In mathematics, “sets, relations and functions” is one of the most important topics of set theory. Sets, relations and functions are three different words having different meaning mathematically but equally important for the preparation of JEE mains and other competitive exam.

It’s a fundamental way to group things of similar properties. Why is this important? Because being able to know how things are related to each other is more powerful (and easier) than just knowing that there are things that have properties. It’s also reflective of how your brain actually works naturally to understand the world. Set Theory is just an explicit formulation of something we all do unconsciously.

You’ll notice that set theoretic phrasing is absolutely everywhere (notice the heavy use of , , ∩, etc.). This is typical—absolutely all math courses and textbooks written above the level of introductory math courses like calculus or linear algebra will assume that the reader is familiar with set theory, and will use its basic symbols liberally.

Set Theory

Set Theory is a branch of mathematics that investigates sets and their properties. The basic concepts of set theory are fairly easy to understand and appear to be self-evident. However, despite its apparent simplicity, set theory turns out to be a very sophisticated subject. In particular, mathematicians have shown that virtually all mathematical concepts and results can be formalized within the theory of sets. This is considered to be one of the greatest achievements of modern mathematics. Given this achievement, one can claim that set theory provides a foundation for mathematics.

Set Notation

The curly brackets { } are sometimes called “set brackets” or “braces”.

The three dots … are called an ellipsis, and mean “continue on”.

But sometimes the “…” can be used in the middle to save writing long lists:

Example: the set of letters:

{a, b, c, …, x, y, z}

In this case it is a finite set (there are only 26 letters, right?)

Numerical Sets

So, what does this have to do with mathematics? When we define a set, all we have to specify is a common characteristic. Who says we can’t do so with numbers?

Set of even numbers: {…, −4, −2, 0, 2, 4, …}

Set of odd numbers: {…, −3, −1, 1, 3, …}

Set of prime numbers: {2, 3, 5, 7, 11, 13, 17, …}

Positive multiples of 3 that are less than 10: {3, 6, 9}

And so on. We can come up with all different types of sets.

We can also define a set by its properties, such as {x |x>0} which means “the set of all x’s, such that x is greater than 0”

Sometimes a set is identified by enclosing a list of its elements by curly brackets; for example, a set of natural numbers A can be identified by the notation

A = {1,2,3,4,5,6,7,8,9}.

More typically, one forms a set by enclosing a particular expression within curly brackets, where the expression identifies the elements of the set. To illustrate this method of identifying a set, we can form a set B of even natural numbers, using the above set A, as follows:

B = {n in A: n text {is even}}. 

which can be read as “the set of n in A such that n is even.” Of course,

{n in A : n is even} = {2,4,6,8}. 

It is difficult to identify the genesis of the set concept. Yet, the idea of a finite collection of objects has existed for as long as the concept of counting. Indeed, mathematicians have been investigating finite sets and methods for measuring the size of finite sets since the beginning of mathematics. For example, the above two sets

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

B= {2,4,6,8} 

are finite sets. As every element in B is an element in A, the set B is said to be a subset of A, denoted by B A. Since there are elements in A that are not in B, we say that B is a proper subset of A. Moreover, the number of elements in B is strictly smaller than the number of elements in A. Thus, one can say, “the whole A is greater in size than its proper part B.”

So for example, A is a set, and a is an element in A. Same with B and b, and C and c.

Also, when we say an element a is in a set A, we use the symbol  to show it.
And if something is not in a set use.

Example: Set A is {1,2,3}. We can see that 1 A, but  A

Equality

Two sets are equal if they have precisely the same members. Now, at first glance they may not seem equal, so we may have to examine them closely!

Example: Are A and B equal where:

  • A is the set whose members are the first four positive whole numbers.
  • B = {4, 2, 1, 3}

Let’s check. They both contain 1. They both contain 2. And 3, And 4. And we have checked every element of both sets, so: Yes, they are equal!

And the equals sign (=) is used to show equality, so we write:

A = B

Example: Are these sets equal?

  • A is {1, 2, 3}
  • B is {3, 1, 2}

Yes, they are equal!

They both contain exactly the members 1, 2 and 3.

It doesn’t matter where each member appears, so long as it is there.

It doesn’t matter where each member appears, so long as it is there.

Subsets

When we define a set, if we take pieces of that set, we can form what is called a subset.

Example: the set {1, 2, 3, 4, 5}

subset of this is {1, 2, 3}. Another subset is {3, 4} or even another is {1}, etc.

But {1, 6} is not a subset, since it has an element (6) which is not in the parent set.

In general:

A is a subset of B if and only if every element of A is in B.

So let’s use this definition in some examples.

Example: Is A a subset of B, where A = {1, 3, 4} and B = {1, 4, 3, 2}?

1 is in A, and 1 is in B as well. So far so good.

3 is in A and 3 is also in B.

4 is in A, and 4 is in B.

That’s all the elements of A, and every single one is in B, so we’re done.

 

Yes, A is a subset of B

Note that 2 is in B, but 2 is not in A. But remember, that doesn’t matter, we only look at the elements in A.

Let’s try a harder example.

Example: Let A be all multiples of 4 and B be all multiples of 2.
Is A a subset of B? And is B a subset of A?

Well, we can’t check every element in these sets, because they have an infinite number of elements. So we need to get an idea of what the elements look like in each, and then compare them.

The sets are:

  • A = {…, −8, −4, 0, 4, 8, …}
  • B = {…, −8, −6, −4, −2, 0, 2, 4, 6, 8, …}

By pairing off members of the two sets, we can see that every member of A is also a member of B, but not every member of B is a member of A:

So:

A is a subset of B, but B is not a subset of A

Proper Subsets

If we look at the definition of subsets and let our mind wander a bit, we come to a weird conclusion.

Let A be a set. Is every element of A in A?

Well, umm, yes of course, right?

So that means that A is a subset of A. It is a subset of itself!

This doesn’t seem very proper, does it? If we want our subsets to be proper, we introduce (what else but) proper subsets:

A is a proper subset of B if and only if every element of A is also in B, and there exists at least one element in B that is not in A.

This little piece at the end is there to make sure that A is not a proper subset of itself: we say that B must have at least one extra element.

Example:

{1, 2, 3} is a subset of {1, 2, 3}, but is not a proper subset of {1, 2, 3}.

Notice that when A is a proper subset of B then it is also a subset of B.

Even More Notation

When we say that A is a subset of B, we write A B.

Or we can say that A is not a subset of B by A  B (“A is not a subset of B”)

When we talk about proper subsets, we take out the line underneath and so it becomes A  B or if we want to say the opposite, A B.

Empty Set and Subsets

So, let’s go back to our definition of subsets. We have a set A. We won’t define it any more than that, it could be any set. Is the empty set a subset of A?

Going back to our definition of subsets, if every element in the empty set is also in A, then the empty set is a subset of A. But what if we have no elements?

It takes an introduction to logic to understand this, but this statement is one that is “vacuously” or “trivially” true.

A good way to think about it is: we can’t find any elements in the empty set that aren’t in A, so it must be that all elements in the empty set are in A.

So, the answer to the posed question is a resounding yes.

Order

No, not the order of the elements. In sets it does not matter what order the elements are in.

Example: {1,2,3,4} is the same set as {3,1,4,2}

When we say order in sets, we mean the size of the set.

Another (better) name for this is cardinality.

A finite set has finite order (or cardinality). An infinite set has infinite order (or cardinality).

For finite sets the order (or cardinality) is the number of elements.

Example: {10, 20, 30, 40} has an order of 4.

For infinite sets, all we can say is that the order is infinite. Oddly enough, we can say with sets that some infinities are larger than others, but this is a more advanced topic in sets.

The empty set is a subset of every set, including the empty set itself.

Summary

 is Union: is in either set or both sets

 is Intersection: only in both sets

 is Difference: in one set but not the other

Ac is the Complement of A: everything that is not in A

Empty Set: the set with no elements. Shown by {}

Universal Set: all things we are interested in

Venn diagram

A Venn diagram is also called a set diagram or a logic diagram showing different set operations such as the intersection of sets, union of sets and difference of sets. It is also used to depict subsets of a set.

Union of three sets in a Venn diagram

Sets and Venn Diagrams

Sets

 set is a collection of things.

For example, the items you wear is a set: these include hat, shirt, jacket, pants, and so on.

You write sets inside curly brackets like this:

{hat, shirt, jacket, pants, …}

You can also have sets of numbers:

 

Ten Best Friends

You could have a set made up of your ten best friends:

  • {alex, blair, casey, drew, erin, francis, glen, hunter, ira, jade}

Each friend is an “element” (or “member”) of the set. It is normal to use lowercase letters for them.

 

Now let’s say that alex, casey, drew and hunter play Soccer:

Soccer = {alex, casey, drew, hunter}

(It says the Set “Soccer” is made up of the elements alex, casey, drew and hunter.)

 

Now let’s say that alex, casey, drew and hunter play Soccer:

Soccer = {alex, casey, drew, hunter}

(It says the Set “Soccer” is made up of the elements alex, casey, drew and hunter.)

And casey, drew and jade play Tennis:

Tennis = {casey, drew, jade

 

We can put their names in two separate circles:

Union

You can now list your friends that play Soccer OR Tennis.

This is called a “Union” of sets and has the special symbol :

Soccer  Tennis = {alex, casey, drew, hunter, jade}

Not everyone is in that set … only your friends that play Soccer or Tennis (or both).

In other words, we combine the elements of the two sets.

We can show that in a “Venn Diagram”:                                       

Venn Diagram: Union of 2 Sets

A Venn Diagram is clever because it shows lots of information:

  • Do you see that alex, casey, drew and hunter are in the “Soccer” set?
  • And that casey, drew and jade are in the “Tennis” set?
  • And here is the clever thing: casey and drew are in BOTH sets!

All that in one small diagram.

Intersection

“Intersection” is when you must be in BOTH sets.

In our case that means they play both Soccer AND Tennis … which is casey and drew.

The special symbol for Intersection is an upside down “U” like this: 

And this is how we write it:

Soccer  Tennis = {casey, drew}

In a Venn Diagram:

Venn Diagram: Intersection of 2 Sets

Which Way Does That “U” Go?

Think of them as “cups”:  holds more water than ∩, right?

So, Union  is the one with more elements than Intersection ∩

Difference

You can also “subtract” one set from another.

For example, taking Soccer and subtracting Tennis means people that play Soccer but NOT Tennis … which is alex and hunter.

And this is how we write it:

Soccer  Tennis = {alex, hunter}

In a Venn Diagram:

Venn Diagram: Difference of 2 Sets

Summary So Far

  •  is Union: is in either set or both sets
  •  is Intersection: only in both sets
  •  is Difference: in one set but not the

Three Sets

You can also use Venn Diagrams for 3 sets.

Let us say the third set is “Volleyball”, which drew, glen and jade play:

Volleyball = {drew, glen, jade}

But let’s be more “mathematical” and use a Capital Letter for each set:

  • S means the set of Soccer players
  • T means the set of Tennis players
  • V means the set of Volleyball players

The Venn Diagram is now like this:

 

Union of 3 Sets: S  T  V

You can see (for example) that:

  • drew plays Soccer, Tennis and Volleyball
  • jade plays Tennis and Volleyball
  • alex and hunter play Soccer, but don’t play Tennis or Volleyball
  • no-one plays only Tennis

          We can now have some fun with Unions and Intersections …

S = {alex, casey, drew, hunter}

This is the Union of Sets T and V

 V = {casey, drew, jade, glen}

This is the Intersection of Sets S and V

 V = {drew}

And how about this …

  • take the previous set S  V
  • then subtract T:

This is the Intersection of Sets S and V minus Set T

(S ∩ V) − T = {}

Hey, there is nothing there!

That is OK, it is just the “Empty Set”. It is still a set, so we use the curly brackets with nothing inside: {}

The Empty Set has no elements: {}

Universal Set

The Universal Set is the set that has everything. Well, not exactly everything. Everything that we are interested in now.

Sadly, the symbol is the letter “U” … which is easy to confuse with the  for Union. You just have to be careful, OK?

In our case the Universal Set is our Ten Best Friends.

U = {alex, blair, casey, drew, erin, francis, glen, hunter, ira, jade}

We can show the Universal Set in a Venn Diagram by putting a box around the whole thing

Now you can see ALL your ten best friends, neatly sorted into what sport they play (or not!).

And then we can do interesting things like take the whole set and subtract the ones who play Soccer:

We write it this way:

U − S = {blair, erin, francis, glen, ira, jade}

Which says “The Universal Set minus the Soccer Set is the Set {blair, erin, francis, glen, ira, jade}”

In other words, “everyone who does not play Soccer”

Complement

And there is a special way of saying “everything that is not“, and it is called “complement”.

We show it by writing a little “C” like this: Sc

Which means “everything that is NOT in S”, like this:

Sc = {blair, erin, francis, glen, ira, jade}

(Exactly the same as the U − S example from above)

Summary

  •  is Union: is in either set or both sets
  •  is Intersection: only in both sets
  •  is Difference: in one set but not the other
  • Ac is the Complement of A: everything that is not in A
  • Empty Set: the set with no elements. Shown by {}
  • Universal Set: all things we are interested in

Cartesian product

In mathematics, specifically set theory, the Cartesian product of two sets A and B, denoted A × B, is the set of all ordered pairs (ab) where a is in A and b is in B. In terms of set-builder notation, that is

A X B = {(a, b) | a ∈ A and b ∈ B}

A table can be created by taking the Cartesian product of a set of rows and a set of columns. If the Cartesian product rows × columns is taken, the cells of the table contain ordered pairs of the form (row value, column value).

Cartesian product A x B of the sets A = {x, y, z) and B = {1, 2, 3}

In set theory and its application to logic, mathematics, and computer science, set builder notation is a mathematical notation for describing a set by enumerating its elements, or starting the properties that its members must satisfy.

In mathematics, an ordered pair is a pair of objects. The order in which the objects appear in the pair is significant: the ordered pair (ab) is different from the ordered pair (ba) unless a = b. (In contrast, the unordered pair {ab} equals the unordered pair {ba}.).

In the ordered pair (ab), the object a is called the first entry, and the object b the second entry of the pair. Alternatively, the objects are called the first and second components, the first and second coordinates, or the left and right projections of the ordered pair.

Cartesian products and binary relations (and hence functions) are defined in terms of ordered pairs

De Morgan’s laws

In propositional logic and Boolean algebra, De Morgan’s laws are a pair of transformation rules that are both valid rules of inference. They are named after Augustus De Morgan, a 19th-century British mathematician. The rules allow the expression of conjunctions and disjunctions purely in terms of each other via negation 

De Morgan’s laws represented with Venn diagrams. In each case, the resultant set is the set of all points in any shade of blue.

The rules can be expressed in English as:

  • The complement of the union of two sets is the same as the intersection of their complements
  • The complement of the intersection of two sets is the same as the union of their complements

 

not (A or B) = (not A) and (not B)

not (A and B) = (not A) or (not B),

where “A or B” is an “inclusive or” meaning at least one of A or B rather than an “exclusive or” that means exactly one of A or B.

In set theory and Boolean algebra, these are written formally as

Where, A and B are sets,

  •  Ac is the complement of A,
  •  is the intersection, and
  •  is the union.

An interval is a set that consists of all real numbers between a given pair of numbers. It can also be thought of as a segment of the real number line. An endpoint of an interval is either of the two points that mark the end of the line segment. An interval comprises the numbers lying between two specific given numbers

Interval Notation

Interval notation is a method to represent an interval on a number line. In other words, it is a way of writing subsets of the real number line.

An interval comprises the numbers lying between two specific given numbers. For example, the set of numbers x satisfying 0 ≤ x ≤ 5 is an interval that contains 0, 5, and all numbers between 0 and 5.

Types of intervals.

The union of two sets is a new set that contains all of the elements that are in at least one of the two sets. The union is written as A U B or “A or B”.

The intersection of two sets is a new set that contains all of the elements that are in both sets. The intersection is written as A∩B or “A and B”.

Property of Set

P-1. Commutative Property of Set

The commutative property states that the numbers on which we operate can be moved or swapped from their position without making any difference to the answer. The property holds for Addition and Multiplication, but not for subtraction and division.

A ∩ B = B ∩ A

For any two sets, the following statements are true

  1. Set union is commutative

 A U B = B U A

  1. Set intersection is commutative

A ∩ B = B ∩ A

Example 1:

A = {-10, 0, 1, 9, 2, 4, 5} and B = {- 1, – 2, 5, 6, 2, 3, 4}, for the sets A and B, verify that

(i) Set union is commutative. Also verify it by using Venn diagram.

(ii) Set intersection is commutative. Also verify it by using Venn diagram.

Solution:

Commutative property:

(i) A U B = B U A

Verifying in Venn diagram:

(ii) A ∩ B = B ∩ A

Verifying in Venn diagram:

(i) A = {-10, 0, 1, 9, 2, 4, 5} and B = {- 1, -2, 5, 6, 2, 3, 4}

A u B = {-10, 0, 1, 9, 2, 4, 5} U {-1,-2, 5, 6, 2, 3, 4} 

= {-10, -2, -1, 0, 1, 2, 3, 4, 5, 6, 9} —-(1)

B U A = {-1, -2, 5, 6, 2, 3, 4} U {-10, 0, 1, 9, 2, 4, 5}

=  {-10, -2, -1, 0, 1, 2, 3, 4, 5, 6, 9} —-(2)

(1) = (2)

 

(ii) A = {-10, 0, 1, 9, 2, 4, 5} and B = {-1,-2, 5, 6, 2, 3, 4}

A ∩B = {-10, 0, 1, 9, 2, 4, 5} ∩ {-1,-2, 5, 6, 2, 3, 4}

= { 2, 4, 5 } —-(3)

 

B n A = {-1,-2, 5, 6, 2, 3, 4} ∩ {-10, 0, 1, 9, 2, 4, 5}

= { 2, 4, 5 } —-(4)

(3) = (4)

 

Example 2 :

A = {1, 2, 3, 4, 5} and B = {3, 4, 5, 6}, for the sets A and B, verify that

(i) set union is commutative. 

(ii) set intersection is commutative. 

Solution :

Commutative property :

(i) A U B = B U A

(ii) A ∩ B = B ∩ A

(i) A = {1, 2, 3, 4, 5} and B = {3, 4, 5, 6} 

A U B = {1, 2, 3, 4, 5} U {3, 4, 5, 6} 

= {1, 2, 3, 4, 5, 6} —-(5)

B U A = {3, 4, 5, 6} U {1, 2, 3,  4, 5}

= {1, 2, 3, 4, 5, 6} —-(6)

(5) = (6)

(i) A = {1, 2, 3,  4, 5} and B = {3, 4, 5, 6} 

A ∩ B = {1, 2, 3, 4, 5} ∩ {3, 4, 5, 6} 

= {3, 4, 5} —-(7)

B ∩ A =  {3, 4, 5, 6} ∩ {1, 2, 3,  4, 5}

= {3, 4, 5} —-(8)

(7) = (8)

P-2. ASSOCIATIVE PROPERTY OF SETS

The associative property used in sets.

For any two two sets, the following statements are true.

(i)  Set union is associative

A U (B U C)  =  (A U B) U C

(i)  Set intersection is associative

A ∩ (B ∩ C)  =  (A ∩ B) ∩ C

Let us look at some example problems based on above properties.

Example 1 :

Given, A = {1, 2, 3, 4, 5}, B = {3, 4, 5, 6} and C = {5, 6, 7, 8}, show that

  1. A U (B U C) = (A U B) U C.
  2.  Verify (i) using Venn diagram.

Solution :

Now,

B U C  =  {3, 4, 5, 6} U {5, 6, 7, 8}  =  {3, 4, 5, 6, 7, 8}

A U (B U C)  =  {1, 2, 3, 4, 5} U { 3, 4, 5, 6, 7, 8}

=  {1, 2, 3, 4, 5, 6, 7, 8} —–(1)

A U B  =  {1, 2, 3, 4, 5} U {3, 4, 5, 6}  =  {1,2,3,4,5,6}

(A U B) U C  =  {1,2,3,4,5,6} U {5,6,7,8}

=  {1, 2, 3, 4, 5, 6, 7, 8}   —–  (2)

(1)  =  (2)

Example 2 :

Let A  =  {a, b, c, d}, B  =  {a, c, e} and C  =  {a, e}.

(i) Show that A ∩ (B ∩ C) = (A ∩ B) ∩ C. (ii) Verify (i) using Venn diagram

Solution :

B ∩ C  =  {a, c, e} ∩ {a, e}  =  {a, e}

A ∩ (B ∩ C)  =  {a, b, c ,d} ∩ {a, e}

=  { a }  —–(1)

A ∩ B  =  {a, b, c, d} ∩ {a, c, e}  =  {a, c}

(A ∩ B) ∩ C  =  {a, c} ∩ {a, e}

  =  {a}  —–(2)

(1)  =  (2)

 

Example 3 :

For A = {x ; x is a prime factor of 42}, B = {x | 5  x ≤ 12, x  N} and C = {1, 4, 5, 6}, verify A U (B U C) = (A U B) U C

Solution :

A = {2, 3, 7}, B = {5, 6, 7, 8, 9, 10, 11, 12}, C = {1, 4, 5, 6}

B U C  =  {5, 6, 7, 8, 9, 10, 11, 12} U {1, 4, 5, 6}

  =  {1, 2, 4, 6, 8, 9, 10, 11, 12}

A U (B U C)  =  {2, 3, 7} U {1, 2, 4, 6, 8, 9, 10, 11, 12}

 =  {1, 2, 3, 4, 6, 7, 8, 9, 10, 11, 12}  —-(1)

A U B  =  {2, 3, 7} U {5, 6, 7, 8, 9, 10, 11, 12}

=  {2, 3, 5, 6, 7, 8, 9, 10, 11, 12}

(A U B) U C  =  {2, 3, 5, 6, 7, 8, 9, 10, 11, 12} U {1, 4, 5, 6}

 =  {1, 2, 3, 4, 6, 7, 8, 9, 10, 11, 12}  —-(2)

(1)  =  (2)

(i)  Involution Law

(AC)C = A

Proof:

Let x be an arbitrary element of (AC)C

i.e., x (AC)C ⟺ x ∉ Ac 

                        ⟺ x ∈ A 

Let ‘U’ = {1, 2, 3, …. 10}

A = {1, 2, 3}

(AC) = {4, 5, 6 …. 10}

(AC)C = {1, 2, 3}

    ⇒ (AC)C = Aproved.

  • Complement Law

(A U Ac) =’U’

Proof:

Let x be an arbitrary element of (A U Ac)

i.e., x (A U Ac) x A or, x Ac 

x A or x ‘U’- A

x A or x ‘U’, x A

x ‘U’

(A U Ac) = ‘U’ Proved.

(iii)

A ∩ Ac = Ø

Proof:

Let x be an arbitrary element of (A ∩ Ac)

i.e., x (A ∩ Ac) x A and (x Ac)

x A and x A

x Ø

A ∩ Ac = Ø, Proved

The formula n (A U B) = n(A) + n(B) – n (A ∩ B) describes how to count elements in two sets that intersect.

let A = {a, b}, and B = {b, c}

n(A) = a + b

n(B) = b + c

n (A U B) = a + b + c

n (A ∩ B) = b

n (A U B) = n(A) + n(B) – n (A ∩ B) Proved

Union of three events (inclusion/exclusion formula): 

n (A ∪ B ∪ C) = n (A) + n (B) + n (C) – n (A ∩ B) – n (A ∩ C) – n (B ∩ C) + n (A ∩ B ∩ C).

Union and intersection of Three sets

Let S = {1,2,3,4,5,6,7,8,9,10} be the universal set. Let sets A, B and C be the subsets of S, where:

Set A = {1,3,8,10}

Set B = {1,4,8}

Set C = {2,4,6,7)

List the elements in the set A ∩ B ∩ C:

A ∩ B ∩ C = {} = Ø,

List the elements in the set (A ∪ B ∪ C):

(A ∪ B ∪ C) = {1,2,3,4,6,7,8,10}

 

Associative properties

A∪(B∪C) =(A∪B) ∪C

A∩(B∩C) =(A∩B) ∩C

Commutative Properties

A∪B=B∪A

A∩B=B∩A

Distributive Properties

A∩(B∪C) =(A∩B) ∪(A∩C)

A∪(B∩C) =(A∪B) ∩(A∪C)

What is a Disjoint Set

A pair of sets which does not have any common element are called disjoint sets. For example, set A= {2,3} and set B= {4,5} are disjoint sets. But set C= {3,4,5} and {3,6,7} are not disjoint as both the sets C and D are having 3 as a common element. Learn more about Disjoint Set here.

The Venn diagram of a disjoint set is given here:

Question 1: 

In a survey of 600 students in a school, 150 students wee found to be taking tea, and 225 taking coffee, 100 were taking both tea and coffee. Find how many students were taking neither tea nor coffee?

Solution:

Let T be the set of students who like tea and C be the set of students who like coffee.

Here n(T) = 150, n(C)= 225, and n(C∩T) =100

We know that

n (CT) =n (C)+n (T)− n(C∩T) 

              = 150+225−100

              =275

Number of students taking either tea or coffee = 275

Number of students taking neither tea nor coffee 

 = 600 − 275

= 325

Hence, number of students taking neither tea or coffee = 325

Alternatively,

Let U be the set of all students who took part in the survey.

Let T be the set of students taking tea.

Let C be the set of students taking coffee.

n(U)=600, n(T)=150, n(C)=225, n(T∩C) =100

To find: Number of students taking neither tea nor coffee i.e., we have to find N (T′ ∩ C′)

Students were taking neither tea nor coffee is a complement of n (T C) = n (T C)’

n (T C )’ = n (U ) – [ n( T – C ) + n (C – T) + n ( T ∩ C)]

We know, n (T′ ∩ C′) =n (T C)′ 

=n(U)−n (TC)

=n(U) − [n(T) + n(C) − n(T∩C)]

=600− [150+225−100]

=600−275

=325

Hence, 325 students were taking neither tea nor coffee.

Question 2:

 In a group of students 100 students know Hindi, 50 know English and 25 know both. Each of the students knows either Hindi or English. How many students are there in the group?

Solution:

To understand it in better way we have to draw venn diagram,

Let E be the set of students who know English, and

H be the set of students who know Hindi.

Number of students who know Hindi = n(H) = 100

Number of students who know English = n(E) = 50

Number of students who know both Hindi or English = n (H ∩ E)

= 25

Since, each of the students know Hindi or English

Number of students in group

= Number of students who know Hindi or English

= n (H ∪ E)

We know that n (H ∪ E) = n(H) + n(E)- n(H ∩ E)

  = 100 + 50 – 25

  = 125

Hence, there are 125 students in the class.

Note: While solving this type of question you need to know that the union of two sets is a new set that contains all of the elements that are in at least one of the two sets. The union is written as A∪B or “A or B” and the intersection of two sets is a new set that contains all of the elements that are in both sets. The intersection is written as A∩B or “A and B”.

Question 3: 

In a survey of 60 people, it was found that 25 people read newspaper ‘H’, 26 read newspaper T, 26 newspaper I, 9 read both H and I, 11 read both H and T, 8 read both T and I, 3 read all three newspapers. Find:

  1. The number of people who read at least one of the 

newspapers,

  1. The number of people who read exactly one newspaper.

 

(i). Let A be the set of people who read newspaper H.

Let B be the set of people who read newspaper T.

Let C be the set of people who read newspaper I.

Given n(A)=25, n(B)=26, and n(C)=26

n(A∩C) =9, n(A∩B) =11, and (B∩C) =8

n(A∩B∩C) =3

Let U be the set of people who took part in the survey.

 

(i) n(ABC) =n(A)+n(B)+n(C)−n(A∩B) −n(B∩C) −n(C∩A) +n(A∩B∩C)

=25+26+26−11−8−9+3

=52

Hence, 52 people read at least one of the newspapers. 

(ii) Let a be the number of people who read newspapers H and T only.

Let b denote the number of people who read newspapers I and H only.

Let c denote the number of people who read newspaper T and I only.

Let d denote the number of people who read all three newspapers.

Accordingly, d=n(A∩B∩C) =3

Now, n(A∩B) =a + d

n(B∩C) =c + d

n(C∩A) =b + d

a+d+c+d+b+d=11+8+9=28

a+b+c+d=28−2d=28−6=22

Hence, (52−22) =30 people read exactly one newspaper.

Question 4:

For any set A and B, show that n(A∩B) = n(A) ∩ n(B)

Proof:

Let x n(A∩B)

Then each element of x is an element of A and B, hence x is also in n(A) and 

n(B) x n(A)∩n(B)

Now

Let y n(A) ∩ n(B)

Then, y n(A) and y n(B)

Therefore, each element of y is an element of A and B. Hence each element of y is in A∩B

⇒y ∈n(A∩B)

X and y are arbitrary; hence we have shown that any set-in n(A∩B) is in n(A) ∩ n(B) and vice versa.

From this we can conclude that the two sets have identical composition and are thus equal.

 

Question 5:

Out of 500 car owners investigated 400 owned car A and 200 car owned B, 50 owned both car A & B. Is this data correct?

Solution:

Let ‘U’ = set of car owners.

⇒ n (‘U’) = 500 

Given, n(A∪B) = 500 ………(i)

We know, N(A∪B) = n(A) + n(B) – n(A∩B)  [Formula]

Given, n(A) = 400

            n(B) = 200

            n(A∩B) = 50

       ∴ n(A∪B) = n(A) + n(B) – n(A∩B)  

                     = 400 + 200 – 50

                           = 600 -50

                           = 550 ……………(ii)

But (i) ≠ (ii)

Hence the given data is not correct.

Question 6:

A college awarded 38 medals in football, 15 in basketball and 20 in cricket. If these medals went to a total of 58 men only and only 3 men got medals in all the three sports, how many received medals in exactly two of the three sports?

In the Venn diagram above, ‘a’ denotes the number of men

who got medals in Football and basketball only, ‘b’ denotes the number

of men who got medals in Football and Cricket only, ‘c’ denotes the number of men who got medals in Basketball and Cricket and ‘d’ denotes the number of men who got medals in all the three.

n(F) = 38,

n(B) = 15,

n(C) = 20,

Thus, n(FBC) = 58, and

n(F∩B∩C) = 3

Again, from Venn diagram we get,

n(F∩B) = a + d

n(F∩C) = b + d

n(B∩C) = d + c

n(F∩B∩C) = d

Given, n(F∩B) + n(B∩C) + n(F∩C) = 18

i.e., a + d + b + d + d + c = 18

⇒ a + b + c + 3d = 18

⇒ a + b + c = 18 – 3d

⇒ a + b + c = 18 – 3*3

a + b + c = 9

As per the Ven diagram the number of people who got medal in exactly two games = a + b + c

Hence, the number of people who got medal in exactly two games = 9

Question:

Let U = {1,2,3,4,5,6}, A = {2,3}, B = {3,4,5}

Find A’∩ B’, A∪B and hence show that A∪B = A’∩ B’

Ans. We know 

A’ = U – A

      = {1,4,5,6}

B’ = U – B

     = {1,2,6}

A∪B = {2,3,4,5}

A’∩B’ = {1,6}

Hence proved