Let M, P and C denote the students of Mathematics, Physics and Chemistry, respectively.
Given,
n(M ∪ P ∪ C) = 50
No. of students passed in Mathematics, n(M) = 37
No, of students passed in Physics, n(P) = 24
No. of students passed in Chemistry, n(C) = 43
No. of students passed in Mathematics and Physics, n(M ∩ P) = 19
No. of students passed in Mathematics and Chemistry, n(M ∩ C) = 29
No. of students passed in Physics and chemistry, n(P ∩ C) = 20
Using identity
n(M ∪ P ∪ C) = n(M) + n(P) + n(C) − n(M ∩ P) − n(M ∩ C) − n(P ∩ C) + n(M ∩ P ∩ C)
∴ n(M ∪ P ∪ C) = n(M ∪ P ∪ C) − n(M) − n(P) − n(C) + n(M ∩ P) + n(M ∩ C) + n(P ∩ C)
= 50 – 37 – 24 – 43 +19 + 29 + 20
= 50 – (37 + 24 + 43) + (19 + 29 + 20)
= 50 – 104 + 68
= 50 – 36
= 14