1. **Problem Statement:** We have 100 students selling fruits: 40 sell apples (A), 46 sell oranges (O), 50 sell mangoes (M). 14 sell both apples and oranges, 15 sell both apples and mangoes, and 10 sell all three fruits. Every student sells at least one fruit. We want to find the number of students selling only apples, only oranges, only mangoes, and those selling exactly two fruits. 2. **Formula and Rules:** Use the principle of inclusion-exclusion for three sets $A$, $O$, and $M$: $$|A \cup O \cup M| = |A| + |O| + |M| - |A \cap O| - |A \cap M| - |O \cap M| + |A \cap O \cap M|$$ Given $|A \cup O \cup M| = 100$ (all students), and the values: $|A|=40$, $|O|=46$, $|M|=50$, $|A \cap O|=14$, $|A \cap M|=15$, $|A \cap O \cap M|=10$. 3. **Find $|O \cap M|$:** Rearranging the formula: $$|O \cap M| = |A| + |O| + |M| - |A \cap O| - |A \cap M| + |A \cap O \cap M| - |A \cup O \cup M|$$ Substitute values: $$|O \cap M| = 40 + 46 + 50 - 14 - 15 + 10 - 100 = 136 - 29 + 10 - 100 = 117 - 100 = 17$$ 4. **Calculate only sellers:** - Only apples: $$|A| - |A \cap O| - |A \cap M| + |A \cap O \cap M| = 40 - 14 - 15 + 10 = 21$$ - Only oranges: $$|O| - |A \cap O| - |O \cap M| + |A \cap O \cap M| = 46 - 14 - 17 + 10 = 25$$ - Only mangoes: $$|M| - |A \cap M| - |O \cap M| + |A \cap O \cap M| = 50 - 15 - 17 + 10 = 28$$ 5. **Calculate exactly two fruits sellers:** - Apples and oranges only: $$|A \cap O| - |A \cap O \cap M| = 14 - 10 = 4$$ - Apples and mangoes only: $$|A \cap M| - |A \cap O \cap M| = 15 - 10 = 5$$ - Oranges and mangoes only: $$|O \cap M| - |A \cap O \cap M| = 17 - 10 = 7$$ **Final answers:** - Only apples: 21 - Only oranges: 25 - Only mangoes: 28 - Exactly two fruits: 4 (A&O), 5 (A&M), 7 (O&M)