Principles of Inclusion-Exclusion

 

Principle of Inclusion-Exclusion

The Principle of Inclusion-Exclusion is a counting method used to compute the cardinality of the union set. 

According to basic Inclusion-Exclusion Principles

For 2 finite sets A and B

     |A U B| = |A| + |B| - |A∩B|

For 3 finite sets A, B and C

     |A U B U C| = |A| + |B| + |C| - |A∩B| - |B∩C| - |A∩C| + |A∩B∩C|

where |A| denotes the cardinality (the number of elements in) of the set A

Properties

It computes the total number of elements that satisfy at least one of several properties.

It prevents the problem of double counting.

 

Examples

1. For three subsets A={2,3,7,9,10}, B={1,2,3,9}, C={2,4,9,10} of S={1,2,…..10}

Compute |A U B U C|.

Solution:

According to basic Inclusion-Exclusion Principles

For 3 finite sets A, B and C

|A U B U C| = |A| + |B| + |C| - |A∩B| - |B∩C| - |A∩C| + |A∩B∩C|

Term

Set

Cardinality

A

{2,3,7,9,10}

5

B

{1,2,3,9}

4

C

{2,4,9,10}

4

A∩B

{2,3,9}

3

B∩C

{2,9}

2

A∩C

{2,9,10}

3

A∩B∩C

{2,9}

2

|A U B U C|=5 + 4 + 4 – 3 – 2 – 3 + 2 = 7

Corresponding to the seven elements

A U B U C = {1,2,3,4,7,9,10}

 

2. Finite sets A, B and C with their corresponding values are given. 

|A| = 2, |B| = 2, |C| = 2,

|A∩B| = 3, |B∩C| = 3, |A∩C| = 3, |A∩B∩C| = 4

Compute |A U B U C|.

Solution:

|A U B U C| = |A| + |B| + |C| - |A∩B| - |B∩C| - |A∩C| + |A∩B∩C|

|A U B U C| = 2 + 2 + 2 – 3 – 3 – 3 + 4 = 1

 

3. How many integers from 1 to 100 are multiples of 2 or 3?

Solution:

Let A be the set of integers from 1 to 100 are multiples of 2, then |A|=50

Let B be the set of integers from 1 to 100 are multiples of 3, then |B|=33

Now, A∩B is the set of integers from 1 to 100 are multiples of both 2 and 3, and hence are multiples of 6, implying |A∩B|=16

Hence by Inclusion-Exclusion Principles

For 2 finite sets A and B

     |AUB|=|A|+|B|-|A∩B|=50+33-16=67

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