Question 11 \( 0 / 2 \) pts \( 3 \rightleftarrows 99 \) Details Let the Universal Set, S, have 131 elements. A and B are subsets of S. Set A contains 68 elements and Set B contains 50 elements. If Sets \( A \) and \( B \) have 20 elements in common, how many elements are in neither A nor B? Answer = elements Question Help: \( \square \) Video \( \square \) Message instructor Submit Question

To find the number of elements that are in neither set A nor set B, we can use the principle of inclusion-exclusion. The principle of inclusion-exclusion states that the number of elements in the union of two sets is equal to the sum of the number of elements in each set minus the number of elements in their intersection. In this case, we want to find the number of elements that are in neither set A nor set B, which is equivalent to finding the number of elements in the complement of the union of set A and set B. The number of elements in the union of set A and set B can be found by adding the number of elements in set A (68) and the number of elements in set B (50), and then subtracting the number of elements in their intersection (20). So, the number of elements in the union of set A and set B is 68 + 50 - 20 = 98. Since the universal set S has 131 elements, the number of elements that are in neither set A nor set B is equal to the number of elements in the complement of the union of set A and set B, which is 131 - 98 = 33. Therefore, there are 33 elements that are in neither set A nor set B.