Question

Let the universal set, s, have 66 elements. a and b are subsets of s. set a contains 19 elements and set b contains 34 elements. if sets a and b have 10 elements in common, how many elements are in neither a nor b?

Answer 1

The cardinality of the union between two sets is given by

$N(A\cup B) = N(A) + N(B) - N(A \cap B)$

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

$N(A \cup B) = 19+34-10 = 43$

This means that the two sets cover, without repetitions, 43 of the 66 elements in the sets. So, there are

$66-43 = 23$

elements which are in neither A nor B