Let \( A=\{ \) raisin, apple, peach, grapefruit \( \}, B=\{o r a n g e \), 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? \( \begin{array}{lll}\text { (a) } \text { (i) } n(A \cup B)=7 \\ \text { (ii) } n(A \cup C)=6 & \\ \begin{array}{lll}\text { (iii) } n(B \cup C)=6 & \text { (i), (ii), and (iii) } \\ \text { (b) In which case is the number of elements in the union not the sum of the number of elements in the individual sets? } \\ \text { (iii) } & \text { (ii) and (iii) } \\ \text { (iii) } & \text { (i) and (ii) } \\ \text { (ii) } & \text { None of the cases } \\ \text { (i) } & \text { (i) }\end{array}\end{array} \).

To solve this problem, let's break it down step by step. First, let's determine the unique elements in the unions for each case: (a) 1. For \( n(A \cup B) \): The sets are: - \( A = \{ \text{raisin, apple, peach, grapefruit} \} \) - \( B = \{ \text{o, r, a, n, g, e, berry, kiwi} \} \) The unique elements when we combine these sets are: - From A: raisin, apple, peach, grapefruit - From B: orange, berry, kiwi Thus, \( A \cup B = \{ \text{raisin, apple, peach, grapefruit, orange, berry, kiwi} \} \). Therefore, \( n(A \cup B) = 7 \). 2. For \( n(A \cup C) \): The sets are: - \( C = \{ \text{apple, tomato, lime} \} \) The unique elements are: - From A: raisin, apple, peach, grapefruit - From C: apple, tomato, lime Therefore, \( A \cup C = \{ \text{raisin, apple, peach, grapefruit, tomato, lime} \} \). Thus, \( n(A \cup C) = 6 \). 3. For \( n(B \cup C) \): - The unique elements are: - From B: orange, berry, kiwi - From C: apple, tomato, lime Therefore, \( B \cup C = \{ \text{o, r, a, n, g, e, berry, kiwi, apple, tomato, lime} \} \). So, \( n(B \cup C) = 6 \). (b) The number of elements in the union is not simply the sum of the number of elements in the individual sets due to the presence of common elements. Specifically, it occurs in cases where the sets share elements. The relevant case here is \( (i) n(A \cup C) \) and \( (iii) n(B \cup C) \), both of which include the element "apple" from set C present in set A. So, to summarize: - \( n(A \cup B) = 7 \), \( n(A \cup C) = 6 \), \( n(B \cup C) = 6 \). - The case of \( (iii) \) is the one where the union does not equal the sum of the elements due to common elements present in both sets.