Practical problems on sets | Part-3

Eleven Standard >> Practical problems on sets | Part-3

Set Theory Problem: Students Taking Different Subjects

In a group of students, each student has taken at least one of the three subjects: Chemistry (C), Mathematics (M), and Physics (P). Given the following data:

Total students = 100

Students taking Chemistry (C) = 60

Students taking Mathematics (M) = 50

Students taking Physics (P) = 40

Students taking both Chemistry and Mathematics = 25

Students taking both Chemistry and Physics = 20

Students taking both Mathematics and Physics = 15

Students taking all three subjects = 5

Find the number of students who have taken:

Only Chemistry

Only Mathematics

Only Physics

Exactly one of the three subjects

Physics and Chemistry but not Mathematics

Mathematics and Physics but not Chemistry

Mathematics and Chemistry but not Physics

Exactly two of the three subjects

At least one of the three subjects

None of the three subjects

Solve:

Let:

n(C) = 60
n(M) = 50
n(P) = 40
n(C ∩ M) = 25
n(C ∩ P) = 20
n(M ∩ P) = 15
n(C ∩ M ∩ P) = 5

Using the principle of inclusion and exclusion:

Total who took at least one subject:
n(C ∪ M ∪ P) = n(C) + n(M) + n(P) − n(C ∩ M) − n(C ∩ P) − n(M ∩ P) + n(C ∩ M ∩ P)
= 60 + 50 + 40 − 25 − 20 − 15 + 5 = 95

Students who took none = 100 − 95 = 5

i) Only Chemistry:

= n(C) − n(C ∩ M) − n(C ∩ P) + n(C ∩ M ∩ P) = 60 − 25 − 20 + 5 = 20

ii) Only Mathematics:

= 50 − 25 − 15 + 5 = 15

iii) Only Physics:

= 40 − 20 − 15 + 5 = 10

iv) Exactly one of the subjects:

= Only C + Only M + Only P = 20 + 15 + 10 = 45

v) Physics and Chemistry but not Mathematics:

= n(C ∩ P) − n(C ∩ M ∩ P) = 20 − 5 = 15

vi) Mathematics and Physics but not Chemistry:

= n(M ∩ P) − n(C ∩ M ∩ P) = 15 − 5 = 10

vii) Mathematics and Chemistry but not Physics:

= n(C ∩ M) − n(C ∩ M ∩ P) = 25 − 5 = 20

viii) Exactly two of the three subjects:

= (C ∩ M − all three) + (C ∩ P − all three) + (M ∩ P − all three) = (25 − 5) + (20 − 5) + (15 − 5) = 20 + 15 + 10 = 45

ix) At least one of the three subjects:

= n(C ∪ M ∪ P) = 95

x) None of the three subjects:

= Total − At least one = 100 − 95 = 5