Let A, B and C denote the set of students who study English, Economics and Mathematics, respectively.
Then, we have,
Total number of students, n (A \(\cup \) B \(\cup \) C) = 40
Number of students who study English, n (A) = 16
Number of students who study Eoconomics, n (B) = 22
Number of students who study Mathematics, n (C) = 26
Number of students who study English and Economics, n (A  \(\cap \) B) = 5
Number of students who study Mathematics and Economics, n( B \(\cap \) C) = 14
and number students who study all subjects,
n (A \(\cup \) B \(\cup \) C) = 2
Clearly, n (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)                       
40 = 16 +22+ 26 - 5- 14 n ( C \(\cap \) A) + 2
40 = 66 - 19 - n ( C \(\cap \) A)
n ( C \(\cap \) A) = 47 - 70
=7
Hence the number of students who study English and Mathematics are 7.

Let H and E respectively doneto the set of persons who read Hindi and English newspapers and let U set of all person in a town.
Then, we have n (U) = 840,
n (H) =450,
n (E) = 300
and  n(H \(\cap \) E) = 200
Clearly, n(H \(\cup \) E) = n(H) + n(E) - n(H \(\cap \) E)
n(H \(\cup \) E) = 450 + 300 - 200
=  750 - 200 = 550
Now, the number of persons who read neither of the newspaper is given by
n(H' \(\cap \)  E' ) =  n(H' \(\cup \) E') [by De Morgan's law]
=  n (U) -  n(H\(\cup \)E)
= 840 - 550 = 290
Hence, 290 Persons read neither of the newspapers 

All rectangles, Rhombus and square are parallelograms because its opposite sides are equal and parallel.
Therefore \({ F }_{ 2 }\subset { F }_{ 1 },{ F }_{ 3 }\subset { F }_{ 1 }\) and \({ F }_{ 4 }\subset { F }_{ 1 }\)
\({ F }_{ 1 }={ F }_{ 2 }\cup { F }_{ 3 }\subset { F }_{ 4 }\)

Let C and T respectively donote the set of students who play Cricket and Tennis, and set U donotes the set of all students in a class.
Then n(C) = 25, n(T) = 20, n(C\(​​\)\(\cap \)T) = 10 and n (U)= 60
We know that, n(C\(​​\)\(\cup \)T) = n(C) + n(T) - n(C\(​​\)\(\cap \)T)
\(         \Rightarrow \) n(C\(​​\)\(\cup \) T) = 25 + 20 - 10 = 45 - 10 = 35
Now the Number of students who play neither game
n(C'\(\cap \)T' )   = n(C'\(\cup \)T' ) 
= n (U) -  n(C\(\cup \)T) = 60 - 35= 25
Hence, 25 students play neither games.