1. **Problem Statement:** We have 96 students in total. The numbers of students participating in different events are: - Long distance races (L): 50 - Short distance races (S): 56 - Field events (F): 15 Also given are the numbers of students participating in the intersections of two events: - Long and short races (L \cap S): 15 - Long races and field events (L \cap F): 12 - Short races and field events (S \cap F): 18 We need to find the number of students who took part in all three events (L \cap S \cap F). 2. **Formula Used:** We use the principle of inclusion-exclusion for three sets: $$|L \cup S \cup F| = |L| + |S| + |F| - |L \cap S| - |L \cap F| - |S \cap F| + |L \cap S \cap F|$$ Since all students are in these events combined, $$|L \cup S \cup F| = 96$$. 3. **Substitute the known values:** $$96 = 50 + 56 + 15 - 15 - 12 - 18 + |L \cap S \cap F|$$ 4. **Simplify the right side:** $$96 = 121 - 45 + |L \cap S \cap F|$$ $$96 = 76 + |L \cap S \cap F|$$ 5. **Solve for the triple intersection:** $$|L \cap S \cap F| = 96 - 76 = 20$$ **Answer:** 20 students took part in all three events. This means 20 students participated in long distance races, short distance races, and field events simultaneously.