Let M be the set of students passing in Mathematics

P be the set of students passing in Physics

C be the set of students passing in Chemistry

Now, n(M ∪ P ∪ C) = 50, n(M) = 37, n(P) = 24, n(C) = 43

n(M ∩ P) ≤ 19, n(M ∩ C) ≤ 29, n(P ∩ C) ≤ 20 (Given)

n(M ∪ P ∪ C) = n(M) + n(P) + n(C) – n(M ∩ P) – n(M ∩ C) – n(P ∩ C) + n(M ∩ P ∩ C) ≤ 50

⇒ 37 + 24 + 43 – 19 – 29 – 20 + n(M ∩ P ∩ C) ≤ 50

⇒ n(M ∩ P ∩ C) ≤ 50 – 36

⇒ n(M ∩ P ∩ C) ≤ 14

Thus, the largest possible number that could have passed all the three examinations is 14.