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Q. From $50$ students taking examinations in Mathematics, Physics and Chemistry, $37$ passed Mathematics, $24$ Physics and $43$ Chemistry. At most $19$ passed Mathematics and Physics, at most $29$ passed Mathematics and Chemistry and at most $20$ passed Physics and Chemistry. The largest possible number that could have passed all three examinations is

Sets

Solution:

Let $M$, $P$ and $C$ be the sets of students taking examinations in Mathematics, Physics and Chemistry respectively.
$\therefore n\left(M \cup P\cup C\right) = 50$,
$n\left(M\right) = 37$, $n\left(P\right) = 24$, $n\left(C\right) = 43$,
$n\left(M \cap P \right)\le 19$, $n \left(M \cap C\right)\le 29$, $n\left(P \cap C\right) \le20$.
We have
$n\left(M \cup P \cup C\right) = n\left(M\right) + n\left(P\right) + n\left(C\right) - n\left(M \cap P\right)$
$- n\left(M \cap C\right) - n\left(P \cap C\right) + n\left(M \cap P \cap C \right)$
$\Rightarrow 50 = 37 + 24 + 43 - n\left(M \cap P\right) - n\left(M \cap C\right) -$
$n\left(P \cap C\right) + n \left(M \cap P \cap C \right)$
$\Rightarrow n \left(M \cap P \cap C \right) = n\left(M \cap P\right) + n \left(M \cap C\right) + n \left(P \cap C \right)-54$
$\Rightarrow n\left(M \cap P \cap C\right) \le 19 + 29 + 20 - 54 = 14$
$\Rightarrow n \left(M \cap P \cap C \right)\le 14$.