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The 49 billion is 3003 (15 choose 5) * 73 (the number of positions you could do that in) squared.
Anybody good at math with a more accurate number?
Also allowing for not all fingers of a hand being used. Since one finger is set, we only look what the remaining 4 can reach:
14 choose 4 + 14 choose 3 + 14 choose 2 + 14 + 1 = 1471
this works for 74 positions for the lowest finger. For the top 14 positions we have to allow all fingers of the hand to be inside the 14, so here we have to add 14 choose 5 = 2002
so 1471 * 74 + (1471 + 2002) = 112327 should be accurate for one hand (or + 1 if we consider the "no fingers of this hand used at all"). That squared is 12.6 billion, but of course there's overlap in there too between the hands. Don't have a clever idea for that one right now that's not just enumerating scenarios...
Some overlap gets reduced by always assuming that the "lowest" finger is choosen to a specific key and seeing how many combinations can be reached from there upwards. Also allowing for not all fingers of a hand being used. Since one finger is set, we only look what the remaining 4 can reach: 14 choose 4 + 14 choose 3 + 14 choose 2 + 14 + 1 = 1471 this works for 74 positions for the lowest finger. For the top 14 posit…
Some overlap gets reduced by always assuming that the "lowest" finger is choosen to a specific key and seeing how many combinations can be reached from there upwards. Also allowing for not all fingers of a hand being used. Since one finger is set, we only look what the remaining 4 can reach: 14 choose 4 + 14 choose 3 + 14 choose 2 + 14 + 1 = 1471 this works for 74 positions for the lowest finger. For the top 14 posit…
Ah clever, totally missed that not all have to be used, thank you for this! I think you meant 12 billion?
Some overlap gets reduced by always assuming that the "lowest" finger is choosen to a specific key and seeing how many combinations can be reached from there upwards. Also allowing for not all fingers of a hand being used. Since one finger is set, we only look what the remaining 4 can reach: 14 choose 4 + 14 choose 3 + 14 choose 2 + 14 + 1 = 1471 this works for 74 positions for the lowest finger. For the top 14 posit…
Some overlap gets reduced by always assuming that the "lowest" finger is choosen to a specific key and seeing how many combinations can be reached from there upwards. Also allowing for not all fingers of a hand being used. Since one finger is set, we only look what the remaining 4 can reach: 14 choose 4 + 14 choose 3 + 14 choose 2 + 14 + 1 = 1471 this works for 74 positions for the lowest finger. For the top 14 posit…
Then, use the above to compute how often there are x keys to the right (edit: left may make computations slightly easier) of the first hand for x in 0, 1, …, 87, repeat the exercise above for those smaller keyboards (derive a formula to speed things up), and multiply/add up everything (for the second hand, _do_ count “no key pressed” as a valid option)
I’ll leave doing the additions as an exercise.
Might also note that when the hand is stretched to its full range you can’t necessarily hit all possible combinations of notes within the range. (That is, the index finger can’t reach everything between the middle of the range and the thumb’s full extent.)