The kth number (counting the first as zero) in the nth row (again counting the first as zero) is known mathematically as "n choose k", or the number of ways of choosing k objects out of a total of n. This mathematical function is known as the "combination" (hence the first word in "Combination Product Set). It is symbolized by

(n)
( )
(k),

except that there should be one big left parenthesis instead of three, and one big right parenthesis instead of three. How does Pascal's triangle work to give you this information? Well, let's say you already know the number of ways of choosing k-1 objects out of n- 1 objects is

(n-1)
(   )
(k-1),

and that the number of ways of choosing k objects out of n-1 objects is

(n-1)
(   )
( k ).

Now let's say you add one new object to your universe, so now you have a total of n. How many ways are there to choose k objects now? Well, you can either choose k objects out of the n-1 you had before, and there are

(n-1)
(   )
( k )

ways of doing that; or you can choose the new object along with k-1 objects out of the n-1 you had before, and there are

(n-1)
(   )
(k-1)

ways of doing that. So the total number of possibilities is the sum of those two numbers; i.e.,

(n)   (n-1)   (n-1)
( ) = (   ) + (   )
(k)   (k-1)   ( k ).

So given the facts that there is only 1 way to choose no objects and only 1 way to choose all the objects; i.e.,

(n)       (n)
( ) = 1,  ( ) = 1,
(0)       (n)

we have the outer "shell" of ones of Pascal's triangle; the previous formula fills out the interior.

Now look at http://www.anaphoria.com/dal16.html again. Note that the figures are arranged in a triangle. Count the number of "dots" in each figure. It's Pascal's triangle!

Okay, so what about the shapes? Let's take the fourth row (n=3) as an example, since it's simple but not so simple as to be "degenerate". The three "objects" are the multiplicative factors A, B, and C. The first "dot" on the left represents the set obtained by multiplying no factors, which Wilson represents with the symbol for the empty set, an "O" with a slash through it. He also writes the symbol

(0)
( )
(3)

to mean the mathematical operation 3-choose-0 (he's got it upside- down relative to me): you're choosing zero elements out of a set of 3. Since 3-choose-0 equals 1 (as you can read off Pascal's triangle), Erv writes a "1" near the dot as well. The result of choosing zero factors out of three is a 1-note "scale", which Erv calls a "monany".

The next shape (representing 3-choose-1, which equals 3) after the dot is a triangle:

   B
  / \
 /   \
A-----C

Choosing 1 element at a time out of 3 leads to three possible products, namely the factors themselves: A, B, and C. This three-note set Erv calls a "Triany" but it's really just a triad. Here Erv has implicitly chosen to represent each of the "consonant" ratios A/C, A/B, and B/C, by horizontal, upward-sloping, and downward-sloping lines, respectively. This choice will hold through the rest of the row.

The next shape is the upside-down triangle for 3-choose-2 = 3:

A*B---B*C
  \   /
   \ /
   A*C

Clearly, each note in this triad comes from choosing two elements out of the set and multiplying them together (thus the second word of "Combination Product Set"). How does the configuration of these notes come about? Well, note that the interval (remember, intervals are _ratios_) between A*B and B*C is

A*B   A
--- = -
B*C   C,

and we have already chosen the horizontal line to represent the interval A/C, with the numerator on the left end of the line and the denominator on the right end. Similarly, the other two, diagonal lines, represent the same interval, in the same orientation, as they did in the previous triange. Check for yourself!

Finally, we come to 3-choose-3, where there is only one note, obtained by multiplying all the elements: A*B*C. Since it is only one note, it is really the same as the 3-choose-0 CPS: there is only one kind of monany, but we've seen two kinds of trianies.

Now let's make this a little more concrete by assigning values to the factors (you can assign any values you like, but I will deal with the most familiar and useful case). A, B, and C will be 1, 3, and 5 (we'll allow factors of 2 for free due to octave equivalence). The monanies are not very interesting, but the trianies are none other than our familiar major and minor triad. The first one is 1:3:5, or 4:6:5, or the JI major triad; the second one is 1*3:1*5:3*5, or 3:5:15, or 12:10:15, or the JI minor triad.

The same logic applies to each row of the Figure 19 (http://www.anaphoria.com/dal16.html). The next row contains the major tetrad and minor tetrad, and the hexany, which I've already discussed quite a bit in the context of the four factors being 1, 3, 5, and 7. The following row shows pentads and dekanies; if we choose 1, 3, 5, 7, and 9 as our factors, the pentads are the complete otonal and utonal chords of 9-limit harmony, and the dekanies are interesting scales that you should work out for yourself as an exercise. Note that in these diagrams and those in the next (and last) row, not all the connecting lines (i.e., consonant intervals) are drawn in by Erv; for example, in the 5-factor case he shows A/E with a line but not B/D. One potentially confusing thing is that the direction of A/E is parallel to that of B/D; hence sometimes these intervals may overlap on the lattice diagram. So by only drawing some of the lines, Erv staves off potentially confusing overlapping lines by omitting lines for certain consonant intervals (fully half of them in the 5-factor case, and two-thirds of them in the 6-factor case -- you will see the omitted lines as dotted lines in certain other figures in the D'Alessandro paper).

Other than that, the orientations of the factors are simply chosen to produce pretty, symmetrical shapes in which at least some of the important intervallic and chordal relationships can be readily seen. Here's a test: in the central diagram on the bottom, the Eikosany, can you find the 30 consonant tetrads that Carl was referring to? Each consonant tetrad will be configured exactly like either four out of the six notes of the major hexad (the 1-out-of-6 CPS), or four out of the six notes of the minor hexad (the 5-out-of-6 CPS). Recopying the figure bigger and drawing in the dotted lines (as suggested by the hexads in http://www.anaphoria.com/dal14.html) may help if you have trouble. (Hint: of all the possible ways of taking four of the six notes of the major hexad, and of all the possible ways of taking four of the six notes of the minor hexad, the Eikosany will have exactly one example of each -- 6-choose-4 is 15, and 15+15=30). If this is too hard, go back to the hexany and show yourself how its 8 consonant triads (the ones that look like triangles with a note at each vertex) are one each of the 4 possible ways of taking a triad out of the major tetrad, and one each of the 4 possible ways of taking a triad out of the minor tetrad. Then come back to the Eikosany. If you want to "cheat", look at the factors that make up the notes of the Eikosany (3 factors for each note) -- each consonant otonal tetrad will have two factors in common to all notes (e.g., A*B*C, A*B*D, A*B*E, A*B*F), and each utonal tetrad will have two factors absent from all notes (e.g., A*B*C, A*B*D, A*C*D, B*C*D).

For many more diagrams and explanations of historical tunings, see my book. If you don't understand my theory or the terms I've used, start here or try some definitions.

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