1. **Problem statement:** We have 50 students, each speaks at least one language: English (E), French (F), or German (G). Given: $|E|=30$, $|F|=15$, $|G|=19$, $|E \cap G|=6$, $|E \cap F|=7$, $|G \cap F|=5$. We want to find $|E \cap F \cap G|$, the number of students who speak all three languages. 2. **Formula used:** For three sets, the inclusion-exclusion principle states: $$|E \cup F \cup G| = |E| + |F| + |G| - |E \cap F| - |E \cap G| - |F \cap G| + |E \cap F \cap G|$$ 3. Since all 50 students speak at least one language, $|E \cup F \cup G| = 50$. 4. Substitute the known values: $$50 = 30 + 15 + 19 - 7 - 6 - 5 + |E \cap F \cap G|$$ 5. Simplify the right side: $$50 = 64 - 18 + |E \cap F \cap G|$$ 6. Calculate $64 - 18$: $$50 = 46 + |E \cap F \cap G|$$ 7. Isolate $|E \cap F \cap G|$: $$|E \cap F \cap G| = 50 - 46$$ 8. Final answer: $$|E \cap F \cap G| = 4$$ **Explanation:** The number of students who speak all three languages is 4. This is found by applying the inclusion-exclusion principle to avoid double counting students who speak multiple languages.