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In a class of $100$ students,$15$ students chose only physics (but not mathematics and chemistry),$3$ chose only chemistry (but not mathematics and physics), and $45$ chose only mathematics(but not physics and chemistry). Of the remaining students, it is found that $23$ have taken physics and chemistry,$20$ have taken physics and mathematics, and $12$ have taken mathematics and chemistry. The number of student who chose all the three subjects is
$6$
$9$
$12$
$15$
Solution
(b)
$n ( P \cap \overline{ M } \cap \overline{ C })=15$
$n ( C \cap \overline{ M } \cap \overline{ P })=3$
$n ( M \cap \overline{ P } \cap \overline{ C })=45$
$n ( P \cap C )=23$
$n ( P \cap M )=20$
$n ( M \cap C )=12$
Let $n(P \cap C \cap M)=x$
$n(P \cup C \cup M)=n(P)+n(C)+n(M)$
$\quad n(P \cap C)-n(C \cap M)-n(P \cap M)$
$+n(C \cap M \cap P)$
$100= 82-2 x \Rightarrow x=9$
