Let \( A= \) \{apple, prune, cantaloupe, pear\}, \( B= \) (peach, lemon\}, and \( C= \) \{peach, kiwi, berry\}. (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) \( n(A \cup B)=6 \) (ii) \( n(A \cup C)=7 \) (iii) \( n(B \cup C)=4 \) (b) In which case is the number of elements in the union not the sum of the number of elements in the individual sets? (iii) (i) (ii) (i) AInd (iii) (i) and (ii) (ii) and (iii) (i). (ii), and (iii) None of the cases
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To find the number of elements in the unions of sets \( A \), \( B \), and \( C \), we first need to determine the elements contained in each. Set \( A = \{ \text{apple, prune, cantaloupe, pear} \} \) has 4 distinct elements. Set \( B = \{ \text{peach, lemon} \} \) has 2 distinct elements. Set \( C = \{ \text{peach, kiwi, berry} \} \) has 3 distinct elements. Now, let's compute the unions: (i) \( A \cup B = \{ \text{apple, prune, cantaloupe, pear, peach, lemon} \} \) Here, 'peach' from set \( B \) is already included in \( A \), so we have: \( n(A \cup B) = 6 \). (ii) \( A \cup C = \{ \text{apple, prune, cantaloupe, pear, peach, kiwi, berry} \} \) Again, 'peach' from \( C \) overlaps with \( A \): \( n(A \cup C) = 7 \). (iii) \( B \cup C = \{ \text{peach, lemon, kiwi, berry} \} \) This time, 'peach' is present in both \( B \) and \( C \): \( n(B \cup C) = 4 \). Now for part (b), here are the number of elements in each case: - In (i), we had 6 elements total in the union, not simply the sum of individual sets (4 + 2 = 6). - In (ii), we had 7 which also does follow this checking out with (4 + 3 - 1 for 'peach'). - In (iii), 4 elements, as 'peach' is counted in both sets B and C. Thus, the answer to part (b) is: None of the cases, because despite overlaps, we still correctly accounted for the total in all scenarios.