1. **State the problem:** We have 68 students in total participating in sports events: field, track, and swimming. Given counts for various intersections and exclusive groups, we need to find: a) Number of students doing swimming only b) Number of students doing track event only c) Number of students doing all three events 2. **Given data:** - Total students, $N = 68$ - Field events, $|F| = 33$ - Track events, $|T| = 40$ - Swimming, $|S| = 23$ - Both field and track, $|F \cap T| = 14$ - Both swimming and field, $|S \cap F| = 8$ - Both swimming and track, $|S \cap T| = 10$ - Field only, $|F \text{ only}| = 15$ 3. **Unknowns:** - Swimming only, $x$ - Track only, $y$ - All three events, $z$ 4. **Use the principle of inclusion-exclusion and set up equations:** - Field total includes field only, field-track only, field-swimming only, and all three: $$|F| = |F \text{ only}| + (|F \cap T| - z) + (|S \cap F| - z) + z$$ Substitute known values: $$33 = 15 + (14 - z) + (8 - z) + z$$ Simplify: $$33 = 15 + 14 - z + 8 - z + z = 37 - z$$ So, $$z = 37 - 33 = 4$$ 5. **Find swimming only $x$:** - Swimming total: $$|S| = x + (|S \cap F| - z) + (|S \cap T| - z) + z$$ Substitute values: $$23 = x + (8 - 4) + (10 - 4) + 4$$ Simplify: $$23 = x + 4 + 6 + 4 = x + 14$$ So, $$x = 23 - 14 = 9$$ 6. **Find track only $y$:** - Track total: $$|T| = y + (|F \cap T| - z) + (|S \cap T| - z) + z$$ Substitute values: $$40 = y + (14 - 4) + (10 - 4) + 4$$ Simplify: $$40 = y + 10 + 6 + 4 = y + 20$$ So, $$y = 40 - 20 = 20$$ 7. **Summary of answers:** - Swimming only = $9$ - Track only = $20$ - All three events = $4$ These values satisfy all given conditions and the total number of students.