In a class 18 students took Physics, 23 students took Chemistry and 24 students took Mathematics of those 13 took both Chemistry and Mathematics, 12 took both Physics and Chemistry and 11 took both Physics and Mathematics. If 6 students offered all the three subjects, find:

(i) The total number of students.

(ii) How many took Mathematics but not Chemistry.

(iii) How many took exactly one of the three subjects.

Solution:

Let P denotes Physics, C denotes Chemistry and M denotes Mathematics.

n(P) = 18

n(C) = 23

n(M) = 24

n(C ⋂ M) = 13

n(P ⋂ C) = 12

n(P ⋂ M) = 11

n(P ⋂ M ⋂ C) = 6

(i) The total number of students = n(P⋃C⋃M) 

= n(P) + n(C) + n(M) – n(P⋂C) – n(C⋂M) – n(P ⋂ M) + n(P ⋂ M ⋂ C)

= 18 + 23 + 24 – 12 – 13 – 11 + 6

= 35

Hence total number of students = 35.

(ii) Number of students who took Mathematics but not Chemistry = n(M-C)

= n(M) – n(M⋂C)

= 24 – 13

= 11

Hence 11 students took Mathematics but not Chemistry.

(iii) Number of students who took exactly one of the three subjects = n(P) + n(C) + n(M) – 2n(P⋂C) – 2n(C⋂M) – 2n(P ⋂ M) + 3n(P ⋂ M ⋂ C)

= 18 + 23 + 24 – (2 × 12) – (2 × 13) – (2 × 11) + (3 × 6)

= 65 – 24 – 26 – 22 + 18

= 11

Hence 11 students took exactly one of the three subjects.

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