Set Theory Formulas

Last Updated : 6 Oct, 2025

In mathematics, a set is simply a collection of well-defined individual objects that form a group. A set can contain any group of items, such as a set of numbers, a day of the week, or a vehicle.

Example: A = { 2, 4, 6, 8 }.

A is a set and  2, 4, 6, and 8 are elements of the set. Elements written in a set can be used in any order, but cannot be repeated.

Set Formulas

A set formula is a formula related to set theory in mathematics. Set knowledge helps you apply set formulas in areas related to statistics, probability, geometry, and sequences.

Set Formulas on the Number of Elements of Sets

If n(A) and n(B) represent the number of elements in two finite sets A and B, respectively, then

  • n (A U B) = n(A) + n(B) - n(A ∩ B)
  • n(A) = n(A U B) + n(A ∩ B) - n(B)
  • n(B) = n(A U B) + n(A ∩ B) - n(A)

If sets A and B are disjoint sets then

  • n(A U B) = n(A) + n(B)
  • A ∩ B = ∅
  • n(A - B) = n(A)

Sets Formulas on Properties of Sets 

Set formulas have almost the same properties as real or natural numbers. Sets also have commutative, associative, and distributive properties. The set formula according to the properties of the set is as follows.

  • Commutativity

A⋂ B = B⋂ A 
A∪ B = B∪ A

  •   Associativity

A⋂ (B⋂ C) = (A⋂ B)⋂ C
A∪ (B∪ C) = (A∪ B)∪ C

  • Distributivity                

A ⋂ (B∪ C) = (A ⋂ B) ∪ (A⋂ C)

  • Idempotent Law

A ⋂ A = A
A ∪ A = A

  • Law of Ø and U

A⋂ Ø = Ø
U ⋂ A = A
A ∪ Ø = A
U ∪ A = U

Sets Formulas of Complement Sets

The set formulas of set complement include the basic law of complement, De Morgan's law, double complement, and the law of empty and universal sets.     

  1. Complement Law                                      A∪A' = U, A⋂A' = Ø and A' = U -  A
  2. De Morgan's Laws                                    (A ∪B)' = A' ⋂B' and (A⋂B)' = A' ∪ B'
  3. Law of Double complementation            (A')' = A
  4. Laws of Empty set and Universal Set       Ø' = U and U' = Ø

Sets Formulas of Difference of Sets

  1. A - A = Ø
  2. B - A = B⋂ A'
  3. B - A = B - (A⋂B)
  4. (A - B) = A if A⋂B =  Ø
  5. (A - B) ⋂ C = (A⋂ C) - (B⋂C)
  6. A ΔB = (A-B) U (B- A) 
  7. n(AUB) = n(A - B) + n(B - A) + n(A⋂B)
  8. n(A - B) = n(A) - n(A⋂B)
  9. n(A') = n(∪) - n(A)

Sample Questions - Set Theory Formulas

Question 1: What is a set in mathematics? Give examples.

Solution:

A set is a collection of individual elements enclosed in braces and separated by commas.  Example: Collecting vegetables, collecting notebooks. Alternatively, the set can be expressed as set A = { 1, 2, 3, 4 } where 1, 2, 3, and 4 are elements of set A.

Question 2: Why do we use sets in math? 

Solution:

The purpose of using sets is to represent a set of related objects in a group. In mathematics, we usually refer to groups of numbers, such as groups of natural numbers, sets of rational numbers, etc.

Question 3: What is the union of sets?

Solution:

The union of two sets A and B contains elements of both sets A and B . It is denoted by the symbol "U". For example, if  set A = { 2, 3 } and  B = { 5, 6 } then AUB = { 2, 3, 5, 6 }. 

Question 4: How can we represent the given set in set-builder form A = {1, 3, 5, 7, 9}.

Solution:

We can represent the given set in set-builder form as A = { x | x is an odd natural number less than 10 }.

Question 5: Given A = { 10, 12, 14, 16, 18} and B = {14, 16}. Find A U B and A ⋂ B and A – B.

Solution:

As, A = { 10, 12, 14, 16, 18 } and B = { 14, 16 }
A U B = { 10, 12, 14, 16, 18 }
A ⋂ B = { 14, 16 }
A – B = { 10, 12, 18 }  

Question 6: What is the formula for the intersection of sets?

Solution:

The set expression for the intersection of sets A and B is denoted by ⋂ and n(A⋂B) denotes the elements common to sets A and B. So,  formula for intersection of sets is given by  n(A⋂B) = n(A) + n(B) - n(A∪B) .

Question 7: What is the application of the set formulas?  

Solution:

Set formulas have broad application in many abstract concepts. 

For example, if R is the set of real numbers and Q is the set of rational numbers, then R - Q is the set of irrational numbers. 

Probability theory adopts set rules.  For example, a sample space is a universal set. If A and B are two mutually exclusive events, then P(A∪B) = P(A) + P(B) - P(A⋂B).

Question 8: There are 240 students in the class, 92 playing badminton, 60  table tennis, 80 rugby, 27 playing badminton and table tennis, 20 playing rugby and table tennis, 16 playing badminton and rugby, and 60 do not play any of the three games. Find  

  1. Number of students who play badminton, table tennis, and rugby.
  2. Number of students who play badminton and not rugby.
  3. Number of students who play badminton and rugby and not table tennis.

Solution:

Consider, that n(U) is the total number of students in class and n(B), n(T), and n(R) is the number of students who play badminton, table tennis, and rugby respectively.

Here, n(B∩T) is the number of students who play both badminton and tennis, n(R∩T) is the number of students who play both rugby and tennis and n(B∩R) is the number of students who play both badminton and rugby. 

We are given that: n(U) = 240 , n(B) = 92 , n(T) = 60 , n(R) = 80 ,  n(B∩T) = 27 , n(R∩T) = 20, n(B∩R) = 16 and n(B’∩T’∩R’) ( number of students which do not play any of the three games ) = 60   

Given  n(B’∩T’∩R’) = 60  ⇒ n(B∪T∪R)’ = 60

So,  n(B∪T∪R) = n(U) -n(B∪T∪R)’ = 240 - 60 = 180

Now , n(B∪T∪R) =  n(B) + n(T) + n(R) - n(B∩T) - n(R∩T) - n(B∩R) + n(B∩T∩R)

180 = 92 + 60 + 80 - 27 - 20 - 16 + n(B∩T∩R)

n(B∩T∩R) = 180 + 63 - 232 = 243 - 232 = 11

  1. Number of students who play badminton, table tennis and rugby = 11.
  2. Number of students who play badminton but not rugby, n(B-R)  =  n(B)-n(B∩R)=92-16=76
  3. Number of students who play badminton and rugby but not table tennis ,n(B∩R∩T’)= n(B∩R)-n(B∩R∩T) =16-11=5.

Question 9: In a survey of 800 students in a school, 250 students were found to be drinking mojito and 500 were drinking juice, 150 were drinking both mojito and juice. Find how many students were drinking neither mojito nor juice.

Solution:

Consider n(U) is the total number of students in school , n (M) is the number of students which are drinking mojito , n (J) is the number of students which are drinking juice , n(M∩J) are the number of students which are drinking both mojito and juice

Given  , n(U) = 800, n(M) = 250, n(J) = 500,n(M∩J) = 150

We have to find number of students which are drinking neither mojito nor juice , n(M’∩J’) =  n(M∪J)’

n(M∪J)’ = n(u)- n(M∪J)
n(M∪J) = n(M) + n(J) - n(M∩J) = 250 + 500 - 150 = 600
 n(M∪J)’ = 800 - 600 = 200.

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