Q: binary combinations/rule of sums I'm currently working on this topic but I'm having a hard time. In an 8-bit string, there are 256 combinations (or is it permutations?). Either way, I know it is 2^8. 
If at least 3 elements must be 1, how many combinations exist? What about exactly 3 1s? At most?
When I use the combinations formula (n! / r! (n-r)! ) on the entire length of the bit string, the answers produce a bell-shaped curve (lowest answer 1, highest [and middle] answer is 70). What am I doing wrong? Is it as simple as "if 3 elements must be 1s, then 2^5 is what's left? Where does that leave my other questions?

A: While the word combination is used loosely as any way of combining things (whether it relates to the binomial coefficient or not), a permutation is specifically a bijective function from a set $S$ to itself.  If the set was ordered, then people will often write out the permutation as a string where the $i^{\text{th}}$ thing that appears is where the $i^{\text{th}}$ term in the unchanged set gets mapped to.  For notation, we can also choose to write it as a matrix where the top row is the original list, and the bottom row is where it gets mapped.
For example, with $S$ being the set $\{A,B,C\}$, you have the following six permutations:
$\begin{pmatrix} A&B&C\\A&B&C\end{pmatrix},\begin{pmatrix} A&B&C\\A&C&B\end{pmatrix},\begin{pmatrix} A&B&C\\B&A&C\end{pmatrix},\begin{pmatrix} A&B&C\\B&C&A\end{pmatrix},\begin{pmatrix} A&B&C\\C&A&B\end{pmatrix},\begin{pmatrix} A&B&C\\C&B&A\end{pmatrix}$
or it can be written more succinctly as $ABC, ACB, BAC, BCA, CAB, CBA$.
The first question on the other hand is not about permutations, but rather is a question of how many functions from $\{1,2,3,\dots,8\}$ there are to the set $\{0,1\}$.
These can be written in a similar manner:
$\begin{pmatrix}1&2&3&4&5&6&7&8\\0&0&0&0&0&0&0&0\end{pmatrix}, \begin{pmatrix}1&2&3&4&5&6&7&8\\0&0&0&0&0&0&0&1\end{pmatrix}, \begin{pmatrix}1&2&3&4&5&6&7&8\\0&0&0&0&0&0&1&0\end{pmatrix},\dots$
or these can be written more succinctly as $00000000, 00000001, 00000010,\dots$
and as you say there are $2^8=256$ of these.
The question of "If at least three elements must be a one" we may approach this in two directions.
In this specific scenario, since there are only two possibilities for things to be mapped to (either zero or one), there are precisely $\binom{8}{k} =\frac{8!}{(8-k)!k!}$ number of binary sequences of length eight which have precisely $k$ ones.  The problem becomes more complicated if the set we were mapping onto had more than two elements.
We may choose to answer as $\#(\text{At least 3 ones}) = \#(\text{Exactly 3 ones})+\#(\text{Exactly 4 ones})+\dots+\#(\text{Exactly 8 ones})$
or we may choose to use fewer arithmetic operations and do the following:
$\#(\text{At least 3 ones}) = \#(\text{Any number of ones}) - \#(\text{Exactly 2 ones}) - \#(\text{Exactly 1 ones}) - \#(\text{Exactly 0 ones})$
Similarly, you have 
$\#(\text{At most 3 ones}) = \#(\text{Exactly 3 ones})+\#(\text{Exactly 2 ones})+\dots+\#(\text{Exactly 0 ones})$
The number $2^5$ would not count how many have exactly $3$ ones, nor would it count how many don't have exactly $3$ ones, but rather it counts how many length 8 binary sequences there are where specifically the first three terms are one and the remaining five terms can be anything (more ones or zeroes or a mix).
