

Let M: Set of students who have taken Maths
P: Set of students who have taken Physics
C: Set of students who have taken Chemistry
Given,
Total students n(U) = 25
n(M) = 15, n(P) = 12, n(C) = 11
n(M ∩ C) = 5, n(P ∩ C) = 4, n(M ∩ P) = 9,
n(M ∩ P ∩ C) = 3
1. Number of students taking only Chemistry = n(C - (M ∪ P))
= n(C) - n(C ∩ (M ∪ P))
= n(C) - [n(C ∩ M) + n(C ∩ P) - n((C ∩ M) ∩ (C ∩ P)) ]
= n(C) - n(C ∩ M) - n(C ∩ P) + n(C ∩ M ∩ P)
= 11 - 5 - 4 + 3
= 14 - 9
= 5
2. Number of students taking only Maths = n(M - (P ∪ C))
= n(M) - n(M ∩ (P ∪ C))
= n(M) - [n(M ∩ P) + n(M ∩ C) - n((M ∩ P) ∩ (M ∩ C)) ]
= n(M) - n(M ∩ P) - n(M ∩ C) + n(M ∩ P ∩ C)
= 15 - 9 - 5 + 3
= 18 - 14
= 4
3. Number of students taking only Physics = n(P - (M ∪ C))
= n(P) - n(P ∩ (M ∪ C))
= n(P) - [n(P ∩ M) + n(P ∩ C) - n((P ∩ M) ∩ (P ∩ C)) ]
= n(P) - n(P ∩ M) - n(P ∩ C) + n(P ∩ M ∩ C)
= 12 - 9 - 4 + 3
= 15 - 13
= 2
4. Number of students taking Physics and Chemistry but not Maths = n((P ∩ C) - M)
= n(P ∩ C) - n(P ∩ M ∩ C)
= 4 - 3
= 1
5. Number of students taking Maths and Physics but not Chemistry = n((M ∩ P) - C)
= n(M ∩ P) - n(P ∩ M ∩ C)
= 9 - 3
= 6
6. Number of students taking only one subject = n((only M) + (only P) + (only C))
= n(only M) + n(only P) + n(only C)
= 4 + 2 + 5
= 11
7. Number of students taking at least one subject = n(M ∪ P ∪ C)
= n M) + n(P) + n(C) - n(M ∩ P) - n(P ∩ C) - n(M ∩ C) + n(M ∩ P ∩ C)
= 15 + 12 + 11 - 9 - 4 - 5 + 3
= 41 - 18
= 23
8. Number of students taking none of three subject = 25 - n(M ∪ P ∪ C)
= 25 - 23
= 2
