Question

In a class of $30$ pupils, $12$ take Chemistry, $16$ take Physics, and $18$ take History. If all the students take at least one subject and no one takes all three, then what is the number of pupils taking 2 subjects?
A. $16$
B. 6
C. 8
D. $20$

Answer 1

Hint: Here, apply the formula of the cardinality of three finite joint sets. Rearrange the formula according to the given conditions. Then substitute the given values to get the required answer.

Formula Used:
The formula of cardinality of finite sets:
Let $A,B$, and $C$ be the finite sets and $A \cap B \cap C \ne \varphi $, then
$n\left( {A \cup B \cup C} \right) = n\left( A \right) + n\left( B \right) + n\left( C \right) - n\left( {A \cap B} \right) - n\left( {B \cap C} \right) - n\left( {A \cap C} \right) + n\left( {A \cap B \cap C} \right)$

Complete step by step solution:
Given: From a class of $30$ pupils, $12$ take Chemistry, $16$ take Physics, and $18$ take History.
Rewrite the given information using the cardinality.
Total number of pupils $ = 30$
Number of pupils taking the Chemistry subject: $n\left( C \right) = 12$
Number of pupils taking the Physics subject: $n\left( P \right) = 18$
Number of pupils taking the History subject: $n\left( H \right) = 16$
It is also given that; all the students take at least one subject and no one takes all three subjects.
Then, we get
$n\left( {A \cup B \cup C} \right) = 30$ and $n\left( {A \cap B \cap C} \right) = 0$

We have to calculate the number of pupils taking 2 subjects.
Apply the formula of the cardinality of three finite joint sets.
$n\left( {C \cup P \cup H} \right) = n\left( C \right) + n\left( P \right) + n\left( H \right) - n\left( {C \cap P} \right) - n\left( {P \cap H} \right) - n\left( {C \cap H} \right) + n\left( {C \cap P \cap H} \right)$
Now rearrange the above formula.
$n\left( {C \cap P} \right) + n\left( {P \cap H} \right) + n\left( {C \cap H} \right) = n\left( C \right) + n\left( P \right) + n\left( H \right) + n\left( {C \cap P \cap H} \right) - n\left( {C \cup P \cup H} \right)$
Substitute the values in the above formula.
$n\left( {C \cap P} \right) + n\left( {P \cap H} \right) + n\left( {C \cap H} \right) = 12 + 18 + 16 + 0 - 30$
$ \Rightarrow n\left( {C \cap P} \right) + n\left( {P \cap H} \right) + n\left( {C \cap H} \right) = 46 - 30$
$ \Rightarrow n\left( {C \cap P} \right) + n\left( {P \cap H} \right) + n\left( {C \cap H} \right) = 16$

Option ‘A’ is correct

Note: The cardinality of a set is the number of unique elements present in the set.
Students often get confused between the symbols of union and intersection.
The symbol $ \cup $ is used to represent the union of sets. It is also called “and”.
The symbol $ \cap $ is used to represent the intersection of sets. It is also called “or”.