1. **Problem Statement:** There are 80 men in a club, each playing at least one game among football, hockey, and baseball. Given: - Football only = 20 - Hockey only = 19 - Baseball only = 22 - Equal number of men play exactly two games only (denote this number as $x$ for each pair). - No men play all three games (assumed 0). Find how many men play baseball. 2. **Formula and Rules:** The total number of men is the sum of men playing only one game, men playing exactly two games, and men playing all three games. $$\text{Total} = \text{Football only} + \text{Hockey only} + \text{Baseball only} + \text{Sum of two-game intersections} + \text{Three-game intersection}$$ Since the three-game intersection is 0, and each two-game intersection has $x$ men, there are 3 such intersections: $$\text{Total} = 20 + 19 + 22 + 3x + 0 = 61 + 3x$$ 3. **Calculate $x$:** Given total men = 80, $$80 = 61 + 3x$$ Subtract 61 from both sides: $$80 - 61 = 3x$$ $$19 = 3x$$ Divide both sides by 3: $$x = \frac{19}{3} \approx 6.33$$ Since number of men must be whole, this suggests a slight inconsistency or rounding, but we proceed with $x = 6.33$ for calculation. 4. **Calculate total baseball players:** Baseball players include: - Baseball only: 22 - Football and Baseball only: $x = 6.33$ - Hockey and Baseball only: $x = 6.33$ - All three games: 0 Total baseball players: $$22 + 6.33 + 6.33 + 0 = 34.66 \approx 35$$ 5. **Summary:** - Number of men playing baseball is approximately 35. - The Venn diagram would show: - Football only: 20 - Hockey only: 19 - Baseball only: 22 - Each two-game intersection: about 6.33 - No men in all three games intersection. This completes the problem.