Principle of inclusion and exclusion for three sets by Sharvari

1. The Principle Of Inclusion
And Exclusion
(For Three Sets)
BY
SHARVARI NAVGIRE

2.  The Number of elements present in a set it is
known as cardinality of sets .
Let ,
A = {a, b, c, d, e}
Then ,
| A | = 5
 And it can be written as |A|
C AR D I N A L I T Y OF S E T S

3. Definition & Formula
|A∪B∪C| = |A| + |B| + |C| - |A∩B| - |A∩C| - |B∩C| + |A∩B∩C|
The Principle of Inclusion and Exclusion for three sets is given by this formula

4. EX.1
Let A = { a, b, c, d, e }
B = { a, b, e, g, h }
C = { b, d, e, g, h, k, m, n }
Solve by using principle of inclusion & exclusion for three sets .
| A ∪ B ∪ C | = | A | + | B | + | C | - | A ∩ B | - | A ∩ C | - | B ∩ C | + | A ∩ B ∩ C |
|A| = 5
|B| = 5
|C| = 8
A ∩ B = {a, b, e }
|A ∩ B| = 3
A ∩ C = {b, e }
|A ∩ C| = 2
B ∩ C = { b, e, g, h }
|B ∩ C| = 4
A ∩ B ∩ C = { b, }
|A ∩ B ∩ C |= 1
A ∪ B ∪ C = { a, b, c, d, e, g, h, k, m, n }
|A∪ B ∪ C|= 10
10 = 5 + 5 + 8 – 3 – 2 – 4 + 1
10 = 10

5. EX.2
If A = { 1, 2 , 3 }
B = { 2, 3 }
C = { 3, 4, 5 }
|A| = 3
|B| = 2
|C| = 3
A ∩ B = { 2, 3 }
|A ∩ B| = 2
A ∩ C = { 3 }
| A ∩ C| = 1
B ∩ C = { 3 }
|B ∩ C| = 1
A ∩ B ∩ C = { 3 }
|A ∩ B ∩ C | = 1
A ∪ B ∪ C = { 1, 2, 3, 4, 5 }
|A ∪ B ∪ C | = 5
|A ∪ B ∪ C | = |A| + |B| + |C| - |A ∩ B| - |A ∩ C| - |B ∩ C| + |A ∩ B ∩ C |
5 = 3 + 2 + 3 – 2 – 1 – 1 + 1
5 = 5

6. EX.3
Let A, B, & C be finite sets with
|A|=6
|B|=8
|C|= 6
then ,
|A ∪ B ∪ C| = 11
|A ∩ B|= 3
| A ∩ C| = 2
|B ∩ C| = 5
Find
|A ∩ B ∩ C| = ?
|A ∪ B ∪ C| = |A| + |B| +|C| - |A ∩ B| - |B ∩ C| - |A ∩ C| + |A ∩ B ∩ C|
11 = 6 + 8 + 6 – 3 – 5 – 2 +|A ∩ B ∩ C|
11 = 20 – 3 – 5 – 2 +|A ∩ B ∩ C|
11 = 20 – 10 + |A ∩ B ∩ C |
11 = 10 +|A ∩ B ∩ C|
|A ∩ B ∩ C| = 10 – 11 = -1
|A ∩ B ∩ C| = - 1

7. T H A N K Y O U