1. Problem: In a competition, medals were awarded in three categories: dance (36 medals), dramatics (12 medals), and music (18 medals). Total persons awarded = 45. Exactly 4 persons received medals in all three categories. Find how many received medals in exactly two categories. 2. Let $D$, $R$, and $M$ be the sets of persons who got medals in dance, dramatics, and music respectively. 3. Given: $$|D|=36, |R|=12, |M|=18, |D \cup R \cup M|=45, |D \cap R \cap M|=4$$ 4. By the principle of inclusion-exclusion: $$|D \cup R \cup M| = |D| + |R| + |M| - |D \cap R| - |R \cap M| - |D \cap M| + |D \cap R \cap M|$$ 5. Plug values: $$45 = 36 + 12 + 18 - (|D \cap R| + |R \cap M| + |D \cap M|) + 4$$ 6. Simplify: $$45 = 66 - (|D \cap R| + |R \cap M| + |D \cap M|) + 4$$ $$45 = 70 - (|D \cap R| + |R \cap M| + |D \cap M|)$$ 7. Thus: $$|D \cap R| + |R \cap M| + |D \cap M| = 70 - 45 = 25$$ 8. The medals received in exactly two categories = sum of all two-category intersections minus thrice the triple intersection (since triple intersection is counted thrice in the sum): $$\text{Exactly two categories} = (|D \cap R| + |R \cap M| + |D \cap M|) - 3 \times |D \cap R \cap M|$$ 9. Substitute values: $$= 25 - 3 \times 4 = 25 - 12 = 13$$ Answer: 13 persons received medals in exactly two categories.