The cardinality of the union between two sets is given by
In fact, if an element is in both A and B, it is counted in both N(A) and N(B), so we're counting it twice. By subtracting the cardinality of the interserction, we're taking back one of these "extra-counted" element, and the count is correct. In your case, the numbers are
This means that the two sets cover, without repetitions, 43 of the 66 elements in the sets. So, there are
elements which are in neither A nor B