1. **Problem statement:** In a class, 45% offer History, 30% offer Physics, and 40% offer Biology. 11% offer History and Physics, 15% offer Biology and Physics, 10% offer History and Biology. 16% offer none of these subjects. Find the percentage who study all three subjects. 2. **Formula and rules:** Use the principle of inclusion-exclusion for three sets $H$, $P$, and $B$: $$|H \cup P \cup B| = |H| + |P| + |B| - |H \cap P| - |P \cap B| - |H \cap B| + |H \cap P \cap B|$$ 3. **Known values:** $|H|=45$, $|P|=30$, $|B|=40$, $|H \cap P|=11$, $|P \cap B|=15$, $|H \cap B|=10$, and $16\%$ offer none, so $|H \cup P \cup B|=100-16=84$. 4. **Substitute values:** $$84 = 45 + 30 + 40 - 11 - 15 - 10 + |H \cap P \cap B|$$ 5. **Simplify:** $$84 = 115 - 36 + |H \cap P \cap B|$$ $$84 = 79 + |H \cap P \cap B|$$ 6. **Solve for triple intersection:** $$|H \cap P \cap B| = 84 - 79 = 5$$ **Answer:** 5% of the students study all three subjects.