How many ways can a flute be fingered? Having trouble with this. Let's consider a flute with 6 holes. There's 1 way to play with no holes covered, 6 ways to play with exactly 1 hole covered, 15 ways to play with exactly 2 holes covered...and that's as far as I can think out.
But I can see that it will be symmetric. So covering 4 holes, there's 15 ways. 5 holes is 6 ways. All 6 holes, obviously there's only one way to do that.
I want to sum up all those ways. Sum = 1 + 6 + 15 + a + 15 + 6 + 1 = 44 + a. a is all the ways to play with 3 holes covered.
I'm sure this has been explored before but I don't know exactly what the combination or permutation or other terminology is. I believe this is a sum of multiple combinations. Anyway, I certainly cannot come up with a mathematical formula for it.
I also want to do this for a 8-hole flute. What is the answer and why?
 A: The answer to your question is a combination. These are represented by binomial coefficents: 
$$\binom{n}{k} = \frac{n!}{k!(n-k)!}$$
$\binom{n}{k}$ is the number of ways to choose $k$ objects (holes in a flute covered) form the $n$ objects in total (total holes in a flute), where you can't choose an object more than once (you can't cover a hole in a flute more than once) and the order in which you choose doesn't matter (it doesn't matter for the sound in what order you put your fingers on the holes).
$\binom{n}{k} = \frac{n!}{k!(n-k)!}$ because there are $n$ ways to choose the first object, $n-1$ to choose the second form the remaining objects, $n-2$ to choose the third form the remaining objects, ... , $n-k+1$ to choose the $k$th form the remaining objects. 
So there are $n(n-1)(n-2)\cdots(n-k+1) = \frac{n!}{(n-k)!}$ possibilities, but we didn't take into account that the order doesn't matter, so we need to divide by $k!$ (the number of ways to order $k$ choices) to get the formula 
$\binom{n}{k} = \frac{n!}{k!(n-k)!}$.

Note that the sum in your question is just missing one value. It should be $1+6+15+a+15+6+1$, with $a = \binom{6}{3} = \frac{6!}{3!3!}  = \frac{720}{6 \cdot 6} = 20$. 
You can also see that, for example, $\binom{6}{2} =  \frac{6!}{4!2!} \frac{720}{24 \cdot 2} = 15$. 
A: You can simply use -
$$\binom nr$$
You can see if cases start reflecting after one point.
Case 1 -
If total number of things are odd then total cases are n+1 including case with 0.
You have to find for $\frac{n+1}2$ cases and multiply by 2.
Case 2 -
If total number of things are even then total cases are n+1 including case with 0.
You have to find for $\frac{n-1}2$ cases and multiply by 2. Then add case $\frac{n+1}2$.
