Given,
n(P) = 18, n(C) = 23, n(M) = 24,
n(C ∩ M) = 13, n(P ∩ C) = 12, n(P ∩ M) =11 and n(P ∩ C ∩ M) = 6
(i) Total no. of students in the class
= n(P ∩ C ∩ M)
= n(P) + n(C) + n(M) − n(P ∩ C)−, n(P ∩ M) − n(C ∩ M) + n(P ∩ C ∩ M)
= 18 + 23 + 24 – 12 – 11 – 13 + 6
= 35
(ii) No. of students who took Mathematics but not chemistry
= n(M − C)
= n(M) − n(M ∩ C)
= 24 – 13
= 11
(iii) No. of students who took exactly one of the three subjects
= n(P) + n(C) + n(M) − 2n(M ∩ P) − 2n(P ∩ C) − 2n(M ∩ C) + 3n(P ∩ C ∩ M)
= 18 + 23 + 24 – 2 x 11 – 2 x 12 – 2 x 13 + 3 x 6
= 65 – 22 – 24 – 26 + 18
= 83 – 72
= 11