Here, you will learn operation of sets i.e intersection and union of two sets with examples.

Let’s begin –

## Basic Operation of Sets – Intersection and Union of Two Sets

**(i) Union of two sets :**

A \(\cup\) B = {x : x \(\in\) A or x \(\in\) B}

Example : A = {1, 2, 3} B = {2, 3, 4} then A \(\cup\) B ={1, 2, 3, 4}

**(ii) Intersection of two sets :**

A \(\cap\) B = {x : x \(\in\) A and x \(\in\) B}

Example : A = {1, 2, 3} B = {2, 3, 4} then A \(\cap\) B ={2, 3}

**(iii) Difference of two sets :**

A – B = {x : x \(\in\) A and x \(\notin\) B}

**(iv) Complement of a set :**

A’ = {x : x \(\notin\) A but x \(\in\) U} = U – A

**(v) De-Morgan Laws :**

(A \(\cup\) B)’ = A’ \(\cap\) B’ ; (A \(\cap\) B)’ = A’ \(\cup\) B’

**(vi) Distributive Laws :**

A \(\cup\) (B \(\cap\) C) = (A \(\cup\) B)\(\cap\) (A \(\cup\) C) ; A \(\cap\) (B \(\cup\) C) = (A \(\cap\) B)\(\cup\) (A \(\cap\) C)

**(vii) Commutative Laws :**

A \(\cup\) B = B \(\cup\) A ; A \(\cap\) B = B \(\cap\) A

**(viii) Associative Laws :**

(A \(\cup\) B) \(\cup\) C = A \(\cup\) (B \(\cup\) C) ; (A \(\cap\) B) \(\cap\) C = A \(\cap\) (B \(\cap\) C)

## Disjoint Sets

If A \(\cap\) B = \(\phi\), then A, B are disjoint.

e.g. if A = {1, 2, 3}, B = {7, 8, 9} then A \(\cap\) B = \(\phi\)

A \(\cap\) A’ = \(\phi\) ] \(\therefore\) A, A’ are disjoint.

## Formula of Sets

If A, B and C are finite sets, and U be the finite universal set, then

(i) n(A \(\cup\) B) = n(A) + n(B) – n(A \(\cap\) B)

(ii) n(A \(\cup\) B) = n(A) + n(B) \(\implies\) A, B are disjoint non-void sets.

(iii) n(A – B) = n(A) – n(A \(\cap\) B)

(iv) n(A \(\triangle\) B) = No. of elements which belong to exactly one of A or B

= n((A – B) \(\cup\) (B – A))

n(A \(\triangle\) B) = n(A – B) + n(B – A)

= n(A) – n(A \(\cap\) B) + n(B) – n(A \(\cap\) B)

n(A \(\triangle\) B) = n(A) + n(B) – 2n(A \(\cap\) B)

(v) \(n(A \cup B \cup C)\) = n(A) + n(B) – n(A \(\cap\) B) – n(B \(\cap\) C) – n(A \(\cap\) C) + \(n(A \cap B \cap C)\)

(vi) Number of elements in exactly two of the sets A, B, C

= n(A \(\cap\) B) + n(B \(\cap\) C) + n(C \(\cap\) A) – \(3n(A \cap B \cap C)\)

(vii) Number of elements in exactly one of the sets A, B, C

= n(A) + n(B) + n(C) – 2n(A \(\cap\) B) – 2n(B \(\cap\) C) – 2n(A \(\cap\) C) + \(3n(A \cap B \cap C)\)

(viii) n(A’ \(\cup\) B’) = n((A \(\cap\) B)’) = n(U) – n(A \(\cap\) B)

(ix) n(A’ \(\cap\) B’) = n((A \(\cup\) B)’) = n(U) – n(A \(\cup\) B)

Example : In a group of 1000 people, there are 750 who can speak Hindi and 400 who can speak Bengali. How many can speak Hindi only? and How many can speak Bengali only? How many can speak both Hindi and Bengali?

Solution : Let A and B be two sets of person who can speak Hindi and Bengali respectively.

then n(A \(\cup\) B) = 1000, n(A) = 750, n(B) = 400

Number of person who can speak both Hindi and Bengali

= n(A \(\cap\) B) = n(A) + n(B) – n(A \(\cup\) B)

= 750 + 400 – 1000 = 150

and Number of persons who can speak hindi only

= n(A – B) = n(A) – n(A \(\cap\) B) = 750 – 150 = 600

Number of persons who can speak Bengali only

= n(B – A) = n(B) – n(A \(\cap\) B) = 400 – 150 = 250

Hope you learn operation of sets i.e. intersection and union of sets and formula of sets.