Question

A class has $175$ students. The following table shows the number of students studying one or more of the following subjects in this class :

Subject	Number of students
Mathematics	$100$
Physics	$70$
Chemistry	$46$
Mathematics and Physics	$30$
Mathematics and Chemistry	$28$
Physics and Chemistry	$23$
Mathematics, Physics and Chemistry	$18$

How many students are enrolled in Mathematics alone, Physics alone and Chemistry alone ? Are there students who have not offered any of these three subjects ?

Answer 1

Ans $60M$, $35P$, $13C$, $22None$.
Here $n\left(U\right)=175$, $n\left(M\right)=100$, $n\left(P\right)=70$
$n\left(C\right)=46$
$n(M\cap P)=30$, $n(M\cap C)=28$
$n(P\cap C)=28$
and $n(M\cap P\cap C)=18$
To find $n(M\cap P^{\prime }\cap C^{\prime })$, $n(P\cap M^{\prime }\cap C^{\prime })$
$n(C\cap M^{\prime }\cap P^{\prime })$
and $n(P^{\prime }\cap M^{\prime }\cap C^{\prime })$
Now
similarly $n(P^{\prime }\cap M^{\prime }\cap C^{\prime })=n\left(P\right)-n(P\cap M)-n(P\cap C)+n(P\cap M\cap C)\}$
$=70-30-23+18=35$
$n(C\cap M^{\prime }\cap P^{\prime })=n\left(C\right)-n(C\cap M)-n(C\cap P)+n(C\cap M\cap P)\}$
$=46-28-23+18=13$
and $n(M^{\prime }\cap P^{\prime }\cap C^{\prime })=n(M\cup P\cup C{)}^{\prime }$
$\overline{z_2}-\overline{z_1}=\left(\overline{z_2-z_1}\right)$
$=n\left(U\right)-\{S_1\cap S_2\cap S_3\}$
$=n\left(U\right)-\left\{n\right(M)+n(P)+n(C)$
$-\left[\right(M\cap P)-n(P\cap C)$
$-(M\cap C)+n(M\cap P\cap C)\}$
$=175-100+70+46-30-23-28+18$
$175-216-81+18=22$