First slide

Operations on Sets

Question

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 number of students who have taken exactly one subject is

Easy

A

6
B

9
C

7
D

All of these

Solution

$n (M) = 23, n (P) = 24, n (C) = 19$
$n (M \cap P) = 12, n (M \cap C) = 9, n (P \cap C) = 7$
$n (M \cap P \cap C) = 4$
We have to find $n (M \cap P^{'} \cap C^{'}), n (P \cap M^{'} \cap C^{'}),$
$n (C \cap M^{'} \cap P^{'})$
Now $n (M \cap P^{'} \cap C^{'}) = n [M \cap {(P \cup C)}^{'}]$
$= n (M) - n (M \cap (P \cup C))$
$= n (M) - n [(M \cap P) \cup (M \cap C)]$
$= n (M) - N (M \cap P) - n (M \cap C) + n (M \cap P \cap C)$
$= 23 - 12 - 9 + 4 = 27 - 21 = 6$
$n (P \cap M^{'} \cap C^{'}) = n [P \cap {(M \cup C)}^{'}]$
$= n (P) - n [P \cap (M \cup C)] = n (P) - n [(P \cap M) \cup (P \cap C)]$
$= n (P) - n (P \cap M) - n (P \cap C) + n (P \cap M \cap C)$
$= 24 - 12 - 7 + 4 = 9$
$n (C \cap M^{'} \cap P^{'}) = n (C) - n (C \cap P) - n (C \cap M) + n (C \cap P \cap M)$
$= 19 - 7 - 9 + 4 = 23 - 16 = 7$

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