1. **Problem statement:** We have 15 students studying mathematics, 8 studying physics, 6 studying chemistry, and 3 studying all three subjects. We want to prove that 27 or more students study none of these subjects. 2. **Formula and principle used:** We use the principle of inclusion-exclusion to find the total number of students studying at least one subject. 3. **Inclusion-exclusion formula:** $$|M \cup P \cup C| = |M| + |P| + |C| - |M \cap P| - |P \cap C| - |M \cap C| + |M \cap P \cap C|$$ 4. **Given values:** - $|M| = 15$ - $|P| = 8$ - $|C| = 6$ - $|M \cap P \cap C| = 3$ 5. **Unknown intersections:** We don't know $|M \cap P|$, $|P \cap C|$, and $|M \cap C|$ individually, but since $|M \cap P \cap C| = 3$, these intersections are at least 3. 6. **Minimum total students studying at least one subject:** Assuming the pairwise intersections are exactly 3 (minimum possible), $$|M \cup P \cup C| \leq 15 + 8 + 6 - 3 - 3 - 3 + 3 = 23$$ 7. **Total students in the group:** The problem implies a larger group, since 15 study mathematics alone. 8. **Students studying none:** If total students are $N$, then students studying none are $$N - |M \cup P \cup C| \geq N - 23$$ 9. **To prove 27 or more study none:** $$N - 23 \geq 27 \implies N \geq 50$$ 10. **Conclusion:** If the total number of students is at least 50, then 27 or more students study none of these subjects. Since the problem states "In a set of 15 students 15 study mathematics," this seems contradictory or incomplete. Possibly the problem means a larger set with these numbers of students studying each subject. Under the assumption total students $N \geq 50$, the statement holds. **Final answer:** At least 27 students study none of these subjects if total students $N \geq 50$.