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In a class of $55$ students, the number of students studying different subjects are $23$ in Mathematics, $24$ in Physics, $19$ in Chemistry, $12$ in Mathematics and Physics, $9$ in Mathematics and Chemistry, $7$ in Physics and Chemistry and $4$ in all the three subjects. The total numbers of students who have taken exactly one subject is
$6$
$9$
$7$
$22$
Solution
(d) $n(M) = 23, n(P) = 24, n(C)= 19$
$n(M \cap P) = 12, n(M \cap C)= 9, n(P \cap C)= 7 ,$
$n(M \cap P \cap C) = 4 $
We have to find $n(M \cap P' \cap C'), n(P \cap M ' \cap C' ),n ( C \cap M' \cap P') $
$= n(M)-n(M \cap (P \cup C))$
$= n(M) – n[(M \cap P) \cup (M \cap C)]$
$= n(M) -n(M \cap P)-n(M \cap C) + n(M \cap P \cap C)$
$= 23 -12 -9 + 4 = 27 -21 = 6$
$= n(P)-n[P \cap (M \cup C)] = $ $n(P) – n[(P \cap M) \cup (P \cap C)]$
$= n(P) -n(P \cap M) -n(P \cap C) + n(P \cap M \cap C) $
$= 24 -12 -7 + 4 = 9$
$n(C \cap M' \cap P') = n(C) – n(C \cap P) – n(C \cap M) + n(C \cap P \cap M)$
$= 19 -7 -9 + 4 = 23 -16 = 7.$
So total number of student is $=6+9+7=22$