Question

In a class of 30 students, 12 students like embroidery, 16 students like physics and 18 students like history. If all students select at least one subject and none of them select all subjects then find the number of students select two subjects.

Answer 1

Hint: Here we use set theory of three different sets to solve for the value of the number of required students. Draw Venn diagram for three sets. Substitute the values in the formula of set theory and calculate the number of students select two subjects.
* If \[P(A)\] denotes the number of elements in A, \[P(B)\] denotes the number of elements in B,\[P(C)\] denotes the number of elements in C, then their intersections \[P(A \cap B),P(B \cap C),P(A \cap C)\] denotes the number of elements in both A and B; in both B and C and in both A and C respectively. The intersection of three sets A, B and C is \[P(A \cap B \cap C)\]. Then the number of elements in A, B or C is given by \[P(A \cup B \cup C)\]. Then the formula for set theory is
\[P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(B \cap C) - P(A \cap C) + P(A \cap B \cap C)\]

Complete step by step solution:
We are given three subjects Embroidery, Physics and history.
Let us assume Embroidery as A, Physics as B and History as C.
We draw Venn diagram

Total number of students in class\[ = 30\]
\[U = 30\]
Number of students who like Embroidery\[ = 12\]
\[P(A) = 12\]
Number of students who like physics\[ = 16\]
\[P(B) = 16\]
Number of students who like history\[ = 18\]
\[P(C) = 18\]
Number of students who like both Embroidery and physics are denoted by \[P(A \cap B)\]
Number of students who like both Physics and History are denoted by \[P(B \cap C)\]
Number of students who like both Embroidery and History are denoted by \[P(A \cap C)\]
Number of students who like all; Embroidery, history and physics are denoted by \[P(A \cap B \cap C)\]
\[P(A \cap B \cap C) = 0\]
Also, we are given all students selected at least on subject. So the union of all three sets is an equal universal set.
Number of students who like either Embroidery, history or physics are denoted by \[P(A \cup B \cup C)\]
\[P(A \cup B \cup C) = 30\]
Now we substitute the values in the formula of set theory i.e.
\[P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(B \cap C) - P(A \cap C) + P(A \cap B \cap C)\]
Put \[P(A \cup B \cup C) = 30;P(A) = 12;P(B) = 16;P(C) = 18;P(A \cap B \cap C) = 0\] in formula
\[ \Rightarrow 30 = 12 + 16 + 18 - P(A \cap B) - P(B \cap C) - P(A \cap C) + 0\]
Add constants in RHS
\[ \Rightarrow 30 = 46 - P(A \cap B) - P(B \cap C) - P(A \cap C) + 0\]
Bring all constants to one side of the equation
\[ \Rightarrow P(A \cap B) - P(B \cap C) - P(A \cap C) = 46 - 30\]
\[ \Rightarrow P(A \cap B) - P(B \cap C) - P(A \cap C) = 16\]
Then the number of students who selected any two subjects is 16.

\[\therefore \]Number of students who selected any two subjects is 16.

Note: Students might get confused from the question and they would attempt to find the initial values of intersections of each set. Keep in mind we have to find the total number of students who selected 2 subjects so that can be calculated as sum, we don’t need to calculate each intersection separately.