1. **Problem Statement:** A group of 22 travellers each had at least one of the following: passport (P), health certificate (H), or convertible currency (C). Given: - $|P|=12$ - $|H|=14$ - $|C|=11$ - $|P \cap H|=6$ - Total travellers $=22$ Find: - (b) Number of travellers with all three items $|P \cap H \cap C|$ - (c) Number of travellers with exactly two of the three items 2. **Step 1: Define variables for unknown intersections** Let: - $x = |P \cap H \cap C|$ (all three items) - $a = |P \cap H| - x = 6 - x$ (passport and health only) - $b = |P \cap C| - x$ (passport and currency only) - $c = |H \cap C| - x$ (health and currency only) 3. **Step 2: Use inclusion-exclusion principle** Total travellers with at least one item: $$ |P \cup H \cup C| = |P| + |H| + |C| - |P \cap H| - |P \cap C| - |H \cap C| + |P \cap H \cap C| $$ Substitute known values: $$ 22 = 12 + 14 + 11 - 6 - (b + x) - (c + x) + x $$ Simplify: $$ 22 = 37 - 6 - b - x - c - x + x $$ $$ 22 = 31 - b - c - x $$ Rearranged: $$ b + c + x = 31 - 22 = 9 $$ 4. **Step 3: Use individual set sizes to find $b$ and $c$** From set $P$: $$ |P| = a + b + x + \text{only P} = 12 $$ From set $H$: $$ |H| = a + c + x + \text{only H} = 14 $$ From set $C$: $$ |C| = b + c + x + \text{only C} = 11 $$ 5. **Step 4: Express only one-item counts in terms of $a,b,c,x$** Let: - $p = $ only P - $h = $ only H - $c_1 = $ only C Then: $$ p = 12 - a - b - x = 12 - (6 - x) - b - x = 12 - 6 + x - b - x = 6 - b $$ $$ h = 14 - a - c - x = 14 - (6 - x) - c - x = 14 - 6 + x - c - x = 8 - c $$ $$ c_1 = 11 - b - c - x $$ 6. **Step 5: Sum all parts to total 22** Sum all disjoint parts: $$ p + h + c_1 + a + b + c + x = 22 $$ Substitute expressions: $$ (6 - b) + (8 - c) + (11 - b - c - x) + (6 - x) + b + c + x = 22 $$ Simplify: $$ 6 - b + 8 - c + 11 - b - c - x + 6 - x + b + c + x = 22 $$ Combine like terms: $$ (6 + 8 + 11 + 6) + (-b - b + b) + (-c - c + c) + (-x - x + x + x) = 22 $$ $$ 31 - b - c = 22 $$ Rearranged: $$ b + c = 31 - 22 = 9 $$ 7. **Step 6: Recall from Step 2: $b + c + x = 9$** From Step 5: $b + c = 9$ From Step 2: $b + c + x = 9$ Subtracting: $$ (b + c + x) - (b + c) = 9 - 9 \Rightarrow x = 0 $$ 8. **Step 7: Final answers** - Number of travellers with all three items: $$ |P \cap H \cap C| = x = 0 $$ - Number with exactly two items: $$ a + b + c = (6 - x) + b + c = 6 + 9 = 15 $$ **Summary:** - (b) Number with all three items = 0 - (c) Number with exactly two items = 15