# Application -Sets

**Application of sets**

Laws of algebra of sets used in solve the problems related to sets

We can use the below formula to solve the problems based on two sets.

n(A ⋃ B) = n(A) + n(B) – n(A ⋂ B)

Venn Diagram of Three Sets

Check the Venn diagram of three sets given below.

The formula used to solve the problems on Venn diagrams with three sets is given below:

n(A ⋃ B ⋃ C) = n(A) + n(B) + n(C) – n(A ⋂ B) – n(B ⋂ C) – n(A ⋂ C) + n(A ⋂ B ⋂ C)

Example

From 50 students taking examinations in Mathematics, Physics and Chemistry, each of the student has passed in at least one of the subject, 37 passed Mathematics, 24 Physics and 43 Chemistry. At most 19 passed Mathematics and Physics, at most 29 Mathematics and Chemistry and at most 20 Physics and Chemistry. What is the largest possible number that could have passed all three examination?

Solution

Let M be the set of students passing in Mathematics

P be the set of students passing in Physics

C be the set of students passing in Chemistry

Now, n(M ∪ P ∪ C) = 50, n(M) = 37, n(P) = 24, n(C) = 43, n(M ∩ P) ≤ 19, n(M ∩ C) ≤ 29, n(P ∩ C) ≤ 20 (Given)

n(M ∪ P ∪ C) = n(M) + n(P) + n(C) – n(M ∩ P) – n(M ∩ C)– n(P ∩ C) + n(M ∩ P ∩ C) ≤ 50

⇒ 37 + 24 + 43 – 19 – 29 – 20 + n(M ∩ P ∩ C) ≤ 50

⇒ n(M ∩ P ∩ C) ≤ 50 – 36

⇒ n(M ∩ P ∩ C) ≤ 14

Thus, the largest possible number that could have passed all the three examinations is 14.