1. **State the problem:** There are 80 men in a club, each playing at least one game: football, hockey, or baseball. Given: 20 play football only, 19 play hockey only, 22 play baseball only, and an equal number play exactly two games. We need to illustrate this with a Venn diagram and find how many play baseball in total. 2. **Define variables:** Let the number of men playing exactly two games be $x$ for each pair: football & hockey, hockey & baseball, and football & baseball. 3. **Total players equation:** Total men = football only + hockey only + baseball only + sum of men playing exactly two games + men playing all three games. Let the number playing all three games be $y$. So, $$20 + 19 + 22 + 3x + y = 80$$ 4. **Simplify:** $$61 + 3x + y = 80$$ $$3x + y = 19$$ 5. **Interpretation:** We have two unknowns $x$ and $y$ but only one equation. Without additional info, we cannot find unique values for $x$ and $y$. 6. **Find total baseball players:** Baseball players include baseball only, baseball & football only, baseball & hockey only, and all three games. Total baseball players = $22 + x + x + y = 22 + 2x + y$ 7. **Express total baseball players in terms of $x$:** From step 4, $y = 19 - 3x$, so $$\text{Baseball total} = 22 + 2x + (19 - 3x) = 41 - x$$ 8. **Conclusion:** The total number of baseball players is $41 - x$, where $x$ is the number of men playing exactly two games in each pair. Without more info, the exact number cannot be determined. **Venn diagram:** Three circles labeled Football, Hockey, Baseball with: - Football only: 20 - Hockey only: 19 - Baseball only: 22 - Each two-game intersection: $x$ - Three-game intersection: $y$ This completes the problem. **Final answer:** Total baseball players = $41 - x$ where $x$ is the equal number playing exactly two games.