JEE Mains Maths · Relations and Functions

Counting Relations and Their Elements

Counting the ordered pairs in a relation, the fewest pairs to add to make it reflexive, symmetric or an equivalence, and the number of relations of a given type on a finite set.

All Relations and Functions notes JEE Mains Maths strategy

Why this matters

Thirty-nine PYQs, the largest page in the chapter, and seventeen of them are numerical answers with no options to check against. The work is careful counting: go element by element, keep order in the pairs, and use the equivalence classes to count what must be added. Three ideas cover the page.

Concept 1 of 3: Counting the pairs in a relation

Take the first element one value at a time and count its partners, then add. When a relation on

A\times B

has a condition linking

a_1

with

b_2

and a separate condition linking

a_2

with

b_1

, count each condition on its own and multiply. When the condition is 'sum on the left = sum on the right', list how often each sum occurs on each side and multiply the matching frequencies.

Definition

$n(R)=\sum_a(\text{number of }b\text{ with }aRb)$ .
Independent conditions: $n(R)=N_1\times N_2$ .
Equal sums: $n(R)=\sum_s(\text{ways to make }s\text{ on the left})\times(\text{ways on the right})$ .
$(a,b)$ and $(b,a)$ are different pairs unless $a=b$ .

Pairs in a relation

n(R)=\sum_{a}\#\{b: aRb\}

Worked example

\{1,2,3,4\}

xRy

x+2y\le8

. Find

n(R)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 22 Jan 2026 Shift 2 · Q53Moderate

Example 1 · Relations and Functions · Counting Relations and Their Elements

The number of elements in the relation

\mathbf{R = \{(x,y)}

\left. \ 4x^{2}+y^{2}< 52,x,y \in Z \right\}

Count both orders

(1,2)

and

(2,1)

are two elements of a relation. A count done over unordered pairs must be doubled, except for pairs with equal entries.

Drill 15 more on counting the pairs in a relation

Concept 2 of 3: Fewest pairs to add: reflexive, symmetric, equivalence

For reflexive, add each missing

(a,a)

. For symmetric, add the reverse of each pair whose reverse is missing. For an equivalence, see which elements the given pairs link into groups; each group becomes a class, unlinked elements are classes of one, and the smallest equivalence has the sum of the squares of the class sizes. Subtract the pairs already there.

Definition

Reflexive: add the missing diagonal pairs.
Symmetric: add the missing reverses.
Equivalence: classes = linked groups; smallest relation has $\sum k_i^2$ pairs.
Added = (size of the smallest relation) − $n(R)$ .

Smallest equivalence containing R

\sum_i k_i^2\ \text{pairs, } k_i=\text{class sizes}

Worked example

R=\{(1,2),(3,4)\}

\{1,2,3,4,5\}

. Fewest pairs to add for an equivalence?

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 23 Jan 2025 · Q61Moderate

Example 2 · Relations and Functions · Counting Relations and Their Elements

Let

\mathbf{R = \{(1,2),(2,3),(3,3)\}}

be a relation defined on the set

\mathbf{\{ 1,2,3,4\}}

. Then the minimum number of elements, needed to be added in R so the R becomes an equivalence relation, is :

Symmetry then transitivity brings the diagonal

Once

(1,2)

and

(2,1)

are both present, transitivity forces

(1,1)

and

(2,2)

. A count for 'symmetric and transitive' must include them.

Drill 16 more on fewest pairs to add: reflexive, symmetric, equivalence

Concept 3 of 3: Counting relations of a given type

A relation on an

n

-element set is any subset of the

n^2

ordered pairs, so there are

2^{n^2}

. Reflexive fixes the

n

diagonal pairs:

2^{n^2-n}

. Symmetric decides each diagonal pair and each unordered off-diagonal pair once:

2^{n(n+1)/2}

. Reflexive and symmetric:

2^{n(n-1)/2}

. Equivalence relations match the ways to split the set into classes: 5 for 3 elements, 15 for 4.

Definition

All relations: $2^{n^2}$ . Reflexive: $2^{n^2-n}$ .
Symmetric: $2^{n(n+1)/2}$ . Reflexive and symmetric: $2^{n(n-1)/2}$ .
Equivalence relations = partitions: $n=3\to5$ , $n=4\to15$ .
Small cases with extra conditions: list them.

Symmetric relations

2^{n(n+1)/2}

Worked example

How many reflexive relations are there on

\{1,2,3,4\}

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 21 Jan 2026 Shift 1 · Q53Moderate

Example 3 · Relations and Functions · Counting Relations and Their Elements

The number of relations, defined on the set

\mathbf{\{ a,b,c,d\}}

, which are both reflexive and symmetric, is equal to:

Equivalences are not a power of 2

Equivalence relations correspond to partitions, not to free yes/no choices. Count the ways to split the set into classes.

Drill 5 more on counting relations of a given type

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Counting the pairs in a relation
Pairs in a relation
$n(R)=\sum_{a}\#\{b: aRb\}$
Fewest pairs to add: reflexive, symmetric, equivalence
Smallest equivalence containing R
$\sum_i k_i^2\ \text{pairs, } k_i=\text{class sizes}$
Counting relations of a given type
Symmetric relations
$2^{n(n+1)/2}$

Watch out for (3)

Count both orders
→ Counting the pairs in a relation
Symmetry then transitivity brings the diagonal
→ Fewest pairs to add: reflexive, symmetric, equivalence
Equivalences are not a power of 2
→ Counting relations of a given type

Test yourself on Relations and Functions

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