Formula Based on Cardinality of Sets

The number of elements in a set is called the cardinality of a set. For example, let A = {h, i, j, k, l}. Then the cardinality of set A is denoted by n(A). There are 5 elements in set A. So n(A) = 5. If two or more sets are combined using the operations on sets, using the formula based on the cardinality of sets, we can calculate the cardinality.

Formulas based on the cardinality of sets are given below.

1. Let A and B be finite sets and (A⋂B) ≠ φ, then

n(A⋃B) = n(A) + n(B) – n(A⋂B). ⋃ denotes union and ⋂ denotes intersection.

To find the cardinality of A⋃B, we subtract the number of common elements of A and B from the sum of the cardinality of A and B.

2. If the sets A and B are disjoint sets (A and B do not have any common elements), n(A⋂B) = 0

n(A⋃B) = n(A) + n(B). The Venn diagram given below gives a clear idea of this formula.

3 . Let A, B and C be finite sets and (A⋂B⋂C) ≠ φ, then

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

4. If A, B and C are disjoint sets, then

n(A⋂B) = 0, n(B⋂C) = 0, n(C⋂A) = 0, n(A⋂B⋂C) = 0

=> n(A⋃B⋃C) = n(A) + n(B) + n(C)

Also Read

Sets Relations and Functions

Solved Examples

Example 1:

In a college, there are 25 teachers who teach physics or mathematics. Of these, 15 teach physics, and 6 teach both physics and mathematics. How many teach mathematics?

a. 15

b. 16

c. 10

d. None of the above

Solution:

Let A denotes the number of teachers who teach physics, and B denotes the number of teachers who teach mathematics.

Total number of teachers who teach physics or maths, n(A⋃B) = 25

Number of physics teachers, n(A) = 15

Number of teachers who teach both mathematics and physics, n(A⋂B) = 6

n(A⋃B) = n(A) + n(B) – n(A⋂B)

=> 25 = 15 + n(B) – 6

=> n(B) = 16

Hence, option b is the answer.

Example 2:

In a group of 200 children, it was found that 120 children liked cricket, 90 liked tennis and 70 football, 40 liked cricket and tennis, 30 liked tennis and football, 50 liked football and cricket, and 20 liked none of these games. Find the number of children who like all three games.

a. 15

b. 16

c. 20

d. None of the above

Solution:

Let cricket = C, Tennis = T, Football = F

Given n(C) = 120

n(T) = 90

n(F) = 70

n(C⋂T) = 40

n(T⋂F) = 30

n(F⋂C) = 50

Given 20 children like none of the games.

n(C⋃T⋃F)’ = n(U) – n(C⋃T⋃F)

20 = 200 – n(C⋃T⋃F)

So n(C⋃T⋃F) = 180

n(C⋃T⋃F) = n(C) + n(T) + n(F) – n(C⋂T) – n(T⋂F) – n(F⋂C) + n(C⋂T⋂F)

180 = 120 + 90 + 70 – 40 – 30 – 50 + n(C⋂T⋂F)

=> n(C⋂T⋂F) = 180 – 160

= 20

So, the number of children who like all three games = 20

Hence, option c is the answer.

Related video