Let \( A=\{ \) raisin, apple, peach, grapefruit \( \}, B=\{ \) orange, berry, kiwi \( \} \), and \( C=\{ \) apple, tomato, lime \( \} \). (a) Find (i) \( n(A \cup B) \), (ii) \( n(A \cup C) \), and (iii) \( n(B \cup C) \). (b) In which case is the number of elements in the union not the sum of the number of elements in the individual sets? (a) (i) \( n(A \cup B)=\square \)

To tackle this problem, let's break down the unions and their cardinalities! First, for part (a): (i) To find \( n(A \cup B) \), we combine both sets \( A \) and \( B \): - \( A = \{ \text{raisin, apple, peach, grapefruit} \} \) has 4 elements. - \( B = \{ \text{orange, berry, kiwi} \} \) has 3 elements. - The union \( A \cup B = \{ \text{raisin, apple, peach, grapefruit, orange, berry, kiwi} \} \) has no overlapping elements. Thus, \( n(A \cup B) = 4 + 3 = 7 \). (ii) For \( n(A \cup C) \): - \( C = \{ \text{apple, tomato, lime} \} \) has 3 elements. - The union \( A \cup C = \{ \text{raisin, apple, peach, grapefruit, tomato, lime} \} \) has overlapping elements (apple). So, \( n(A \cup C) = 4 + 3 - 1 = 6 \) (subtracting the overlapping apple). (iii) For \( n(B \cup C) \): - The union \( B \cup C = \{ \text{orange, berry, kiwi, apple, tomato, lime} \} \), where there are no overlaps. Therefore, \( n(B \cup C) = 3 + 3 = 6 \). (b) The number of elements in the union is not the sum of the individual sets in cases where there are overlapping elements, as seen in \( n(A \cup C) \) where we had to subtract 1 for the overlapping apple. So, for part (a): (a) (i) \( n(A \cup B) = 7 \)