How would I determine all possible rhythmic combinations given certain limitations? If I have:
1) kick drum
2) hihat
3) snare drum
And I can play
1) One measure of 4/4 time
2) any note values between sixteenth and whole EDIT: (excluding tuplets)
How many possible drum beats could I form, starting with a measure of complete silence and ending with all three instruments playing a full-measure stream of 16th notes at the same time?
How many could I form that exclude simultaneity, i.e. no two drums can strike at the same time?
I'm a music person and not a math person, not sure where to start with this at all.
 A: In how many ways can we either hit a drum or rest in a 16 beat measure? This is an example of computation of a Power Set. We construct a 16 element set where each element represents a possible position for a note. The Power Set is the set of all possible subsets (i.e. exactly "possible organizations") of these elements. As a matter of convention, when constructing such a possible subset, we say that we "include" the positions where we'd place a note and "exclude" those where we'd place a rest. There are therefore 2 possibilities for what to do with each element. By the fundamental principle of counting (see: "Rule of product"), for a 16 element set, there are $2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2$ or $2^{16}$ subsets in the power set of 16 elements.
So if on any given beat we can have any combination from the set {snare, kick, hi-hat}, we need to find out how many combinations of these exist. Well we've already done something just like this when we found the Power Set of the set of beats in the measure above! In how many ways can we choose a musical action for a beat? $2^3 = 8$ (2 because we either hit some drum or don't, 3 because there are 3 drums). If we have 8 possible actions for each beat and 16 beats in a measure, we have $8^{16} = 281474976710656$ combinations. (this is the answer to your question)
Now, if we can only hit at most one drum during any given measure how many combinations do we have? Well, for one measure, how many choices do we have? Snare, kick, hi-hat or rest (i.e. no drum). 
To understand better, let's talk about "$k$-combination". $k$-combination answers the question "in how many ways can I choose $k$ elements out of a set with $n$ elements"? The answer is "$n$ choose $k$" or $\binom{n}{k}$ (which gives a "binomial coefficient"). $\binom{n}{k} = \frac{n!}{k!(n-k)!}$
Okay, so back to our problem. We want to find the sum of $1$-combinations (1 drum) and $0$-combinations (no drum). So for $1$-combinations, let $k=1$ & $n=3$. That is, there are 3 kinds of drum hits and we want to know how many different ways there are to choose 1 of them. $\frac{3!}{1!(3-1)!} = \frac{3!}{2!} = 3$. This is of, course, pretty obvious in our case here. How many $0$-combinations? Well, in how many way can we not hit a drum? 1, no drum or what would be formally called the "empty set". So for some beat, we have $3+1=4$ options.
4 options per beat, 16 beats. $4^{16} = 4294967296$. Though crucially, this answer doesn't hold if we can consider drum hits to last longer than 1/16th of a measure.
A: But there are 16 combinations for each 1/16th of the measure for two sticks and two feet. Therefore, your answer is 16^16
