1. **Problem Statement:** In a country club of 144 people, 61 play football, 65 play baseball, and 72 play hockey. 22 play all three games, and 11 play none. We need to find: (i) How many play only two games? (ii) How many play only football? 2. **Known values:** Total people, $N = 144$ Football players, $F = 61$ Baseball players, $B = 65$ Hockey players, $H = 72$ Play all three, $F \cap B \cap H = 22$ Play none, $N_0 = 11$ 3. **Formula and approach:** Using the principle of inclusion-exclusion for three sets: $$|F \cup B \cup H| = |F| + |B| + |H| - |F \cap B| - |B \cap H| - |F \cap H| + |F \cap B \cap H|$$ Also, total people who play at least one game: $$N - N_0 = 144 - 11 = 133$$ Let the number of people who play exactly two games be $x$ for each pair (equal number given): $$|F \cap B| = |B \cap H| = |F \cap H| = x + 22$$ Because the intersection of two games includes those who play all three (22) plus those who play exactly two games. 4. **Calculate $x$:** Substitute values into inclusion-exclusion: $$133 = 61 + 65 + 72 - 3(x + 22) + 22$$ Simplify: $$133 = 198 - 3x - 66 + 22$$ $$133 = 154 - 3x$$ $$3x = 154 - 133 = 21$$ $$x = 7$$ So, exactly two games players per pair is 7. 5. **Calculate only football players:** Number who play only football: $$|F| - |F \cap B| - |F \cap H| + |F \cap B \cap H|$$ $$= 61 - (7 + 22) - (7 + 22) + 22$$ $$= 61 - 29 - 29 + 22 = 61 - 58 + 22 = 25$$ **Final answers:** (i) Number who play only two games (each pair) = 7 (ii) Number who play only football = 25