1. **State the problem:** Given three sets A, B, and C with sizes $n(A)=40$, $n(B)=30$, and $n(C)=35$, and a Venn diagram with intersections labeled as follows: - $12$ in $A$ only - $9$ in $B$ only - $10$ in $C$ only - $a$ in $A \cap B$ only - $b$ in $A \cap C$ only - $c$ in $B \cap C$ only - $8$ in $A \cap B \cap C$ We need to find $n(A \cup B \cup C)$. 2. **Formula used:** The formula for the union of three sets is: $$ n(A \cup B \cup C) = n(A) + n(B) + n(C) - n(A \cap B) - n(B \cap C) - n(A \cap C) + n(A \cap B \cap C) $$ 3. **Express intersections:** From the Venn diagram: - $n(A) = 12 + a + b + 8 = 40$ - $n(B) = 9 + a + c + 8 = 30$ - $n(C) = 10 + b + c + 8 = 35$ 4. **Set up equations:** - $12 + a + b + 8 = 40 \implies a + b = 40 - 20 = 20$ - $9 + a + c + 8 = 30 \implies a + c = 30 - 17 = 13$ - $10 + b + c + 8 = 35 \implies b + c = 35 - 18 = 17$ 5. **Solve for $a$, $b$, and $c$:** From the three equations: $$ \begin{cases} a + b = 20 \\ a + c = 13 \\ b + c = 17 \end{cases} $$ Subtract second from first: $$ (a + b) - (a + c) = 20 - 13 \implies b - c = 7 $$ Add this to $b + c = 17$: $$ (b - c) + (b + c) = 7 + 17 \implies 2b = 24 \implies b = 12 $$ Then from $b + c = 17$: $$ 12 + c = 17 \implies c = 5 $$ From $a + b = 20$: $$ a + 12 = 20 \implies a = 8 $$ 6. **Calculate intersections:** - $n(A \cap B) = a + 8 = 8 + 8 = 16$ - $n(B \cap C) = c + 8 = 5 + 8 = 13$ - $n(A \cap C) = b + 8 = 12 + 8 = 20$ - $n(A \cap B \cap C) = 8$ 7. **Calculate union:** $$ \begin{aligned} n(A \cup B \cup C) &= n(A) + n(B) + n(C) - n(A \cap B) - n(B \cap C) - n(A \cap C) + n(A \cap B \cap C) \\ &= 40 + 30 + 35 - 16 - 13 - 20 + 8 \\ &= 105 - 49 + 8 = 64 \end{aligned} $$ **Final answer:** $$n(A \cup B \cup C) = 64$$