Finding Cardinality Of A Set

The number of elements in a set up is called the cardinality of a prepare. For example, permit A = {h, i, j, one thousand, l}. And then the cardinality of set A is denoted by due north(A). There are 5 elements in set up A. So due north(A) = v. If ii or more sets are combined using the operations on sets, using the formula based on cardinality of sets, we can calculate the cardinality.

Formulas based on cardinality of sets are given below.

i. Let A and B exist finite sets and (A⋂B) ≠ φ, then

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

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

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

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

Cardinality Of Disjoint Sets

iii . Let A, B and C exist finite sets and (A⋂B⋂C) ≠ φ, and then

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

Cardinality Of Finite Sets

four. If A, B and C are disjoint sets, so

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

=> due north(A⋃B⋃C) = n(A) + n(B) + n(C)

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Sets Relations and Functions

Solved Examples

Instance 1:

In a college at that place are 25 teachers who teach physics or mathematics. Of these, fifteen teach physics and 6 teach both physics and mathematics. How many teach mathematics?

a. 15

b. sixteen

c. 10

d. None of these

Solution:

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

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

Number of physics teachers, n(A) = fifteen

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

n(A⋃B) = n(A) + north(B) – n(A⋂B)

=> 25 = 15 + n(B) – 6

=> n(B) = 16

Hence option b is the answer.

Case 2:

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

a. fifteen

b. sixteen

c. 20

d. None of these

Solution:

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

Given due north(C) = 120

n(T) = xc

north(F) = 70

due north(C⋂T) = 40

due north(T⋂F) = 30

north(F⋂C) = fifty

Given 20 children like none of the games.

n(C⋃T⋃F)' = northward(U) – n(C⋃T⋃F)

20 = 200 – north(C⋃T⋃F)

So n(C⋃T⋃F) = 180

n(C⋃T⋃F) = n(C) + n(T) + n(F) – northward(C⋂T) – north(T⋂F) – due north(F⋂C) + n(C⋂T⋂F)

180 = 120 + 90 + seventy – 40 – thirty – 50 + n(C⋂T⋂F)

=> n(C⋂T⋂F) = 180 – 160

= twenty

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

Hence, choice c is the answer.

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Ofttimes Asked Questions

Define the Cardinality of sets.

The number of elements in a set is called the cardinality.

What is A⋂B, if A and B are disjoint sets?

If A and B are disjoint sets, then A⋂B = 0.

If A and B are finite sets and (A⋂B) ≠ φ, then what is the equation for n(A⋃B)?

If A and B are finite sets and (A⋂B) ≠ φ, due north(A⋃B) = northward(A) + n(B) – n(A⋂B).

If A, B and C are disjoint sets, and then what is north(A⋃B⋃C) ?

If A, B and C are disjoint sets, then northward(A⋃B⋃C) = n(A) + north(B) + n(C).