Consider the sets *A* = {7, 9, 5} and *B* = {4, 6, 8}

**What will we obtain, if we write all the elements of the sets ****A**** and ****B**** together in another set?**

We will obtain the set {4, 5, 6, 7, 8, 9}.

This set, which we have obtained by writing all the elements of the sets *A* and *B* together, is the **union **of the two given sets *A* and *B*.

The union of two sets is defined as follows:

**The union of two sets A and B is the set that consists of all the elements of A, all the elements of B, and the common elements taken only once.**

The symbol ‘∪’ is used for denoting the union. For example, if *X* = {2, 4, 6, 8, 10} and *Y *= {4, 8, 12}, then the union of *X* and *Y* is given by *X* ∪ *Y* = {2, 4, 6, 8, 10, 12}

There are some properties of union of two sets:

*A* ∪ *B* = *B* ∪ *A *(Commutative Law)
*A* ∪ Φ = *A * (Law of identity element Φ )
*A* ∪ *A* = *A * (Idempotent Law)
- (
*A* ∪ *B*) ∪ *C* = *A* ∪ (*B* ∪ *C*) (Associative Law)
*U* ∪ *A* = *U* (Law of universal set, *U*)

Now, consider the sets *A* = {4, 5, 9, 14} and *B* = {2, 4, 8, 10, 12, 14}

**Are there any elements, which are common to both the sets ****A**** and ****B****?**

We can observe that the elements 4 and 14 are common to both the sets *A* and *B*. The set, which consists of the common elements i.e., the set {4, 14}, is the **intersection** of the sets *A* and *B*.

The intersection of two sets is defined as follows:

**The intersection of sets A and B is the set of all elements that are common to both A and B.**

The symbol ‘∩’ is used for denoting the intersection. For example, if *X* = {A, E, I, O, U} and *Y* = {A, B, C, D, E}, then the intersection of the sets *X* and *Y* is given by *X* ∩ *Y* = {A, E}

The properties of the intersection of two sets are given as follows:

*A* ∩ *B* = *B* ∩ *A *(Commutative Law)
- Φ ∩
*A* = Φ (Law of identity element Φ )
*A* ∩ *A* = *A *(Idempotent Law)
- (
*A* ∩ *B*) ∩ *C* = *A* ∩ (*B* ∩ *C*) (Associative law)
*U* ∩ *A* = *A* (Law of *U*)
*A* ∩ (*B* ∪ *C*) = (*A* ∩ *B*) ∪ (*A* ∩ *C*) (Distributive law)

**Have you ever observe some properties involving both union and intersection.**

**Some **properties are given below.

**Distributive law of union over intersection of sets:**

*A* ∪ (*B* ∩ *C*) = (*A* ∪ *B*) ∩ (*A* ∪ *C*)

**Distributive law of intersection over union of sets**

*A* ∩ (*B* ∪ *C*) = (*A* ∩ *B*) ∪ (*A* ∩ *C*)

**These can be verified by taking any three sets.**

Now, let us again consider the above discussed sets *X* = {A, E, I, O, U} and *Y* = {A, B, C, D, E}. We found that the elements A and E are common to both these sets. Such type of sets, which have one or more elements in common, are called **overlapping sets**.

**Two sets are called overlapping (or joint) sets, if they have at least one element in common.**

Now consider the sets {1, 2, 3} and {4, 5}. **Is there any element common to these sets?**

We can observe that there is no element common to these sets. These types of sets are called by a special name, which is **disjoint sets**.

Disjoint sets are defined as follows:

**If two sets ****A**** and ****B**** are such that ****A ****∩ ****B**** = Φ i.e., they have no element in common, then ****A**** and ****B****are called disjoint sets.**

Now let us study some formulae used in set theory.

For any two finite sets *A* and *B*, *n* (*A* ∪ *B*) = *n* (*A*) + *n* (*B*) − *n* (*A* ∩ *B*)

If *A* and *B* are disjoint sets, i.e., *A* ∩ *B* = Φ then *n* (*A* ∪ *B*) = *n* (*A*) + *n* (*B*).

We know that if *A* and *B* are two sets, then there can be the following relationships between *A* and *B*:

*A* and *B* can be overlapping.
*A* and *B* can be disjoint.
- Either of the sets
*A* or *B* will be contained in the other set.

The Venn diagrams representing the intersection and union of the sets *A* and *B* in the above cases can be shown as follows:

- When the sets are overlapping, they can be shown by two intersecting closed figures.

When the sets* A* and *B* are overlapping, the Venn diagram representing *A* ∪ *B* can be shown as:

When the sets* A* and *B* are overlapping, the set *A* ∩ *B* is the shaded portion of the following the Venn diagram.

- When the sets are disjoint, they can be shown by two separate figures drawn side by side.

When the sets* A* and *B* are disjoint, the Venn diagrams representing *A* ∪ *B* can be shown as:

When the sets* A* and *B* are disjoint, the Venn diagrams representing *A* ∩ *B* can be shown as:

- When all the elements of one set are present in the second set, they can be represented by drawing one circle inside the other.

When set* B* is fully contained in set *A*, the Venn diagrams representing *A* ∪ *B* can be shown as:

When set* B* is fully contained in set *A*, the Venn diagrams representing *A* ∩ *B* can be shown as:

Now, if three sets *A*, *B* and *C* are given, then **how will we represent the union and the intersection of these three sets?**

Let’s see.

The union of the three sets *A*, *B* and *C*, i.e., *A* ∪ *B* ∪ *C*, is represented by the shaded portion of the following Venn diagram.

The intersection of the three sets *A*, *B* and *C*, i.e., *A* ∩ *B* ∩ *C* is represented by the shaded portion of the following Venn diagram.

**Example 1:**

**Three sets ****A****, ****B****, and ****C**** are defined as ****A**** = {3, 6, 8, 2, 11, 13, 12}, ****B**** = {7, 9, 3, 2, 10, 14, 15} and ****C****= {1, 2, 3, 6, 8, 10, 11}. Find ****A ****∩ (****B ****∪ ****C****). ****Also, prove the associative law of intersection and union using these sets. **

**Solution:**

We have to find *A* ∩ (*B* ∪ *C*). Hence, let us first find the union of *B* and *C*, and then its intersection with *A*.

*B* = {7, 9, 3, 2, 10, 14, 15}

*C* = {1, 2, 3, 6, 8, 10, 11}

*B* ∪ *C* = {1, 2, 3, 6, 7, 8, 9, 10, 11, 14, 15}

Let us represent the set *B* ∪ *C* as *D.*

*D* = {1, 2, 3, 6, 7, 8, 9, 10, 11, 14, 15}

Now, we have to find *A* ∩ *D.*

*A* = {3, 6, 8, 2, 11, 13, 12}

∴ *A* ∩ *D* = {3, 6, 8, 2, 11}

Hence, we have

*A* ∩ (*B* ∪ *C*) = {3, 6, 8, 2, 11}

Now, let us prove the associative laws:

We have,

*A* = {3, 6, 8, 2, 11, 13, 12}

*B* = {7, 9, 3, 2, 10, 14, 15}

*C* = {1, 2, 3, 6, 8, 10, 11}

*A* ∪ *B* = {2, 3, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15}

*A* ∩ *B* = {2, 3}

*B* ∪ *C* = {1, 2, 3, 6, 7, 8, 9, 10, 11, 14, 15}

*B* ∩ *C* = {2, 3, 10}

Now, (*A* ∩ *B*) ∩ *C *= {2, 3} ∩ {1, 2, 3, 6, 8, 10, 11} = {2, 3}

and *A* ∩ (*B* ∩ *C*) = {3, 6, 8, 2, 11, 13, 12} ∩ {2, 3, 10} = {2, 3}

Thus, **(****A**** ****∩**** ****B****) ∩**** ****C**** ****=**** ****A**** ****∩ (****B**** ****∩**** ****C****) **

Also, (*A* ∪ *B*) ∪ *C* = {2, 3, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15} ∪ {1, 2, 3, 6, 8, 10, 11}

={1, 2, 3, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15}

and *A* ∪ (*B* ∪ *C*)* =* {3, 6, 8, 2, 11, 13, 12} ∪ {1, 2, 3, 6, 7, 8, 9, 10, 11, 14, 15}

= {1, 2, 3, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15}

Thus,** ****(****A**** ****∪**** ****B****) ∪**** ****C**** ****= ****A**** ****∪ (****B**** ****∪**** ****C****)**

**Example 2:**

**Three sets ****A,**** ****B ****and**** C**** are defined as ****A**** = {1, 2, 3, 4, 5, 6, 7, 8}, ****B**** = {4, 8, 12, 16, 20, 24} and ***C* = {1, 4, 12, 15}. Their universal set is given as *U* = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 15, 16, 18, 20, 22, 24}.** **

**Verify the following results:**

**U**** ****∩**** ****A**** ****=**** ****A**** and ****U**** ∪ ****B**** ****=**** ****U**
**n****(****A ****∪ ****B****) = ****n****(****A****) + ****n****(****B****) − ****n****(****A ****∩ ****B****)**
**n****(****A ****∪ ****B ****∪ ***C*) =** n****(****A****) + ****n ****(****B****) + ****n****(****C****) − ****n****(****A ****∩ ****B****) − ****n****(****B ****∩ ****C****) − ****n****(****C ****∩ ****A****) + ****n****(****A ****∩ ****B ****∩ ****C****)**

**Solution:**

**1.**

*A* = {1, 2, 3, 4, 5, 6, 7, 8}, *B* = {4, 8, 12, 16, 20, 24} and

*U* = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 15, 16, 18, 20, 22, 24}

∴ *U *∩ *A* = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 15, 16, 18, 20, 22, 24} ∩ {1, 2, 3, 4, 5, 6, 7, 8}

= {1, 2, 3, 4, 5, 6, 7, 8}

= *A*

Similarly,

*U *∪ *B* = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 15, 16, 18, 20, 22, 24} ∪ {4, 8, 12, 16, 20, 24}

= {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 15, 16, 18, 20, 22, 24}

= *U*

**2.**

*A* = {1, 2, 3, 4, 5, 6, 7, 8} and *B* = {4, 8, 12, 16, 20, 24}

Therefore, *n*(*A*) = 8 and *n*(*B*) = 6

We have:

*A *∪ *B* = {1, 2, 3, 4, 5, 6, 7, 8, 12, 16, 20, 24}

*A *∩ *B* = {4, 8}

∴ *n*(*A* ∪ *B*) = 12 and *n*(*A*∩*B*) = 2

Now, *n*(*A*) + *n*(*B*) − *n*(*A*∩*B*) = 8 + 6 − 2 = 12 = *n*(*A *∪ *B*)

Thus, *n*(*A* ∪ *B*) = *n*(*A*) + *n*(*B*) − *n*(*A *∩ *B*)

**3. **

We have

*A* = {1, 2, 3, 4, 5, 6, 7, 8}, *B* = {4, 8, 12, 16, 20, 24} and *C* = {1, 4, 12, 15}

Therefore, *n*(*A*) = 8, *n*(*B*) = 6 and *n*(*C*) = 4

Also, *A *∩ *B* = {4, 8}, *B *∩ *C* = {4, 12}, *C *∩ *A *= {1, 4}, *A *∪ *B *∪ *C *= {1, 2, 3, 4, 5, 6, 7, 8, 12, 15 16, 20, 24} and *A *∩ *B *∩ *C* = {4}

Therefore, *n*(*A *∪ *B *∪ *C*) = 13, *n*(*A *∩ *B*) = 2, *n*(*B *∩ *C*) = 2, *n* (*C *∩ *A*) = 2 and *n* (*A *∩ *B *∩ *C*) = 1

Now,

*n*(*A*) + *n*(*B*) + *n*(*C*) − *n*(*A *∩ *B*) − *n*(*B *∩ *C*) − *n* (*C *∩ *A*) + *n* (*A *∩ *B *∩ *C*)

= 8 + 6 + 4 − 2 − 2 − 2 + 1

= 13

**= ***n*(*A *∪ *B *∪ *C*)

Thus, *n*(*A *∪ *B *∪ *C*) = *n*(*A*) + *n*(*B*) + *n*(*C*) − *n*(*A *∩ *B*) − *n*(*B *∩ *C*) − *n* (*C *∩ *A*) + *n* (*A *∩ *B *∩ *C*)

**Example 3:**

**Classify the following sets as disjoint or overlapping sets:**

**A**** = {****x**** : 17 < ****x****≤ 30} and ****B**** = {Prime numbers lying less than 20}**
**X**** = {****x**** : ****x**** is a letter in the word COMPUTER} and ****Y**** = {****y**** : ****y**** is the letter in the word BAG}**

**Solution:**

- We have
*A* = {18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30} and

*B* = {2, 3, 5, 7, 11, 13, 17, 19}

*A* ∩ *B* = {19}

Therefore, the sets *A* and *B* have one element in common. Hence, they are overlapping sets.

- We have
*X* = {C, O, M, P, U, T, E, R} and *Y* = {B, A, G}

The sets *X* and *Y* have no element in common. Hence, they are disjoint sets.

**Example 4:**

Let A = {1, 2, 4, 6, 8}, B = {1, 3, 6, 9, 12} and C = {1, 3, 5, 7, 9, 11}.

Verify the following.

(a) D**istributive law of union over intersection of sets:**

**(b) Distributive law of intersection over union of sets**

**Solution:**

**(a)**

B ∩ C = {1, 3, 6, 9, 12} ∩ {1, 3, 5, 7, 9, 11} = {1, 3, 9}

A ∪ (B ∩ C) = {1, 2, 4, 6, 8} ∪ {1, 3, 9} = {1, 2, 3, 4, 6, 8, 9}

A ∪ B = {1, 2, 4, 6, 8} ∪ {1, 3, 6, 9, 12} = {1, 2, 3, 4, 6, 8, 9, 12}

A ∪ C = {1, 2, 4, 6, 8}∪{1, 3, 5, 7, 9, 11} = {1, 2, 3, 4, 5, 6, 7, 8, 9, 11}

(A ∪ B) ∩ (A ∪ C) = {1, 2, 3, 4, 6, 8, 9, 12} ∩ {1, 2, 3, 4, 5, 6, 7, 8, 9, 11} = {1, 2, 3, 4, 6, 8, 9}

Thus, A ∪ (B ∩ C) = (A ∪B) ∩ (A ∪C)

Hence, d**istributive law of union over intersection of sets** is verified.

**(b)**

B ∪ C = {1, 3, 6, 9, 12} ∪ {1, 3, 5, 7, 9, 11} = {1, 3, 5, 6, 7, 9, 11, 12}

A ∩ (B ∪ C) = {1, 2, 4, 6, 8} ∩ {1, 3, 5, 6, 7, 9, 11, 12} = {1, 6}

A ∩ B = {1, 2, 4, 6, 8} ∩ {1, 3, 6, 9, 12} = {1, 6}

A ∩ C = {1, 2, 4, 6, 8} ∩ {1, 3, 5, 7, 9, 11} = {1}

∴ (A ∩ B) ∪(A ∩ C) = {1, 6} ∪{1} = {1, 6}

A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)

Hence, distributive **law of intersection over union of sets **is verified.

**Example 5:**

**If ****A**** and ****B**** are two disjoint sets such that ****n**** (****A****) = 17 and ****n**** (****A****∪****B****) = 25, then what is the cardinal number of the set ****B****?**

**Solution:**

Since *A* and *B* are disjoint sets,

*n* (*A* ∪ *B*) =* n* (*A*) + *n *(*B*)

⇒ 25 = 17 + *n *(*B*)

⇒ *n *(*B*) = 25 − 17 = 8

Therefore, the cardinal number of set *B* is 8.

**Example 6:**

**In a group of 60 people, 38 people prefer to take coffee and 29 people prefer to take tea. If a person can choose at least one between tea and coffee, then how many people prefer to take both tea and coffee?**

**Solution:**

Let the sets *A* and *B* denote

*A*= {people who prefer to take coffee}

*B* = {people who prefer to take tea}

Clearly, *A* ∪ *B* = {all people}

*A* ∩ *B* = {people who prefer both tea and coffee}

Now, we have

*n* (*A* ∪ *B*) = 60, *n* (*A*) = 38, *n *(*B*) = 29

We know that,

*n* (*A* ∪ *B*) =* n* (*A*) + *n *(*B*) − *n* (*A* ∩ *B*)

⇒ 60 = 38 + 29 − *n* (*A* ∩ *B*)

⇒ 60 = 67 − *n* (*A* ∩ *B*)

⇒ 60 − 67 = −*n* (*A* ∩ *B*)

⇒ −7 = −*n* (*A* ∩ *B*)

⇒ *n* (*A* ∩ *B*) = 7

Therefore, 7 people prefer to take both tea and coffee.

**Example 7:**

**In a survey, 150 people liked winter, 200 liked summer and 50 liked both summer and winter. Find the number of people who liked**

**Winter but not summer**
**Winter or summer**

**Solution:**

1) Let *A* denote the set of people who like winter and *B* denote the set of people who like summer.

*n* (*A*) = 150 *n* (*B*) = 200 *n *(*A* ∩ *B*) = 50

1) From the Venn diagram, we have

*A* = (*A* − *B*) ∪ (*A* ∩ *B*)

*n* (*A*) = *n *(*A* − *B*) + *n* (*A* ∩ *B*)

*n *(*A* − *B*) = *n* (*A*) − *n* (*A* ∩ *B*)

= 150 − 50

= 100

Thus, the number of people who like winter but not summer is 100.

2) The number of people who like winter or summer is simply the union of the two sets *A* and *B*.

*n* (*A* ∪ *B*) = *n* (*A*) + *n* (*B*) − *n* (*A* ∩ *B*)

** = **150 + 200 − 50

= 300

Thus, the number of people who like winter or summer is 300.

**Example 8:**

**The Department of Non-Conventional Energy Resources conducted a survey of 1110 families to know the number of families who use different types of fuels as the means for cooking food. In the survey, it was reported that 600 families use LPG, 400 families use coal and 300 families use wood. 20 families use all the three types of fuels, 100 families use both LPG and coal, and 80 families use both LPG and wood. Also, each family uses at least one of the three fuels. How many families use both coal and wood as a fuel for cooking food?**

**Solution:**

Let *A*, *B*, *C* denote the sets of families who use LPG, coal, and wood respectively as a fuel for cooking food. Accordingly, we have

*n* (*A* ∪ *B* ∪ *C*) = 1110, *n* (*A*) = 600, *n* (*B*) = 400, *n *(*C*) = 300, *n* (*A* ∩ *B* ∩ *C*) = 20,

*n* (*A* ∩ *B*) = 100, *n* (*A* ∩ *C*) = 80

We know that

*n* (*A* ∪ *B* ∪ *C*) = *n* (*A*) +* n *(*B*) + *n* (*C*) − *n* (*A* ∩ *B*) − *n* (*A* ∩ *C*) − *n* (*B* ∩ *C*) + *n* (*A* ∩ *B* ∩ *C*)

⇒ 1110 = 600 + 400 + 300 − 100 − 80 − *n* (*B *∩ *C*) + 20

⇒ 1110 = 1140 − *n* (*B* ∩ *C*)

⇒ *n* (*B* ∩ *C*) = 30

Thus, 30 families use both coal and wood as a fuel for cooking food.