Principle of inclusion and exclusion can be defined as- Let we have A and B are finite sets. Then A ∪B and A ∩B are finite and n(A ∪B) = n(A) + n(B) −n(A ∩B)

That is, we find the number of elements in A or B (or both) by first adding n(A) and n(B) (inclusion) and then Subtracting n(A ∩B) (exclusion) since its elements were counted twice

Here n(A) is the cardinality of set A, n(B) is the cardinality of set B and n(A⋂B) is the cardinality of (A⋂B).

For three sets-

n(A ∪B ∪C) = n(A) + n(B) + n(C) −n(A ∩B) −n(A ∩C) −n(B ∩C) + n(A ∩B ∩C)

## Example:

In college, 49 study Physics , 37 study English and 21 study Biology. If 9 n(P⋂E) = 9

n(E⋂B) = 5

n(P⋂B) = 4

n(P⋂E⋂B) = 3

therefore, by using the formula we get

n(P⋃E⋃B) = n(P) + n(E) + n(B) – n(P⋂E) – n(E⋂B) – n(P⋂B) + n(P⋂E⋂B)

= 49 + 37 + 21 – 9 – 5 – 4 + 3

= 92

Hence, the number of students in the college is 92.

of these students study Maths Physics and English, 5 study English and Biology, 4 study Physics and Biology and 3 study Physics, English and Biology, find the number of students in the college.

Solution:

Suppose P represents the number of students who study Physics, E represents the number of students who study English and B represent the number of students who study Biology.

As we know that-

Number of students in the group = n(P⋃E⋃B)

Here it is given that- n(P) = 49, n(E) = 37, n(B) = 21

**Example: Find the number of students at a college choosing at least one of the subject from mathematics, physics and statistics.**

**The data is given as-**

**65 study maths, 45 study physics, 42 study statistics**

**20 study maths and physics, 25 study maths and statistics, 15 study physics and statistics and 8 study all the three subjects.**

Sol.

Here we need to find , where

M maths, P = physics, S = statistics

Hence, by using Inclusion-Exclusion principle-

Which gives-

Therefore, we find that 100 students study at least one of the three subjects.

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