones. In our current consideration, the possible "sub"-combinations of "2-out-of-3" common-tones are:

A A         B B
A A         C C
B B         C C

Finally, each of these 3 sub-combinations 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 pitch-class assignments for (N, K, D), where U = K-D, can be arrived at by multiplying together the four preceding terms:
 
   K!        x       N!        x    (N-D)!   x    B
D! x U!         (N-D)!
 
After canceling the (N-D)! 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].
 
Step-5:
 
Since there are D = 2 held common-tones, and N-D = 3 non-common-tones plus unheld common-tones, then
 
(1) When we insist that common-tones be held in the same octave, there are 2 x 2 x 2 x 1 x 1 = 8 "up-or-down" options. The general case is that there are 2(N-D) x 1D = 2(N-D) "up-or-down" options. So, for [5, 3, 2], the 1,440 permutations of pitch-class assignments each have 8 "up-or-down" options for a semi-grand total of 11,520 possible voicings among the 5 instruments.
 
(2) If common-tones 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 "up-or-down-or-same" options. The general case is that there are 2(N-D) x 3D "up-or-down-or-same" options. So, for [5, 3, 2], the 1,440 permutations of pitch-class assignments each have 72 "up-or-down-or-same" options for a semi-grand total of 103,680 possible voicings among the 5 instruments.
 
Step-6:
 
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 semi-grand 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)
"up-or-down"
options
Case (1) semi-
grand total of
voicings
Case (2) "up-
or-down-or-
same" options
Case (2) semi-
grand total of
voicings
0 7,680 25 = 32 245,760 25x30 = 32 245,760
1 5,040 24 = 16 80,640 24x31 = 48 241,920
2 1,440 23 = 8 11,520 23x32 = 72 103,680
3 240 22 = 4 960 22x33 = 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 lower-magnitude number of combinations, Y2, that are possible when [ N2 = N-D; K2 = K-D; D2 = D-D = 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 pitch-class assignments can be made for two-note chords and for three-note chords. For now, the following chart shows only those cases when D = 0, or, in other words, when potential common-tones are not actually held by any one instrument.
 

  N K D Y
C0 2 0 0 2
C1 2 1 0 1
C2 2 2 0 1
D0 3 0 0 6
D1 3 1 0 4
D2 3 2 0 3
D3 3 3 0 2


Now, notice that the Y-difference between D0 and D1 equals C0. The Y-difference between D1 and D2 equals C1. The Y-difference between D2 and D3 equals C2. Let's make one more test. We will examine how many combinations of pitch-class assignments can be made for three-note chords and for four-note chords.
 

  N K D Y
D0 3 0 0 6
D1 3 1 0 4
D2 3 2 0 3
D3 3 3 0 2
E0 4 0 0 24
E1 4 1 0 18
E2 4 2 0 14
E3 4 3 0 11
E4 4 4 0 9


Again, notice the Y-difference between E0 and E1 equals D0. The Y-difference between E1 and E2 equals D1. The Y-difference between E2 and E3 equals D2, and the Y-difference between E3 and E4 equals D3. 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 E2 (Y = 14) in the above chart. E2 = E1 - D1. Also, E1 = E0 - D0; and D1 = D0 - C0. Furthermore, E0 = 4! = 24, D0 = 3! = 6, C0 = 2! = 2. So, to substitute terms, E2 = (E0 - D0) - (D0 - C0). Thus, E2 = (1)E0 - (2)D0 + (1)C0 = (1)4! - (2)3! + (1)2! = 14. The co-efficients 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 YN when (N, K-1, 0) and Y(N-1) when (N-1, K-1, 0).
 
We can conclude from this that, when D=0, the value of B is: [ (C1)(N)! - (C2)(N-1)! + (C3)(N-2)! ± . . .], where the co-efficients, Cx, are determined by the (K-D)th line of "Pascal's Triangle," and the entire expression will contain K-D+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 (K-D)!   0! x K!    


 
If for [N, K, D] D>0, then B = Y2 when [N2=(N-D); K2=(K-D); D2=(D-D)=0 ] We can, therefore, for all cases, restate the series of products of decreasing factorials and Pascal co-efficient as: B = [ (C1)(N-D)! - (C2)(N-D-1)! + (C3)(N-D-2)! ± . . .]
 
The formula for determining X permutations of pitch-class assignments for ALL cases of [N, K, D], and exclusively in terms of N, K, and D, is:
 
X  =  [ ( N! x K! )  /  ( D! x (K-D)! ) ]  x  { (C1)(N-D)! - (C2)(N-D-1)! + (C3)(N-D-2)! ± . . .}
 
The numerator's bracketed {} portion will contain K-D+1 terms. The Coefficients, Cx, are determined by the (K-D)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 "up-or-down" options. V1 will equal the total voicings in the case where we insist that common-tones be held in the same octave. V2 will equal the total voicings in the case when common-tones 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:
 
V1 = X x 2(N-D)
 
and
 
V2 = X x 2(N-D) x 3D
 
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 semi-grand-totals (all the V1's, or all the V2's) for the various possible values of D.
 
A small but simple detail remains, though. Since music uses nowadays (mostly) a 12-pitch-class-per-octave equal-tempered system, we must realize that, if the chords to be played have more than 6 pitch-classes (N > 6), then the minimal number of common-tones, Kmin, 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 Grand-Totals for N, when N < 7, is N+1, and the number of semi-grand-totals for (N, K) is K+1. However, if N > 6, then K may only range from Kmin = 2N-12 to N, though D is still free to range from 0 to K. In particular:
 

For
N =
the minimum value of K
Kmin = (2N - 12) =
7 2
8 4
9 6
10 8
11 10
12 12


 
This will only effect the number of "Grand-Totals" that must be found for a given N. When N > 6, the number of Grand-Totals will be (N+1) - (2N - 12) = (13 - N), though the number of semi-grand-totals for (N, K) is still K+1. The process itself, for finding the values of the "semi-grand-totals" 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 voice-leading 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 viiº7 - I progression (where, say, the I-chord is specified to have a doubled root), then there are 42 x 32 x 22 x 12 = 576 pitch-class assignments. Since there are no common-tones in this progression, the 4 instruments have 24 = 16 "up-or-down" options. Thus, when playing these two chords, our brave quartet has 576 x 16 = 9,216 different voice-leading 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 pitch-class sets (creating "incomplete" chords with arbitrary numbers of doublings), then the number of possibilities again increases. Consider our wind trio playing a "C-major chord" followed by a "D-major 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 pitch-classes 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 pitch-class-or-rest 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 pitch-class-or-rest assignments. This does not take into account yet the 8 "up-or-down" options, and before you try calculating that, remember that you have to "deduct" the "up-or-down" 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 equal-tempered scales that divide the octave into less or more than 12 pitch-classes! (Just kidding: the only thing that is effected then is the minimal number of common-tones 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 twelve-tone 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 12-tone 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 pitch-class possibilities of music, it should also be quite a relief to composers to realize that, if three instruments playing a single pair of 3-note 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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