From $50$ students taking examination in Mathematics, Physics and Chemistry, each of the students has passed in at least one of the subject, $37$passed Mathematics, $24$ Physics and $43$ Chemistry. At most $19$passed Mathematics and Physics, at most $29$ Mathematics and Chemistry and at most $20$ Physics and Chemistry. Then, the largest numbers that could have passed all three examinations, are

Let M be the set of students passing in Mathematics, P be the set of students passing in Physics and C be the set of students passing in Chemistry.

Now, $n(M\cup P\cup C)=50,n\left(M\right)=37,n\left(P\right)=24n\left(C\right)=43$

$n(M\cap P)\le 19,n(M\cap C)\le 29,n(P\cap C)\le 20$                        [given]

$$\begin{gathered} n(M\cup P\cup C)=n\left(M\right)+n\left(P\right)+n\left(C\right)-n(M\cap P)-n(M\cap C)-n(P\cap C)+n(M\cap P\cap C)\le 50 \\ \begin{array}{l}\Rightarrow 37+24+43-19-29-20+\mathrm{n}(\mathrm{M}\cap \mathrm{P}\cap \mathrm{C})\le 50\\ \Rightarrow \mathrm{n}(\mathrm{M}\cap \mathrm{P}\cap \mathrm{C})\le 50-36\\ \Rightarrow \mathrm{n}(\mathrm{M}\cap \mathrm{P}\cap \mathrm{C})\le 14\end{array} \end{gathered}$$

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