*Originally posted by Cheshire Cat*

**I could be wrong but one reason that this doesn't work is that for the first note there are 13 choices since a rest would be possible. Taking that into consideration, the second sixteenth note would have 13 possibilities as well. Th ...[text shortened]... d 29 more spaces and what do you have? That's where I get stuck.**

Nuathala did not include the rest. But let me explain what she meant.

*Every 13 mention in this proof is the number of possible notes and rest.*
presume we are at the beginning of the melody, looking at only a three sixteenth bar.

If we place three sixteenth notes we have 13^3 possibilities. If we take an eight note and then a sixteenth note, we have 2*13^2 posibilities. 13 for each position and 2 because we can reverse oreder. If we take a three sixteenth note we have only 13 posibilities. Total possible combinations: 13^3 + 2*13^2 + 13 = 13*(13+1)^2=13*14^2.

But as Nuathala says, one can see an eight note as a combination of an sixteenth note with an enlenghtment. If you write your note bar with only sixteenth notes and $'s if you lengthen the note, (then an eighth note is writen as a sixteenth note and a $ and a three sixteenth note would be writen as an sixteenth note with 2 $s) then one sees that on each position there can be 14 symbols (twelve notes and the $). Therefor there are 14^32 possible combinations.

Unless (I don't know much of this) it is not allowed the lenghten the the last note of a bar into the next bar. Then there are only 13*14^15 possibilities per bar. which makes 13^2*14^30=24,201,432,355,484,595,421,941,037,243,826,300 ~= 24*10^33

*Or am I missing something?*