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) \( n(A \cup B)=7 \) (ii) \( n(A \cup C)=\square \)

We begin by recalling that for any two sets \( X \) and \( Y \), \[ n(X\cup Y)=n(X)+n(Y)-n(X\cap Y). \] **(a) (i) \( n(A\cup B) \)** The sets are: \[ A=\{\text{raisin, apple, peach, grapefruit}\},\quad n(A)=4, \] \[ B=\{\text{orange, berry, kiwi}\},\quad n(B)=3. \] Since \( A \) and \( B \) have no common elements, \[ n(A\cap B)=0. \] Thus, \[ n(A\cup B)=4+3-0=7. \] **(a) (ii) \( n(A\cup C) \)** The sets are: \[ A=\{\text{raisin, apple, peach, grapefruit}\},\quad n(A)=4, \] \[ C=\{\text{apple, tomato, lime}\},\quad n(C)=3. \] Notice the common element in \( A \) and \( C \) is \(\text{apple}\), so \[ n(A\cap C)=1. \] Then, \[ n(A\cup C)=4+3-1=6. \] **(a) (iii) \( n(B\cup C) \)** The sets are: \[ B=\{\text{orange, berry, kiwi}\},\quad n(B)=3, \] \[ C=\{\text{apple, tomato, lime}\},\quad n(C)=3. \] Since \( B \) and \( C \) have no elements in common, \[ n(B\cap C)=0. \] So, \[ n(B\cup C)=3+3-0=6. \] **(b)** The union of the number of elements is not equal to the sum of the individual numbers when the sets have a common element (i.e., they are not disjoint). In this problem, \( A \) and \( C \) share the element \(\text{apple}\). Hence in the case of \( A\cup C \), the number of elements in the union is not the sum of \( n(A) \) and \( n(C) \). Thus, the answer for (a)(ii) is \( n(A\cup C)=6 \), and it is for \( A\cup C \) that the union is not simply the sum of the individual numbers of elements.