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. 

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.
