Combinations of an alphabet of letters $a\ldots f$ which contain $a$ or $b$. 
Let $S = \{a, b, c, d, e, f\}$.  Find the number of subsets of $S$ which include the elements $a$ or $b$.

I physically counted it out and found it to be $27$, however I'd like to know how to solve this using more a more elegant method.  It's been awhile since I've done math like this, but I was thinking I need to use unions or intersections. 
 A: The number of subsets of S is equal to $2^6=64$, and the number of subsets which do not contain a or b is equal to $2^4=16$, so there are $64-16=48$ such subsets.
A: You can use inclusion exclusion.  The number of subsets of the language including 'a' is $2^5$ this is because each other letter may either appear or not (2 options for each).  Similarly, 'b' is in $2^5$ strings.  But, you've over counted the strings with both.  There are $2^4$ strings with both 'a' and 'b'.  So the total is:
$$2^5 + 2^5 - 2^4 = 32+32-16 = 48$$
A: Approach 1:
$5 \choose 2$ for those containing a,
plus $5 \choose 2$ for those containing b,
minus $4 \choose 1$ for those containing both a and b 
(as we're counted them twice so far).     
That's the count of the k=3 element sub-sets.
If you generalize this (for all k), you'll get a sum.
Be careful with k=1, the others seem straightforward. 
Approach 2:
Alternatively, you can count those sets which contain neither a, nor b.
This approach is easier. These are 2^4. Then we subtract that number
from the full count which is 2^6 and we get $2^6-2^4$.  
