In a college, 60 students enrolled in chemistry, 40 in physics, 30 in biology, 15 in chemistry and physics, 10 in physics and biology, 5 in biology and chemistry. No one enrolled in all the three. Find how many are enrolled in at least one of the subjects.

Let’s consider that student enrolled in chemistry, physics and Biology be C, P, B respectively

We are given the number of students enrolled per subject so can write,

Number of students enrolled in chemistry n(C) = 60

Number of students enrolled in Physics n(P) = 40

Number of students enrolled in biology n(B) = 30

Number of students enrolled in chemistry and Physics n(C ⋂ P) = 15

Number of students enrolled in Physics and Biology n(P ⋂ B) = 10

Number of students enrolled in biology and chemistry n(B ⋂ C) = 5

Number of students enrolled in Physics Biology and

Chemistry n(P ⋂ B ⋂ C) = 0

Here to find the number of student enrolled in at least on subjects first we have to find the of sum student with all three subject, with two subjects only and with only one subject

To find the no of student enrolled to a single subject only we have to find the differential value of each subjects.

So,

Student enrolled in Physics only

n(P’) = n(P) – n(C ∩ P) – n(P ∩ B) – n(P ∩ B ∩ C)

= 40 – 15 – 10 – 0 = 15

Student enrolled in Chemistry only,

n(C’) = n(C) – n(C ∩ P) – n(C ∩ B) – n(P ∩ B ∩ C)

= 60 – 15 – 5 – 0

= 40

Student enrolled in Biology only,

n(B’) = n(B) – n(P ∩ B) – n(B ∩ C) – n(P ∩ B ∩ C)

= 30 – 10 – 5 – 0

= 15

So number of student enroll for at least one subject,

n(P U B U C) = n(P’) + n(B’) + n(C’) – n(P ∩ B) – n(B ∩ C) – n(P ∩ C) + n(P ∩ B ∩ C)

= 15 + 40 + 15 + 15 + 10 + 5 + 0

= 100