Question

19. In a class of 42 students, every students has taken at least one subject. 14 have taken
Mathematics, 20 have taken Physics, and 24 have selected English subject, 3 students have
taken both Mathematics and Physics, 2 have taken both English and Physics and no student has
taken all three subject. Find the number of students who have taken only Mathematics. [Ans. O]
then find​

Answer 1

Given :-

• Total students in class = 42

• Students having maths = 14

• Students having physics = 20

• Student having English = 24

• Students having both maths and physics = 3

• Students having both English and physics = 2

• Students having all three subjects = 0

To Find :-

• Students having only mathematics .

Solution :-

$\boxed{\sf{\implies a \: \cup \: b \: \cup \: c = n(a) + n(b) + n(c) - n(a \cap \: b) - n (b \cap \: c) - n(a \cap \: c) + n(a \cap \: b \cap \: c) }}$

${\sf{\implies M \: \cup \: P \: \cup \: E = n(M) + n(P) + n(E) - n(M \cap \: P) - n (P \cap \: E) - n(E \cap \: M) + n(M \cap \: P \cap \: E) }}$

${\sf{\implies 42= 14 + 20 + 24 - 3 - 2 - n(M \cap \: E) + 0 }}$

${\sf{\implies 42= 14 + 20 + 24 - 3 - 2 - n(M \cap \: E) + 0 }}$

${\sf{\implies 42= 58 - 5 - n(M \cap \: E) }}$

${\sf{\implies 42= 53 - n(M \cap \: E) }}$

${\sf{\implies 53 - 42= n(M \cap \: E) }}$

• $\underline{\sf{\implies n(M \cap \: E) = 11 }}$

Students having only maths = $\sf{n(M) \: - [n(M \cap \: E) + n(M \cap \: P)]}$

$\sf{ 14 \: - [11 + 3 ]}$

$\sf{ 14 \: - 14 = 0 }$

So number of student having only maths is zero