Question

In a class of \[30\] pupils , \[12\] take needlework , \[16\] take physics and \[18\] take history. If all the \[30\] students take at least one object and no one takes all three then the number of pupils taking 2 subjects is
\[\left( 1 \right){\text{ }}16\]
\[\left( 2 \right){\text{ }}6\]
\[\left( 3 \right){\text{ 8}}\]
\[\left( 4 \right){\text{ 20}}\]

Answer 1

Hint: For better understanding first we have to make the Venn diagram according to the given information. This question is based on the Euler Venn diagram and on the addition theorem of sets or inclusion and exclusion principle. In short, substitute the given values in the formula \[n\left( {X \cup Y \cup Z} \right) = n\left( X \right) + n\left( Y \right) + n\left( Z \right) - n\left( {X \cap Y} \right) - n\left( {Y \cap Z} \right) - n\left( {Z \cap X} \right) + n\left( {X \cap Y \cap Z} \right)\] to find number of people taking 2 subjects.

Complete step-by-step answer:
By reading the statement of the question we can easily understand that this question is based on the concepts of set theory, so we have to keep in mind the algorithms and basics of set theory. First and basic method to solve this question is given by Leonhard Euler and John Venn which is popularly known as Venn Diagrams. So, by making the Venn diagram of the given statement we can easily find out the answer to the asked question. But we have a second method to solve this question, but for this we have to remember that formula that is known as addition theorem of sets or inclusion and exclusion principle. So now let’s start the solution,
Let needlework be represented by \[N\] , Physics be represented by \[P\] and history be represented by \[H\] .
In the question, it is given that total number of pupils in a class \[ = {\text{ 30}}\] , from which
Number of pupils who take needlework \[ = \] \[n\left( N \right) = 12\] ,
Number of pupils who take physics \[ = \] \[n\left( P \right) = 16\] ,
Number of pupils who take history \[ = \] \[n\left( H \right) = 18\] ,
Number of pupils who do not take any of the three subjects \[ = n\left( {N \cap P \cap H} \right) = 0\] and
Number of pupils who take at least one subject or we can say number of students who take all the three subjects ; needlework, physics and history \[ = \] \[n\left( {N \cup P \cup H} \right) = 30\]
Number of pupils who take both needlework and physics \[ = {\text{ }}n\left( {N \cap P} \right)\]
Number of pupils who take both physics and history \[ = {\text{ }}n\left( {P \cap H} \right)\]
Number of pupils who take both history and needlework \[ = {\text{ }}n\left( {H \cap N} \right)\]
By applying the formula,
\[n\left( {N \cup P \cup H} \right) = n\left( N \right) + n\left( P \right) + n\left( H \right) - n\left( {N \cap P} \right) - n\left( {P \cap H} \right) - n\left( {H \cap N} \right) + n\left( {N \cap P \cap H} \right)\]
By substituting all the values in the above equation we get ,
\[30 = 12 + 16 + 18 - n\left( {N \cap P} \right) - n\left( {P \cap H} \right) - n\left( {H \cap N} \right) + 0\]
\[ \Rightarrow n\left( {N \cap P} \right) + n\left( {P \cap H} \right) + n\left( {H \cap N} \right) = 16 + 12 + 18 - 30\]
Further simplifying we get,
\[ \Rightarrow n\left( {N \cap P} \right) + n\left( {P \cap H} \right) + n\left( {H \cap N} \right) = 16 + 30 - 30\]
\[ \Rightarrow n\left( {N \cap P} \right) + n\left( {P \cap H} \right) + n\left( {H \cap N} \right) = 16\]
Therefore, the number of pupils taking 2 subjects is 16 .
Hence, the correct option is \[\left( 1 \right){\text{ }}16\] .
So, the correct answer is “Option 1”.

Note: For any three sets we use the formula \[n\left( {X \cup Y \cup Z} \right) = n\left( X \right) + n\left( Y \right) + n\left( Z \right) - n\left( {X \cap Y} \right) - n\left( {Y \cap Z} \right) - n\left( {Z \cap X} \right) + n\left( {X \cap Y \cap Z} \right)\] . Keep in mind that there is no need to calculate each intersection separately , we just have to find the total number of students who take at least 2 subjects . \[n\] denotes the number of elements in a set .