Maths Formulas

Sets Formula

The concept of set serves as a fundamental part of the general mathematics of day-to-day needs. Today this concept is being used in many branches of mathematics. Further sets are used to define the concepts of relations and functions in mathematics. Also, the study of geometry, sequences, probability, etc. requires the knowledge of sets. Set theory has its own various notations and symbols that may seem sometimes unusual for many. In this topic, we will discuss the Sets Formula with examples. It is a very interesting concept!

Concept of Set Theory

A collection of objects is known as a Set. The theory of sets was given by German mathematician Georg Cantor (1845-1918). He first encountered the sets while working with problems on the trigonometric series.

Sets Formula

In our life, we often speak of collections of objects of some kind, like, a pack of cards, a crowd of people, a cricket team, etc. In mathematics also, we come across the collections of natural numbers, points, prime numbers, etc. We can examine the following collections as:

  1. Collection of odd natural numbers less than 12, i.e., 1, 3, 5, 7, 9,11
  2. The rivers in India
  3. The vowels in the English alphabet, which are a, e, i, o, u
  4. Various kinds of the triangles
  5. The solution of the quadratic equation: \(x^2 – 5x + 6 = 0\), which are, 2 and 3.

As we can see that these are the well-defined collection of objects. Therefore, set theory is a branch of mathematics that deals with the study of sets of the collection of similar objects. It is one of the most fundamental branches of mathematics. But it is also very complex if you try to analyze the large collection of sets.

We will also see a few more examples of sets used particularly in mathematics. The symbols for the special sets given above will be referred to as the syllabuses of many courses.

Sets Formula

Following are some basic formulas from the set theory:

(A) For a group of two sets A, B

  1. If A and P are overlapping set, \(n( A\cup P ) = n(A) +n(P) – n (A\cap P) \)
  2. If A and B are disjoint set, \(n(A\cup B) = n(A) +n(B)\)
  3. \(n(U) = n(A) +n(B) – n(A\cap B) + n ((A\cup B)^c )\)
  4. \(n(A\cup B) = n(A-B) +n(B-A) + n (A\cap B)\)
  5. \(n(A-B) = n(A\cap B)- n(B)\)
  6. \(n(A-B)= n(A)- n(A\cap B)\)
  7. \(n(A^c) = n(U)- n(A)\)

(B) For a group of three sets P, Q, C

\(n(P \cup Q \cup C) = n(P)+n( Q )+n( C )-n( P\cap Q )-n(Q \cap C)-n(C\cap P)+n(P \cap Q\cap C)\)

Solved Examples for Sets Formula

Q.1: In a class, there are 120 students, 35 like drawing and 45 like music. 20 like both. Determine how many of them like either of them or neither of them?


  • Total number of students, n(U) = 120
  • Number of drawing students, n(D) = 35
  • Number of music students, n(M) = 45

The Number of students who like both, \(n(D\cap M) = 20\)

Number of students who like either of them,

\(n(D\cup M) = n(D) + n(M) – n(D\cap M)\)

\(n(D\cup M) = 4520 = 60\)

Number of students who like neither,

\(n (U) – n (D \cup M) = 120 – 60 = 60\)

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One response to “Equation Formula”

  1. KUCKOO B says:

    I get a different answer for first example.
    I got Q1 as 20.5
    median 23 and
    Q3 26

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