Combinations of $8$ digit binary strings We are composing $8$ digit strings that are composed solely of $1$'s and $0$'s. For example, $10011011$ would be such a string. 
I would like to find how many strings have the property such that their second and fourth digits are $1$'s?
Would it simply be $2 \cdot 1 \cdot 2 \cdot 1 \cdot 2 \cdot 2 \cdot 2 \cdot 2$ or $64$? With the ones corresponding to only having $1$ choice for the second and fourth digits having to be $1$'s?
 A: For the second question, how many strings have either the second or fourth bit $1$, you can follow your logic for the first.  Three of four choices for the second and fourth bits have at least one $1$, so it is $\frac 34$ of all the eight bit strings. $\frac 34 \cdot 256=192$.  Another way is that the only way to fail is to have $0$ in both the second and fourth positions.  This is the same $64$ as you have calculated for having $1$ in both positions.  Subtract those from all strings and you have $256-64=192$.  Each approach is easier for some problems.
A: Your solution to the problem of determining the number of eight digit binary strings such that their second and fourth digits are $1$'s is correct.
Let $A$ be the set of binary strings of length $8$ that have the property that their second digit is a $1$.  Let $B$ be the set of binary strings of length $8$ that have the property that their fourth digit is a $1$.  Then the set of strings that have the property that their second or fourth digit is a $1$ is $A \cup B$.  Thus, we need to calculate $|A \cup B|$.  
$|A| = 2^7$ since there are two choices for each digit save the second, which can be chosen in one way.  Similarly, $|B| = 2^7$ since there are two choices for each digit save the fourth, which can be chosen in one way.  
Observe that if we simply add the number of elements in set $B$ to the number of elements in set $A$, then we will have counted the number of elements in their intersection twice.  Hence, we must subtract the number of elements in the intersection from $|A| + |B|$ to find the number of elements in the union, that is
$$|A \cup B| = |A| + |B| - |A \cap B|$$
That intersection consists of the strings that have the property that both their second and fourth digits are $1$'s.  You correctly calculated that $|A \cap B| = 2^6$.  Hence, the number of eight digit binary strings that have a $1$ in the second or fourth digits is 
$$|A \cup B| = |A| + |B| - |A \cap B| = 2^7 + 2^7 - 2^6 = 2 \cdot 2^7 - 2^6 = 2^8 - 2^6$$
Note that this calculation is based on the Inclusion-Exclusion Principle.
