Sets and Relations for JEE: Venn Counting, Types and Equivalence

A set is a well defined collection of objects, and a relation is a set of ordered pairs. These two ideas carry the whole language of class 11 and 12 mathematics: domains and ranges are sets, solution sets of inequalities are sets, and every function is a special relation. JEE Main asks this chapter in almost every shift, usually as a counting question: count the elements of a relation, count the pairs to add, count subsets or count equivalence relations.

This note covers sets, subsets and power sets, the operations and their laws, counting with Venn diagrams, the cartesian product, relations and their types, counting relations, the smallest number of pairs to add, and equivalence relations with their classes. Every worked example is a JEE Main 2025 or 2026 or JEE Advanced 2026 question unless it says otherwise, and every answer was re-derived from scratch.

Sets and their representation

A set is a well defined collection of distinct objects, called its elements. "Well defined" means that for any object we can say without argument whether it belongs. Write $x\in A$ when $x$ belongs to $A$ and $x\notin A$ when it does not.

There are two ways to write a set.

In roster form the order does not matter and repeats do not count: $\{1,2,2\}=\{2,1\}$.

The standard number sets lie one inside the next.

Five nested rounded rectangles labelled R, Q, Z, W and N from the outside in; the numbers 1, 2, 3 lie in the innermost box N, 0 lies in W outside N, minus 3 lies in Z outside W, two sevenths lies in Q outside Z, and root 2 and pi lie in R outside Q

Types of sets.

$n(A)$ denotes the number of elements of a finite set $A$. Equal sets are equivalent; equivalent sets need not be equal.

Example 1 (IITianForum). (a) Write in roster form: $A=\{x\in\mathbb{N}:x^2<20\}$ and $B=\{x\in\mathbb{Z}:\lvert 2x-1\rvert\le5\}$. (b) Write $C=\{1,4,9,16,25\}$ in set-builder form.

▸ SHOW SOLUTION

Step 1. Set A

$x^2<20$ with $x\in\mathbb{N}$: $x=1,2,3,4$, since $5^2=25$.

$$A=\{1,2,3,4\}$$

Step 2. Set B

$$-5\le2x-1\le5\ \Rightarrow\ -2\le x\le3$$$$B=\{-2,-1,0,1,2,3\}$$

Step 3. Set C

The elements are the squares of $1$ to $5$.

$$C=\{x^2:x\in\mathbb{N},\ x\le5\}$$

Answer: $A=\{1,2,3,4\}$, $B=\{-2,-1,0,1,2,3\}$, $C=\{x^2:x\in\mathbb{N},\ x\le5\}$

Example 2 (IITianForum). Classify each set as empty, singleton, finite or infinite.

$$P=\{x\in\mathbb{R}:x^2+1=0\}$$ $$Q=\{x\in\mathbb{N}:2<x<3\}$$ $$S=\{x\in\mathbb{Z}:x^2=4,\ x>0\}$$ $$T=\{\varnothing\},\qquad V=\{x\in\mathbb{Q}:0<x<1\}$$

▸ SHOW SOLUTION

Step 1. P and Q

$x^2=-1$ has no real root, so $P=\varnothing$. No natural number lies strictly between $2$ and $3$, so $Q=\varnothing$.

Step 2. S and T

$S=\{2\}$, a singleton. $T$ has one element, the empty set itself, so $T$ is a singleton and is not empty.

Step 3. V

$\frac12,\frac13,\frac14,\dots$ all lie in $V$, so $V$ is infinite.

Answer: $P,Q$ empty; $S,T$ singletons; $V$ infinite

Subsets and power set

$A$ is a subset of $B$, written $A\subseteq B$, when every element of $A$ is an element of $B$. If also $A\ne B$, then $A$ is a proper subset, $A\subset B$.

• $\varnothing\subseteq A$ and $A\subseteq A$ for every set $A$.
• $A=B$ exactly when $A\subseteq B$ and $B\subseteq A$.
• $\in$ joins an element to a set; $\subseteq$ joins a set to a set.

A rectangle labelled U containing a large ellipse labelled A, and inside it a smaller shaded ellipse labelled B, showing B as a subset of A

Power set

The power set $P(A)$ is the set of all subsets of $A$. Building a subset means deciding "in or out" once for each element, so

$$n(A)=m\ \Rightarrow\ n\big(P(A)\big)=2^m$$

Example 3 (IITianForum). Let $A=\{1,\{2\},\{1,2\}\}$. (a) Decide true or false: $1\in A$; $2\in A$; $\{2\}\in A$; $\{2\}\subseteq A$; $\{1\}\subseteq A$; $\{\{2\}\}\subseteq A$; $\{1,2\}\subseteq A$. (b) Find $n(P(A))$ and $n(P(P(A)))$.

▸ SHOW SOLUTION

Step 1. The elements

$A$ has three elements: $1$, $\{2\}$ and $\{1,2\}$. The number $2$ is not one of them.

Step 2. Membership

$1\in A$ true; $2\in A$ false; $\{2\}\in A$ true.

Step 3. Subsets

A set is a subset of $A$ when each of its elements is an element of $A$.

• $\{2\}\subseteq A$ needs $2\in A$: false.
• $\{1\}\subseteq A$: true. $\{\{2\}\}\subseteq A$: true.
• $\{1,2\}\subseteq A$ needs $2\in A$: false. (But $\{1,2\}\in A$ is true.)

Step 4. Power sets

$$n(P(A))=2^3=8$$$$n(P(P(A)))=2^8=256$$

Answer: (a) T, F, T, F, T, T, F; (b) $8$ and $256$

Example 4 (IITianForum). Finite sets $A$ and $B$ have $m$ and $n$ elements. The number of subsets of $A$ is $112$ more than the number of subsets of $B$. Find $m+n$.

▸ SHOW SOLUTION

Step 1. Equation

$$2^m-2^n=112$$$$2^n\left(2^{m-n}-1\right)=2^4\cdot7$$

Step 2. Match the factors

$2^{m-n}-1$ is odd, so it must be $7$ and $2^n=16$.

$$n=4,\qquad 2^{m-n}=8\ \Rightarrow\ m=7$$

Check: $128-16=112$.

Answer: $m+n=11$

Example 5 (JEE Main 2026, 22 January Shift 2). Let $S$ be the set of the first $11$ natural numbers. Then the number of elements in $A=\{B\subseteq S:n(B)\ge2\text{ and the product of all elements of }B\text{ is even}\}$ is ______.

▸ SHOW SOLUTION

Step 1. Subsets with at least two elements

$$2^{11}-1-11=2036$$

Step 2. Remove the odd products

The product is odd exactly when every element is odd. $S$ has $6$ odd numbers $1,3,5,7,9,11$; their subsets with at least two elements number

$$2^6-1-6=57$$

Step 3. Count

$$2036-57=1979$$

Answer: $1979$

Example 6 (JEE Main 2025, 7 April Shift 1). For $n\ge2$, let $S_n$ denote the set of all subsets of $\{1,2,\dots,n\}$ with no two consecutive numbers. For example $\{1,3,5\}\in S_6$, but $\{1,2,4\}\notin S_6$. Then $n(S_5)$ is equal to ______.

▸ SHOW SOLUTION

Step 1. Count by size

A subset of $\{1,\dots,5\}$ with $k$ elements and no two consecutive: choose $k$ numbers so that gaps of at least one remain. The count is ${}^{6-k}C_k$.

$$k=0:\ 1,\qquad k=1:\ 5$$$$k=2:\ {}^4C_2=6,\qquad k=3:\ {}^3C_3=1$$

For $k=3$ the only subset is $\{1,3,5\}$. No subset with $4$ elements exists.

Step 2. Add

$$1+5+6+1=13$$

The empty set is counted; it has no two consecutive numbers.

Answer: $13$

Example 7 (JEE Main 2026, 24 January Shift 2). Let $S$ be a set of $5$ elements and $P(S)$ denote the power set of $S$. Let $E$ be the event of choosing an ordered pair $(A,B)$ from the set $P(S)\times P(S)$ such that $A\cap B=\varnothing$. If the probability of the event $E$ is $\dfrac{3^p}{2^q}$, where $p,q\in\mathbb{N}$, then $p+q$ is equal to ______.

▸ SHOW SOLUTION

Step 1. Total pairs

$$n\big(P(S)\times P(S)\big)=2^5\cdot2^5=2^{10}$$

Step 2. Disjoint pairs

Each element of $S$ has three choices: in $A$ only, in $B$ only, or in neither. It cannot be in both. So there are $3^5$ disjoint pairs.

Step 3. Probability

$$P(E)=\frac{3^5}{2^{10}}\ \Rightarrow\ p=5,\ q=10$$

Answer: $p+q=15$

Example 8 (JEE Main 2025, 24 January Shift 1). Let $S=\{p_1,p_2,\dots,p_{10}\}$ be the set of first ten prime numbers. Let $A=S\cup P$, where $P$ is the set of all possible products of distinct elements of $S$. Then the number of all ordered pairs $(x,y)$, $x\in S$, $y\in A$, such that $x$ divides $y$, is ______.

▸ SHOW SOLUTION

Step 1. Elements of A as subsets

Every element of $A$ is the product of a non-empty subset of $S$, and different subsets give different products. So $A$ matches the $2^{10}-1$ non-empty subsets of $S$.

Step 2. Fix x

$x$ divides $y$ exactly when the subset of $y$ contains $x$. Subsets of $S$ containing a fixed $x$ number $2^9=512$.

Step 3. Count

$$10\times512=5120$$

Answer: $5120$

Operations on sets

All sets in a question are taken inside a universal set $U$. A Venn diagram draws $U$ as a rectangle and the sets as circles. Two sets cut $U$ into $4$ regions; three sets cut it into $8$ regions.

Two Venn diagrams side by side: on the left two overlapping circles A and B in a rectangle U with regions labelled p in A only, q in both, r in B only and s outside both; on the right three overlapping circles A, B and C with regions labelled a, b, c in one set only, d, e, f in exactly two, g in all three and h outside all

$A$ and $B$ are disjoint when $A\cap B=\varnothing$.

Three Venn diagrams of two overlapping circles A and B inside U; from left to right the shaded part is the union of A and B, then only the lens common to both, then the part of A outside B

Two Venn diagrams of two overlapping circles A and B inside U; on the left both circles are shaded except their common lens, showing the symmetric difference; on the right everything outside circle A is shaded, showing the complement of A

Three facts used again and again:

$$A-B=A\cap B'$$ $$A\,\triangle\,B=(A\cup B)-(A\cap B)$$ $$n(A-B)=n(A)-n(A\cap B)$$

Example 9 (IITianForum). Let $U=\{1,2,\dots,10\}$, $A=\{1,2,3,4,5\}$ and $B=\{2,4,6,8,10\}$. Find $A\cup B$, $A\cap B$, $A-B$, $B-A$, $A\,\triangle\,B$, $A'$ and $(A\cup B)'$.

▸ SHOW SOLUTION

Step 1. Union and intersection

$$A\cup B=\{1,2,3,4,5,6,8,10\}$$$$A\cap B=\{2,4\}$$

Step 2. Differences

$$A-B=\{1,3,5\},\quad B-A=\{6,8,10\}$$$$A\,\triangle\,B=\{1,3,5,6,8,10\}$$

Step 3. Complements

$$A'=\{6,7,8,9,10\},\quad (A\cup B)'=\{7,9\}$$

Answer: as above; note $n(A\,\triangle\,B)=n(A-B)+n(B-A)=6$

Example 10 (JEE Main 2026, 21 January Shift 2). Let $A=\{x:\lvert x^2-10\rvert\le6\}$ and $B=\{x:\lvert x-2\rvert>1\}$. Then

▸ SHOW SOLUTION

Step 1. Set A

$$-6\le x^2-10\le6\ \Rightarrow\ 4\le x^2\le16$$$$A=[-4,-2]\cup[2,4]$$

Step 2. Set B

$x-2>1$ or $x-2<-1$:

$$B=(-\infty,1)\cup(3,\infty)$$

Step 3. Test each option

• $A\cup B=(-\infty,1)\cup[2,\infty)$; option (1) is wrong at $x=1$ and $x=2$.
• $A-B=A\cap[1,3]=[2,3]$; option (2) misses $3$.
• $A\cap B=[-4,-2]\cup(3,4]$; option (3) wrongly includes $3$.
• $B-A$: remove $A$ from $B$. This leaves $(-\infty,-4)$, $(-2,1)$ and $(4,\infty)$. Option (4) is correct.

Answer: option (4)

Example 11 (JEE Main 2026, 2 April Shift 1). Let $A$ be the set of first $101$ terms of an A.P. whose first term is $1$ and common difference is $5$, and let $B$ be the set of first $71$ terms of an A.P. whose first term is $9$ and common difference is $7$. Then the number of elements in $A\cap B$ which are divisible by $3$ is

▸ SHOW SOLUTION

Step 1. The two sets

$$A=\{1,6,11,\dots,501\}$$$$B=\{9,16,23,\dots,499\}$$

Step 2. Common terms

The first common term is $16$. Common terms form an A.P. with difference $\mathrm{LCM}(5,7)=35$:

$$16+35k\le499\ \Rightarrow\ k\le13$$

So $A\cap B$ has $14$ elements, $k=0,1,\dots,13$.

Step 3. Divisible by 3

$$16+35k\equiv1+2k\pmod 3$$

This is $0$ when $k\equiv1\pmod3$: $k=1,4,7,10,13$.

The numbers are $51,156,261,366,471$.

Answer: $5$, option (2)

Example 12 (JEE Main 2025, 4 April Shift 1). Let $A=\{1,6,11,16,\dots\}$ and $B=\{9,16,23,30,\dots\}$ be the sets consisting of the first $2025$ terms of two arithmetic progressions. Then $n(A\cup B)$ is:

▸ SHOW SOLUTION

Step 1. Last terms

$$a_{2025}=1+5\cdot2024=10121$$$$b_{2025}=9+7\cdot2024=14177$$

Step 2. Common terms

Common terms are $16,51,86,\dots$ with difference $35$, and they cannot pass $10121$:

$$16+35k\le10121\ \Rightarrow\ k\le288$$

So $n(A\cap B)=289$.

Step 3. Union

$$n(A\cup B)=2025+2025-289=3761$$

Answer: option (3)

Example 13 (JEE Main 2025, 7 April Shift 2). Let $A=\{(\alpha,\beta)\in\mathbb{R}\times\mathbb{R}:\lvert\alpha-1\rvert\le4\text{ and }\lvert\beta-5\rvert\le6\}$ and $B=\{(\alpha,\beta)\in\mathbb{R}\times\mathbb{R}:16(\alpha-2)^2+9(\beta-6)^2\le144\}$. Then:

▸ SHOW SOLUTION

Step 1. Set A is a rectangle

$$-3\le\alpha\le5,\qquad -1\le\beta\le11$$

Step 2. Set B is an ellipse with its inside

$$\frac{(\alpha-2)^2}{9}+\frac{(\beta-6)^2}{16}\le1$$

Each fraction is at most $1$, so for every point of $B$,

$$-1\le\alpha\le5,\qquad 2\le\beta\le10$$

Step 3. Compare

Every point of $B$ satisfies the rectangle conditions, so $B\subseteq A$. The corner $(-3,-1)$ lies in $A$ but not in $B$, so $B\ne A$. Also $A\cup B=A$, and $A$ has $-3\le\alpha\le5$, so option (2) is false.

Answer: $B\subset A$, option (1)

A rectangle A with corners at x equals minus 3 and 5 and y from minus 1 to 11, drawn on axes, with a shaded ellipse B centred at the point 2 comma 6 lying wholly inside it and touching the right side

Algebra of sets

These laws hold for all sets inside $U$. Each can be checked on the Venn regions.

Three Venn diagrams of two circles A and B inside U; the first shades everything outside A, the second shades everything outside B, and the third shades only the part outside both circles, which is the common part of the first two shadings

The picture shows the first law of De Morgan: the region common to $A'$ and $B'$ is exactly the region outside $A\cup B$. The same laws hold with differences:

$$A-(B\cup C)=(A-B)\cap(A-C)$$ $$A-(B\cap C)=(A-B)\cup(A-C)$$

Example 14 (IITianForum). Simplify $(A-B)\cup(A\cap B)\cup(A'\cap B')$.

▸ SHOW SOLUTION

Step 1. First two pieces

$$(A\cap B')\cup(A\cap B)=A\cap(B'\cup B)=A$$

Step 2. Add the third piece

$$A\cup(A'\cap B')=(A\cup A')\cap(A\cup B')$$$$=U\cap(A\cup B')=A\cup B'$$

In regions: $p,q$ and $s$, everything except $r$.

Answer: $A\cup B'$, which equals $(B-A)'$

Example 15 (IITianForum). Simplify $\big[(A\cup B)'\cup(A'\cap B)\big]'$.

▸ SHOW SOLUTION

Step 1. De Morgan inside

$$(A\cup B)'\cup(A'\cap B)$$$$=(A'\cap B')\cup(A'\cap B)$$

Step 2. Take out A'

$$A'\cap(B'\cup B)=A'\cap U=A'$$

Step 3. Complement

$$(A')'=A$$

Answer: $A$

Example 16 (IITianForum). Let $A=\{1,2,3,5\}$ and $B=\{2,4,5,6\}$. Find the set $X$ with $A\,\triangle\,X=B$. Is it unique?

▸ SHOW SOLUTION

Step 1. Take the symmetric difference with A on both sides

$\triangle$ is associative, $A\,\triangle\,A=\varnothing$ and $\varnothing\,\triangle\,X=X$. So

$$A\,\triangle\,(A\,\triangle\,X)=A\,\triangle\,B\ \Rightarrow\ X=A\,\triangle\,B$$

Step 2. Compute

$$X=\{1,3\}\cup\{4,6\}=\{1,3,4,6\}$$

Check: $A\,\triangle\,X=\{2,5\}\cup\{4,6\}=B$.

Answer: $X=\{1,3,4,6\}$, and it is unique

Counting with sets

For finite sets,

$$n(A\cup B)=n(A)+n(B)-n(A\cap B)$$

For three sets write $S_1=n(A)+n(B)+n(C)$, $S_2=n(A\cap B)+n(B\cap C)+n(C\cap A)$ and $S_3=n(A\cap B\cap C)$. Then

$$n(A\cup B\cup C)=S_1-S_2+S_3$$

Example 17 (IITianForum). If $n(A)=25$, $n(B)=18$ and $n(A\cup B)=35$, find $n(A\cap B)$, $n(A-B)$ and $n(A\,\triangle\,B)$.

▸ SHOW SOLUTION

Step 1. Intersection

$$n(A\cap B)=25+18-35=8$$

Step 2. The rest

$$n(A-B)=25-8=17$$$$n(B-A)=18-8=10$$$$n(A\,\triangle\,B)=17+10=27$$

Answer: $8$, $17$, $27$

Example 18 (IITianForum). In a class of $100$ students, $45$ study Mathematics, $40$ Physics and $35$ Chemistry; $15$ study M and P, $12$ study P and C, $10$ study M and C, and $5$ study all three. Find the number who study none, exactly one, exactly two and at least two subjects.

▸ SHOW SOLUTION

Step 1. The sums

$$S_1=120,\quad S_2=37,\quad S_3=5$$

Step 2. At least one and none

$$n(M\cup P\cup C)=120-37+5=88$$

None: $100-88=12$.

Step 3. Exactly one, exactly two, at least two

$$\text{exactly two}=37-3\cdot5=22$$$$\text{exactly one}=120-74+15=61$$$$\text{at least two}=37-10=27$$

Check: $61+22+5=88$.

Answer: none $12$, exactly one $61$, exactly two $22$, at least two $27$

A three-set Venn diagram with circles M, P and C inside U; the regions carry 25 in M only, 18 in P only, 18 in C only, 10 in M and P only, 7 in P and C only, 5 in M and C only, 5 in all three, and 12 outside all circles

The picture fills the regions from the centre outwards: first $5$ in all three, then $15-5=10$, $12-5=7$, $10-5=5$ in exactly two, then $45-10-5-5=25$ and so on in one set only.

Bounds on the overlap

When only $n(A)$, $n(B)$ and $n(U)$ are known, the overlap is not fixed, but it is bounded:

$$n(A\cap B)\ge n(A)+n(B)-n(U)$$ $$n(A\cap B)\le\min\{n(A),n(B)\}$$

The overlap is also never below $0$. Both ends can occur.

Example 19 (IITianForum). In a group of $200$ people, $130$ read Hindi and $110$ read English. Find the least and the greatest number who read both, and the range of the number who read at least one.

▸ SHOW SOLUTION

Step 1. Least overlap

$n(H\cup E)\le200$, so

$$n(H\cap E)\ge130+110-200=40$$

Step 2. Greatest overlap

The overlap cannot exceed the smaller set: $n(H\cap E)\le110$, reached when every English reader also reads Hindi.

Step 3. At least one

$$n(H\cup E)=240-n(H\cap E)$$

As the overlap runs from $40$ to $110$, the union runs from $200$ down to $130$.

Answer: both: from $40$ to $110$; at least one: from $130$ to $200$

Example 20 (IITianForum). Of $100$ students, $70$ like tea, $75$ like coffee and $80$ like juice. Find the least possible number who like all three.

▸ SHOW SOLUTION

Step 1. Count those who miss something

$30$ do not like tea, $25$ do not like coffee and $20$ do not like juice. A student outside $T\cap C\cap J$ misses at least one drink, so

$$n\big((T\cap C\cap J)'\big)\le30+25+20=75$$

Step 2. Least value

$$n(T\cap C\cap J)\ge100-75=25$$

It is reached when the three groups of "misses" have no student in common.

Answer: $25$. In general the least triple overlap is $n(A)+n(B)+n(C)-2n(U)$ when this is positive.

Cartesian product

An ordered pair $(a,b)$ has a first and a second entry: $(a,b)=(c,d)$ exactly when $a=c$ and $b=d$. The cartesian product is

$$A\times B=\{(a,b):a\in A,\ b\in B\}$$

Example 21 (IITianForum). Let $A=\{1,2\}$ and $B=\{x,y,z\}$. Write $A\times B$, and find $n(A\times B)$ and $n(A\times B\times A)$. Is $A\times B=B\times A$?

▸ SHOW SOLUTION

Step 1. List

$A\times B$ consists of the six pairs $(1,x)$, $(1,y)$, $(1,z)$, $(2,x)$, $(2,y)$, $(2,z)$.

Step 2. Sizes

$$n(A\times B)=2\cdot3=6$$$$n(A\times B\times A)=2\cdot3\cdot2=12$$

Step 3. Order

$(1,x)\in A\times B$ but $(1,x)\notin B\times A$, since the first entry of a pair in $B\times A$ is a letter.

Answer: $6$, $12$; $A\times B\ne B\times A$

Example 22 (IITianForum). $A\times B$ has $6$ elements, three of which are $(-1,2)$, $(0,3)$ and $(1,2)$. Find $A$, $B$ and the remaining elements.

▸ SHOW SOLUTION

Step 1. What is known

First entries give $\{-1,0,1\}\subseteq A$; second entries give $\{2,3\}\subseteq B$.

Step 2. Sizes

$n(A)\,n(B)=6$ with $n(A)\ge3$ and $n(B)\ge2$ forces $n(A)=3$, $n(B)=2$. So

$$A=\{-1,0,1\},\qquad B=\{2,3\}$$

Step 3. The rest

$$(-1,3),\ (0,2),\ (1,3)$$

Answer: $A=\{-1,0,1\}$, $B=\{2,3\}$; remaining $(-1,3),(0,2),(1,3)$

Example 23 (IITianForum). Let $A=\{1,2,3\}$ and $B=\{2,3,4\}$. Find $(A\times B)\cap(B\times A)$.

▸ SHOW SOLUTION

Step 1. Use the two-product rule

$$(A\times B)\cap(B\times A)$$$$=(A\cap B)\times(B\cap A)$$

Step 2. Compute

$A\cap B=\{2,3\}$, so the answer is $\{2,3\}\times\{2,3\}$.

$$\{(2,2),(2,3),(3,2),(3,3)\}$$

Answer: $4$ elements, as listed

Two lattice diagrams; on the left six dots form a 2 by 3 grid with 1 and 2 on the horizontal axis and x, y, z on the vertical axis, showing A cross B for Example 21; on the right solid dots mark the points of A cross B for A equal to 1, 2, 3 and B equal to 2, 3, 4, hollow rings mark the points of B cross A, and a shaded square holds the four points 2 comma 2, 2 comma 3, 3 comma 2 and 3 comma 3 that carry both a dot and a ring

The left picture is Example 21. In the right picture, solid dots are $A\times B$ and rings are $B\times A$; the shaded square, where both appear, is $(A\cap B)\times(A\cap B)$ of Example 23.

Relations

A relation $R$ from $A$ to $B$ is any subset of $A\times B$. If $(a,b)\in R$ we write $a\,R\,b$. A relation from $A$ to $A$ is a relation on $A$.

• Domain of $R$: the set of first entries.
• Range of $R$: the set of second entries.
• Inverse: $R^{-1}=\{(b,a):(a,b)\in R\}$, a relation from $B$ to $A$.

Every subset of $A\times B$ is a relation, so for $n(A)=m$ and $n(B)=n$,

$$\text{relations from }A\text{ to }B=2^{mn}$$

On a set with $n$ elements there are $2^{n^2}$ relations, including the void relation $\varnothing$ and the universal relation $A\times A$.

Example 24 (IITianForum). On $A=\{1,2,3,4,5,6\}$ let $R=\{(x,y):y=2x-1\}$. Write $R$ in roster form, its domain, its range and $R^{-1}$.

▸ SHOW SOLUTION

Step 1. List the pairs

$x=1,2,3$ give $y=1,3,5$; $x=4$ gives $7\notin A$.

$$R=\{(1,1),(2,3),(3,5)\}$$

Step 2. Domain, range, inverse

$$\text{Dom}\,R=\{1,2,3\}$$$$\text{Range}\,R=\{1,3,5\}$$$$R^{-1}=\{(1,1),(3,2),(5,3)\}$$

Answer: as above; the range is a subset of the co-domain $A$

An arrow diagram with two ovals, each holding the numbers 1 to 6 of the set A; arrows go from 1 to 1, from 2 to 3 and from 3 to 5, while 4, 5 and 6 on the left send no arrow

Example 25 (IITianForum). Let $n(A)=3$ and $n(B)=2$. Find the number of relations (a) from $A$ to $B$; (b) on $A$; (c) from $A$ to $B$ with exactly two pairs; (d) non-empty relations on $A$.

▸ SHOW SOLUTION

Step 1. Using the subset count

$n(A\times B)=6$ and $n(A\times A)=9$.

Step 2. Each part

$$\text{(a) }2^6=64,\qquad \text{(b) }2^9=512$$$$\text{(c) }{}^6C_2=15,\qquad \text{(d) }2^9-1=511$$

Answer: $64$, $512$, $15$, $511$

Relations given by a rule

Most JEE questions define $R$ by a formula and ask for $n(R)$. Fix the first entry, count the second entries that work, and add.

Example 26 (JEE Main 2025, 29 January Shift 2). Let $S=\mathbb{N}\cup\{0\}$. Define a relation $R$ from $S$ to $\mathbb{R}$ by $R=\left\{(x,y):\log_e y=x\log_e\dfrac25,\ x\in S,\ y\in\mathbb{R}\right\}$. Then the sum of all the elements in the range of $R$ is equal to:

▸ SHOW SOLUTION

Step 1. Solve for y

$$y=\left(\frac25\right)^x,\quad x=0,1,2,\dots$$

Step 2. Range

$$\text{Range}=\left\{1,\ \frac25,\ \frac4{25},\ \dots\right\}$$

Step 3. Sum of the G.P.

$$1+\frac25+\frac4{25}+\cdots=\frac{1}{1-\frac25}=\frac53$$

Answer: option (2)

Example 27 (JEE Main 2026, 22 January Shift 2). The number of elements in the relation $R=\{(x,y):4x^2+y^2<52,\ x,y\in\mathbb{Z}\}$ is:

▸ SHOW SOLUTION

Step 1. Possible x

$4x^2<52$ gives $x^2<13$, so $x\in\{0,\pm1,\pm2,\pm3\}$.

Step 2. Count y for each x

$x$	condition	$y$ values
$0$	$y^2<52$	$\lvert y\rvert\le7$: $15$
$\pm1$	$y^2<48$	$\lvert y\rvert\le6$: $13$ each
$\pm2$	$y^2<36$	$\lvert y\rvert\le5$: $11$ each
$\pm3$	$y^2<16$	$\lvert y\rvert\le3$: $7$ each

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Step 3. Add

$$15+2(13+11+7)=77$$

Answer: option (1)

Example 28 (JEE Main 2026, 2 April Shift 2). Let $A=\{2,3,4,5,6\}$. Let $R$ be a relation on the set $A\times A$ given by $(x,y)\,R\,(z,w)$ if and only if $x$ divides $z$ and $y\le w$. Then the number of elements in $R$ is ______.

▸ SHOW SOLUTION

Step 1. Split the condition

The condition on $(x,z)$ and the condition on $(y,w)$ are independent, so

$$n(R)=(\text{pairs }x\mid z)\times(\text{pairs }y\le w)$$

Step 2. Pairs with x dividing z

$x=2$: $z=2,4,6$; $x=3$: $z=3,6$; $x=4,5,6$: $z=x$ only.

$$3+2+1+1+1=8$$

Step 3. Pairs with y at most w

$$5+4+3+2+1=15$$

Answer: $8\times15=120$

Example 29 (JEE Main 2026, 24 January Shift 1). Let $R$ be a relation defined on the set $\{1,2,3,4\}\times\{1,2,3,4\}$ by $R=\{((a,b),(c,d)):2a+3b=3c+4d\}$. Then the number of elements in $R$ is:

▸ SHOW SOLUTION

Step 1. Range of values

$2a+3b$ runs from $5$ to $20$. So only those $(c,d)$ with $3c+4d\le20$ can match.

Step 2. Table

For each such $(c,d)$, list the $(a,b)$ with $2a+3b=3c+4d$:

$(c,d)$	$3c+4d$	$(a,b)$
$(1,1)$	$7$	$(2,1)$
$(1,2)$	$11$	$(1,3),(4,1)$
$(1,3)$	$15$	$(3,3)$
$(2,1)$	$10$	$(2,2)$
$(2,2)$	$14$	$(1,4),(4,2)$
$(2,3)$	$18$	$(3,4)$
$(3,1)$	$13$	$(2,3)$
$(3,2)$	$17$	$(4,3)$
$(4,1)$	$16$	$(2,4)$
$(4,2)$	$20$	$(4,4)$

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The other six $(c,d)$ give values above $20$.

Step 3. Count

$$8\cdot1+2\cdot2=12$$

Answer: option (3)

Example 30 (JEE Main 2026, 5 April Shift 2). Let $A=\{1,4,7\}$ and $B=\{2,3,8\}$. Then the number of elements in the relation $R=\{((a_1,b_1),(a_2,b_2))\in(A\times B)\times(A\times B):a_1+b_2\text{ divides }a_2+b_1\}$ is ______.

▸ SHOW SOLUTION

Step 1. Re-pair the entries

Choosing $((a_1,b_1),(a_2,b_2))$ is the same as choosing the two pairs $(a_1,b_2)$ and $(a_2,b_1)$ freely from $A\times B$. So count ordered pairs $(s,t)$ of sums $a+b$ with $s\mid t$.

Step 2. The nine sums

$$1+\{2,3,8\}=\{3,4,9\}$$$$4+\{2,3,8\}=\{6,7,12\}$$$$7+\{2,3,8\}=\{9,10,15\}$$

The list is $3,4,6,7,9,9,10,12,15$, with $9$ twice.

Step 3. Count divisors in the list

$t$	sums $s$ dividing $t$	count
$3$	$3$	$1$
$4$	$4$	$1$
$6$	$3,6$	$2$
$7$	$7$	$1$
$9$ (twice)	$3,9,9$	$3+3$
$10$	$10$	$1$
$12$	$3,4,6,12$	$4$
$15$	$3,15$	$2$

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$$1+1+2+1+6+1+4+2=18$$

Answer: $18$

Example 31 (IITianForum). One relation is chosen at random from all non-empty relations on $A=\{1,2,3\}$. Find the probability that it is a function from $A$ to $A$, and the probability that it is a one-one function.

▸ SHOW SOLUTION

Step 1. Total

$$2^9-1=511$$

Step 2. Functions

A function gives each of $1,2,3$ exactly one image: $3^3=27$ relations. One-one functions: $3!=6$.

Answer: $\dfrac{27}{511}$ and $\dfrac{6}{511}$

Example 32 (JEE Main 2025, 2 April Shift 2). Let $A=\{1,2,3,\dots,100\}$ and $R$ be a relation on $A$ such that $R=\{(a,b):a=2b+1\}$. Let $(a_1,a_2),(a_2,a_3),(a_3,a_4),\dots,(a_k,a_{k+1})$ be a sequence of $k$ elements of $R$ such that the second entry of an ordered pair is equal to the first entry of the next ordered pair. Then the largest integer $k$, for which such a sequence exists, is equal to:

▸ SHOW SOLUTION

Step 1. Run the chain backwards

$a_i=2a_{i+1}+1$, so the numbers grow as we go back. To make the chain long, start the end at the smallest value $a_{k+1}=1$:

$$1\to3\to7\to15\to31\to63\to127$$

Step 2. Stay inside A

$127>100$, so the longest chain uses $63,31,15,7,3,1$: six numbers.

$$(63,31),(31,15),(15,7),(7,3),(3,1)$$

Answer: $k=5$, option (3)

Types of relations

Let $R$ be a relation on a set $A$.

On a grid of points $(x,y)$, a reflexive relation contains every point of the diagonal $y=x$, and a symmetric relation is its own mirror image in that diagonal.

Two 4 by 4 grids of points with the diagonal y equals x dashed; the left grid has all four diagonal points filled together with 1 comma 3, 2 comma 4 and 4 comma 1, a reflexive relation; the right grid has filled points 1 comma 2 and 2 comma 1, 1 comma 4 and 4 comma 1, 2 comma 3 and 3 comma 2, and 3 comma 3, each off-diagonal point matched by its mirror, a symmetric relation

Example 33 (IITianForum). On $A=\{1,2,3\}$ decide whether each relation is reflexive, symmetric or transitive.

$$R_1=\{(1,1),(2,2),(3,3),(1,2)\}$$ $$R_2=\{(1,2),(2,1)\}$$ $$R_3=\{(1,2),(2,3),(1,3)\},\qquad R_4=\varnothing$$

▸ SHOW SOLUTION

Step 1. R1

Reflexive: all three diagonal pairs. Not symmetric: $(2,1)$ is missing. Transitive: the only chain is through $(1,2)$ with diagonal pairs, which gives nothing new.

Step 2. R2

Not reflexive. Symmetric. Not transitive: $(1,2),(2,1)$ need $(1,1)$.

Step 3. R3

Not reflexive, not symmetric. Transitive: the only chain $(1,2),(2,3)$ needs $(1,3)$, which is present.

Step 4. R4

Not reflexive; symmetric and transitive, with nothing to test.

Answer: $R_1$: R, T; $R_2$: S; $R_3$: T; $R_4$: S, T

Example 34 (IITianForum). On $\mathbb{Z}$, decide which properties hold for $aRb$ defined by (a) $5$ divides $a-b$; (b) $a\le b$; (c) $\lvert a-b\rvert\le1$.

▸ SHOW SOLUTION

Step 1. Divisibility by 5

$5\mid0$; $5\mid(a-b)\Rightarrow5\mid(b-a)$; $5\mid(a-b)$ and $5\mid(b-c)$ give $5\mid(a-c)$. An equivalence relation.

Step 2. Less than or equal

Reflexive and transitive. Not symmetric: $1\le2$ but $2\not\le1$. Anti-symmetric: $a\le b$ and $b\le a$ give $a=b$.

Step 3. Distance at most 1

Reflexive and symmetric. Not transitive: $(0,1)$ and $(1,2)$ are in $R$, but $\lvert0-2\rvert=2$.

Answer: (a) equivalence; (b) R, T, anti-symmetric; (c) R, S only

Example 35 (IITianForum). On $\mathbb{R}$, $aRb$ holds whenever $a\le\lvert b\rvert$. Then $R$ is

▸ SHOW SOLUTION

Step 1. Reflexive

$a\le\lvert a\rvert$ for every real $a$. Reflexive.

Step 2. Symmetric

$1\le\lvert3\rvert$, so $(1,3)\in R$; but $3\le\lvert1\rvert$ is false. Not symmetric.

Step 3. Transitive

$(3,-3)\in R$ since $3\le3$, and $(-3,0)\in R$ since $-3\le0$. But $(3,0)\notin R$. Not transitive.

Answer: option (A)

Example 36 (IITianForum). On $\mathbb{R}$, $aRb$ holds whenever $a-b$ is irrational. Which of reflexive, symmetric, transitive hold?

▸ SHOW SOLUTION

Step 1. Reflexive

$a-a=0$ is rational. Not reflexive.

Step 2. Symmetric

If $a-b$ is irrational, so is $b-a=-(a-b)$. Symmetric.

Step 3. Transitive

$(\sqrt2,0)$ and $(0,\sqrt2)$ are in $R$, since $\pm\sqrt2$ is irrational. But $\sqrt2-\sqrt2=0$, so $(\sqrt2,\sqrt2)\notin R$. Not transitive.

Answer: symmetric only

Example 37 (JEE Main 2025, 29 January Shift 1). Define a relation $R$ on the interval $\left[0,\dfrac\pi2\right)$ by $x\,R\,y$ if and only if $\sec^2x-\tan^2y=1$. Then $R$ is:

▸ SHOW SOLUTION

Step 1. Rewrite

$$1+\tan^2x-\tan^2y=1$$$$\tan^2x=\tan^2y$$

Step 2. Use the interval

On $\left[0,\frac\pi2\right)$, $\tan$ is non-negative and increasing, so $\tan^2x=\tan^2y$ gives $x=y$. Thus $R$ is the identity relation on the interval, which is reflexive, symmetric and transitive.

Answer: option (1)

Example 38 (JEE Main 2025, 2 April Shift 1). Let $A$ be the set of all functions $f:\mathbb{Z}\to\mathbb{Z}$ and $R$ be a relation on $A$ such that $R=\{(f,g):f(0)=g(1)\text{ and }f(1)=g(0)\}$. Then $R$ is:

▸ SHOW SOLUTION

Step 1. Reflexive

$fRf$ needs $f(0)=f(1)$. For $f(x)=x$ this fails. Not reflexive.

Step 2. Symmetric

$f(0)=g(1)$ and $f(1)=g(0)$ say the same as $g(0)=f(1)$ and $g(1)=f(0)$. Symmetric.

Step 3. Transitive

Take $f(x)=x$ and $g(x)=1-x$. Then $f(0)=0=g(1)$ and $f(1)=1=g(0)$, so $fRg$, and by symmetry $gRf$. Transitivity would give $fRf$, which is false. Not transitive.

Answer: option (2)

Example 39 (JEE Main 2026, 8 April Shift 2). Consider the relation $R$ on the set $\{-2,-1,0,1,2\}$ defined by $(a,b)\in R$ if and only if $1+ab>0$. Among the statements: I. The number of elements in $R$ is $17$. II. $R$ is an equivalence relation.

▸ SHOW SOLUTION

Step 1. Pairs that fail

$1+ab\le0$ means $ab\le-1$: one entry negative, one positive.

$$(-2,1),(-2,2),(-1,1),(-1,2)$$

and their reverses: $8$ pairs.

$$n(R)=25-8=17$$

So I is true.

Step 2. Equivalence?

$1+a^2>0$, so $R$ is reflexive, and it is symmetric. But $(-2,0)$ and $(0,2)$ are in $R$ while $(-2,2)$ is not. Not transitive, so II is false.

Answer: option (1)

Example 40 (JEE Main 2025, 8 April Shift 2). Let $A=\{0,1,2,3,4,5\}$. Let $R$ be a relation on $A$ defined by $(x,y)\in R$ if and only if $\max\{x,y\}\in\{3,4\}$. Then among the statements $(S_1)$: The number of elements in $R$ is $18$, and $(S_2)$: The relation $R$ is symmetric but neither reflexive nor transitive:

▸ SHOW SOLUTION

Step 1. Count

Pairs with maximum exactly $3$: both entries in $\{0,\dots,3\}$ but not both in $\{0,1,2\}$.

$$4^2-3^2=7,\qquad 5^2-4^2=9$$

So $n(R)=16$ and $S_1$ is false.

Step 2. Properties

$\max\{x,y\}=\max\{y,x\}$, so $R$ is symmetric. $(0,0)\notin R$: not reflexive. $(0,3),(3,0)\in R$ but $(0,0)\notin R$: not transitive. So $S_2$ is true.

Answer: option (3) :::## Types of relations

Let $R$ be a relation on a set $A$.

Type	Condition
reflexive	$(a,a)\in R$ for every $a\in A$
symmetric	$(a,b)\in R\Rightarrow(b,a)\in R$
transitive	$(a,b),(b,c)\in R\Rightarrow(a,c)\in R$
anti-symmetric	$(a,b),(b,a)\in R\Rightarrow a=b$
identity	$R=\{(a,a):a\in A\}$
equivalence	reflexive, symmetric and transitive

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On a grid of points $(x,y)$, a reflexive relation contains every point of the diagonal $y=x$, and a symmetric relation is its own mirror image in that diagonal.

Two 4 by 4 grids of points with the diagonal y equals x dashed; the left grid has all four diagonal points filled together with 1 comma 3, 2 comma 4 and 4 comma 1, a reflexive relation; the right grid has filled points 1 comma 2 and 2 comma 1, 1 comma 4 and 4 comma 1, 2 comma 3 and 3 comma 2, and 3 comma 3, each off-diagonal point matched by its mirror, a symmetric relation

How to decide

To prove a property, argue for all elements. To disprove it, one counterexample is enough. Symmetric and transitive hold automatically for the void relation, because there is no pair to test; it is not reflexive on a non-empty set.

Example 33 (IITianForum). On $A=\{1,2,3\}$ decide whether each relation is reflexive, symmetric or transitive.

$$R_1=\{(1,1),(2,2),(3,3),(1,2)\}$$ $$R_2=\{(1,2),(2,1)\}$$ $$R_3=\{(1,2),(2,3),(1,3)\},\qquad R_4=\varnothing$$

▸ SHOW SOLUTION

Step 1. R1

Reflexive: all three diagonal pairs. Not symmetric: $(2,1)$ is missing. Transitive: the only chain is through $(1,2)$ with diagonal pairs, which gives nothing new.

Step 2. R2

Not reflexive. Symmetric. Not transitive: $(1,2),(2,1)$ need $(1,1)$.

Step 3. R3

Not reflexive, not symmetric. Transitive: the only chain $(1,2),(2,3)$ needs $(1,3)$, which is present.

Step 4. R4

Not reflexive; symmetric and transitive, with nothing to test.

Answer: $R_1$: R, T; $R_2$: S; $R_3$: T; $R_4$: S, T

Example 34 (IITianForum). On $\mathbb{Z}$, decide which properties hold for $aRb$ defined by (a) $5$ divides $a-b$; (b) $a\le b$; (c) $\lvert a-b\rvert\le1$.

▸ SHOW SOLUTION

Step 1. Divisibility by 5

$5\mid0$; $5\mid(a-b)\Rightarrow5\mid(b-a)$; $5\mid(a-b)$ and $5\mid(b-c)$ give $5\mid(a-c)$. An equivalence relation.

Step 2. Less than or equal

Reflexive and transitive. Not symmetric: $1\le2$ but $2\not\le1$. Anti-symmetric: $a\le b$ and $b\le a$ give $a=b$.

Step 3. Distance at most 1

Reflexive and symmetric. Not transitive: $(0,1)$ and $(1,2)$ are in $R$, but $\lvert0-2\rvert=2$.

Answer: (a) equivalence; (b) R, T, anti-symmetric; (c) R, S only

Example 35 (IITianForum). On $\mathbb{R}$, $aRb$ holds whenever $a\le\lvert b\rvert$. Then $R$ is

▸ SHOW SOLUTION

Step 1. Reflexive

$a\le\lvert a\rvert$ for every real $a$. Reflexive.

Step 2. Symmetric

$1\le\lvert3\rvert$, so $(1,3)\in R$; but $3\le\lvert1\rvert$ is false. Not symmetric.

Step 3. Transitive

$(3,-3)\in R$ since $3\le3$, and $(-3,0)\in R$ since $-3\le0$. But $(3,0)\notin R$. Not transitive.

Answer: option (A)

Example 36 (IITianForum). On $\mathbb{R}$, $aRb$ holds whenever $a-b$ is irrational. Which of reflexive, symmetric, transitive hold?

▸ SHOW SOLUTION

Step 1. Reflexive

$a-a=0$ is rational. Not reflexive.

Step 2. Symmetric

If $a-b$ is irrational, so is $b-a=-(a-b)$. Symmetric.

Step 3. Transitive

$(\sqrt2,0)$ and $(0,\sqrt2)$ are in $R$, since $\pm\sqrt2$ is irrational. But $\sqrt2-\sqrt2=0$, so $(\sqrt2,\sqrt2)\notin R$. Not transitive.

Answer: symmetric only

Example 37 (JEE Main 2025, 29 January Shift 1). Define a relation $R$ on the interval $\left[0,\dfrac\pi2\right)$ by $x\,R\,y$ if and only if $\sec^2x-\tan^2y=1$. Then $R$ is:

▸ SHOW SOLUTION

Step 1. Rewrite

$$1+\tan^2x-\tan^2y=1$$$$\tan^2x=\tan^2y$$

Step 2. Use the interval

On $\left[0,\frac\pi2\right)$, $\tan$ is non-negative and increasing, so $\tan^2x=\tan^2y$ gives $x=y$. Thus $R$ is the identity relation on the interval, which is reflexive, symmetric and transitive.

Answer: option (1)

Example 38 (JEE Main 2025, 2 April Shift 1). Let $A$ be the set of all functions $f:\mathbb{Z}\to\mathbb{Z}$ and $R$ be a relation on $A$ such that $R=\{(f,g):f(0)=g(1)\text{ and }f(1)=g(0)\}$. Then $R$ is:

▸ SHOW SOLUTION

Step 1. Reflexive

$fRf$ needs $f(0)=f(1)$. For $f(x)=x$ this fails. Not reflexive.

Step 2. Symmetric

$f(0)=g(1)$ and $f(1)=g(0)$ say the same as $g(0)=f(1)$ and $g(1)=f(0)$. Symmetric.

Step 3. Transitive

Take $f(x)=x$ and $g(x)=1-x$. Then $f(0)=0=g(1)$ and $f(1)=1=g(0)$, so $fRg$, and by symmetry $gRf$. Transitivity would give $fRf$, which is false. Not transitive.

Answer: option (2)

Example 39 (JEE Main 2026, 8 April Shift 2). Consider the relation $R$ on the set $\{-2,-1,0,1,2\}$ defined by $(a,b)\in R$ if and only if $1+ab>0$. Among the statements: I. The number of elements in $R$ is $17$. II. $R$ is an equivalence relation.

▸ SHOW SOLUTION

Step 1. Pairs that fail

$1+ab\le0$ means $ab\le-1$: one entry negative, one positive.

$$(-2,1),(-2,2),(-1,1),(-1,2)$$

and their reverses: $8$ pairs.

$$n(R)=25-8=17$$

So I is true.

Step 2. Equivalence?

$1+a^2>0$, so $R$ is reflexive, and it is symmetric. But $(-2,0)$ and $(0,2)$ are in $R$ while $(-2,2)$ is not. Not transitive, so II is false.

Answer: option (1)

Example 40 (JEE Main 2025, 8 April Shift 2). Let $A=\{0,1,2,3,4,5\}$. Let $R$ be a relation on $A$ defined by $(x,y)\in R$ if and only if $\max\{x,y\}\in\{3,4\}$. Then among the statements $(S_1)$: The number of elements in $R$ is $18$, and $(S_2)$: The relation $R$ is symmetric but neither reflexive nor transitive:

▸ SHOW SOLUTION

Step 1. Count

Pairs with maximum exactly $3$: both entries in $\{0,\dots,3\}$ but not both in $\{0,1,2\}$.

$$4^2-3^2=7,\qquad 5^2-4^2=9$$

So $n(R)=16$ and $S_1$ is false.

Step 2. Properties

$\max\{x,y\}=\max\{y,x\}$, so $R$ is symmetric. $(0,0)\notin R$: not reflexive. $(0,3),(3,0)\in R$ but $(0,0)\notin R$: not transitive. So $S_2$ is true.

Answer: option (3)

Counting relations of each type

For a set with $n$ elements, the $n^2$ pairs split into $n$ diagonal pairs $(a,a)$ and $\frac{n(n-1)}2$ off-diagonal partner sets $\{(a,b),(b,a)\}$.

Example 41 (IITianForum). For $A=\{1,2,3\}$, find the number of relations that are (a) reflexive; (b) symmetric; (c) reflexive and symmetric; (d) anti-symmetric.

▸ SHOW SOLUTION

Step 1. The pairs

$3$ diagonal pairs and $3$ partner sets $\{(1,2),(2,1)\}$, $\{(1,3),(3,1)\}$, $\{(2,3),(3,2)\}$.

Step 2. Reflexive and symmetric counts

Reflexive: diagonal fixed, the other $6$ pairs free.

$$\text{(a) }2^6=64$$

Symmetric: each diagonal pair free and each partner set wholly in or out.

$$\text{(b) }2^{3+3}=64,\qquad\text{(c) }2^3=8$$

Step 3. Anti-symmetric

Each diagonal pair is free. Each partner set may take neither pair, the first or the second, but not both: $3$ choices.

$$\text{(d) }2^3\cdot3^3=216$$

Answer: $64$, $64$, $8$, $216$

Example 42 (JEE Main 2026, 21 January Shift 1). The number of relations defined on the set $\{a,b,c,d\}$ which are both reflexive and symmetric is

▸ SHOW SOLUTION

Step 1. Diagonal

Reflexive: all $4$ diagonal pairs are in, $1$ way.

Step 2. Partner sets

The $12$ off-diagonal pairs form $6$ partner sets; symmetric means each set is wholly in or out.

$$2^6=64$$

Answer: option (4)

Example 43 (JEE Main 2025, 22 January Shift 2). Let $A=\{1,2,3\}$. The number of relations on $A$, containing $(1,2)$ and $(2,3)$, which are reflexive and transitive but not symmetric, is ______.

▸ SHOW SOLUTION

Step 1. Pairs forced in

Reflexive: $(1,1),(2,2),(3,3)$. Given: $(1,2),(2,3)$. Transitive: $(1,3)$. Call this set $R_0$; it is transitive.

Step 2. Pairs left to decide

$(2,1)$, $(3,2)$ and $(3,1)$.

• Adding $(3,1)$: then $(3,1),(1,2)$ need $(3,2)$ and $(2,3),(3,1)$ need $(2,1)$. That is all of $A\times A$, which is symmetric.
• Adding $(2,1)$ and $(3,2)$: $(3,2),(2,1)$ need $(3,1)$, back to $A\times A$.
• Adding only $(2,1)$, or only $(3,2)$: each stays transitive, and neither is symmetric.

Step 3. Count

$R_0$, $R_0\cup\{(2,1)\}$ and $R_0\cup\{(3,2)\}$.

Answer: $3$

Smallest number of pairs to add

To make a given $R$ have a property with the fewest new pairs:

For an equivalence relation with classes of sizes $n_1,n_2,\dots$ the number of pairs is

$$n_1^2+n_2^2+\cdots$$

Example 44 (JEE Main 2026, 22 January Shift 1). Let the relation $R$ on the set $M=\{1,2,3,\dots,16\}$ be given by $R=\{(x,y):4y=5x-3,\ x,y\in M\}$. Then the minimum number of elements required to be added in $R$ to make it symmetric is

▸ SHOW SOLUTION

Step 1. List R

$y=\dfrac{5x-3}4$ is an integer when $x=3,7,11,15$, giving $y=3,8,13,18$. $18\notin M$.

$$R=\{(3,3),(7,8),(11,13)\}$$

Step 2. Missing reverses

$(3,3)$ is its own reverse. Add $(8,7)$ and $(13,11)$.

Answer: $2$, option (2)

Example 45 (JEE Main 2026, 21 January Shift 2). Let $A=\{2,3,5,7,9\}$ and $R$ be the relation on $A$ defined by $xRy$ iff $2x\le3y$. Let $\ell$ be the number of elements in $R$ and $m$ the minimum number of elements to add to make $R$ symmetric. Then $\ell+m$ is

▸ SHOW SOLUTION

Step 1. Count R

$y\ge\frac{2x}3$:

$x$	$y$	count
$2,3$	all five	$5$ each
$5,7$	$5,7,9$	$3$ each
$9$	$7,9$	$2$

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$$\ell=5+5+3+3+2=18$$

Step 2. Missing reverses

$(x,y)\in R$ with $(y,x)\notin R$ means $2y>3x$:

$$(2,5),(2,7),(2,9),(3,5)$$$$(3,7),(3,9),(5,9)$$

So $m=7$.

Answer: $\ell+m=25$, option (2)

Example 46 (JEE Main 2026, 23 January Shift 1). Let $A=\{-2,-1,0,1,2,3,4\}$. Let $R$ be a relation on $A$ defined by $xRy$ if and only if $2x+y\le2$. Let $l$ be the number of elements in $R$. Let $m$ and $n$ be the minimum number of elements required to be added in $R$ to make it reflexive and symmetric relations respectively. Then $l+m+n$ is equal to:

▸ SHOW SOLUTION

Step 1. Count R

$y\le2-2x$:

$x$	$y$	count
$-2,-1$	all seven	$7$ each
$0$	$y\le2$	$5$
$1$	$y\le0$	$3$
$2$	$y=-2$	$1$
$3,4$	none	$0$

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$$l=7+7+5+3+1=23$$

Step 2. Reflexive

$(a,a)\in R$ iff $3a\le2$. Missing: $(1,1),(2,2),(3,3),(4,4)$, so $m=4$.

Step 3. Symmetric

$(x,y)\in R$ but $2y+x>2$:

$$(-2,3),(-2,4),(-1,2)$$$$(-1,3),(-1,4),(0,2)$$

So $n=6$.

Answer: $23+4+6=33$, option (3)

Example 47 (JEE Main 2026, 6 April Shift 2). Let $R=\{(x,y)\in\mathbb{N}\times\mathbb{N}:\log_e(x+y)\le2\}$. Then the minimum number of elements required to be added in $R$ to make it a transitive relation, is ______.

▸ SHOW SOLUTION

Step 1. R in plain terms

$x+y\le e^2\approx7.39$, so $x+y\le7$. For $x+y=s$ there are $s-1$ pairs, $s=2,\dots,7$:

$$n(R)=1+2+3+4+5+6=21$$

Step 2. What transitivity forces

Every $a\in\{1,\dots,6\}$ satisfies $(a,1)\in R$, and $(1,c)\in R$ for every $c\le6$. So $(a,c)$ must be added for all $a,c\in\{1,\dots,6\}$.

Step 3. The full set works

$\{1,\dots,6\}\times\{1,\dots,6\}$ is transitive and contains $R$, with $36$ pairs.

$$36-21=15$$

Answer: $15$

Example 48 (JEE Main 2025, 23 January Shift 1). Let $R=\{(1,2),(2,3),(3,3)\}$ be a relation defined on the set $\{1,2,3,4\}$. Then the minimum number of elements needed to be added in $R$ so that $R$ becomes an equivalence relation, is:

▸ SHOW SOLUTION

Step 1. Classes

$(1,2)$ and $(2,3)$ put $1,2,3$ in one class. $4$ is linked to nothing, so it is a class by itself.

Step 2. Pairs needed

$$3^2+1^2=10$$

Step 3. Add

$R$ already has $3$ of them.

$$10-3=7$$

Answer: option (4)

Equivalence relations and classes

Let $R$ be an equivalence relation on $A$. The equivalence class of $a$ is

$$[a]=\{x\in A:(a,x)\in R\}$$

• Every element lies in its own class, since $aRa$.
• Two classes are either equal or disjoint.
• The classes together cover $A$.

So the classes form a partition of $A$: disjoint non-empty blocks whose union is $A$. Conversely every partition gives an equivalence relation, "same block". Hence

$$\text{equivalence relations}\ \leftrightarrow\ \text{partitions}$$

Example 49 (JEE Main 2025, 28 January Shift 1). The relation $R=\{(x,y):x,y\in\mathbb{Z}\text{ and }x+y\text{ is even}\}$ is:

▸ SHOW SOLUTION

Step 1. Same parity

$x+y$ is even exactly when $x$ and $y$ are both even or both odd.

Step 2. Conclude

"Same parity" is of the form $f(x)=f(y)$ with $f(x)$ the remainder on division by $2$. So $R$ is an equivalence relation with two classes, the even integers and the odd integers.

Answer: option (3)

Example 50 (JEE Main 2026, 23 January Shift 2). Let $A=\{0,1,2,\dots,9\}$. Let $R$ be a relation on $A$ defined by $(x,y)\in R$ if and only if $\lvert x-y\rvert$ is a multiple of $3$. Given two statements: Statement I: $n(R)=36$. Statement II: $R$ is an equivalence relation.

▸ SHOW SOLUTION

Step 1. Statement II

$3\mid(x-y)$ means $x$ and $y$ leave the same remainder on division by $3$. This is of the form $f(x)=f(y)$, so $R$ is an equivalence relation.

Step 2. Classes

$$\{0,3,6,9\},\quad\{1,4,7\},\quad\{2,5,8\}$$

Step 3. Statement I

$$n(R)=4^2+3^2+3^2=34\ne36$$

Answer: option (2)

A rounded box holding the numbers 0 to 9 split into three blocks: 0, 3, 6, 9 in the first block, 1, 4, 7 in the second and 2, 5, 8 in the third

Example 51 (IITianForum). On $\mathbb{R}$, $xRy$ holds whenever $x^2+x=y^2+y$. Show that $R$ is an equivalence relation, and find the classes of $2$ and of $-\frac12$.

▸ SHOW SOLUTION

Step 1. Equivalence

With $f(x)=x^2+x$, $xRy$ means $f(x)=f(y)$. So $R$ is reflexive, symmetric and transitive.

Step 2. Class of 2

$$y^2+y=6\ \Rightarrow\ (y-2)(y+3)=0$$$$[2]=\{2,-3\}$$

Step 3. Class of minus one half

$$y^2+y=-\frac14\ \Rightarrow\ \left(y+\frac12\right)^2=0$$$$\left[-\tfrac12\right]=\left\{-\tfrac12\right\}$$

Answer: $[2]=\{2,-3\}$; $\left[-\frac12\right]$ has one element. In general $[a]=\{a,-1-a\}$.

Example 52 (JEE Main 2025, 23 January Shift 2). Let $X=\mathbb{R}\times\mathbb{R}$. Define a relation $R$ on $X$ as $(a_1,b_1)\,R\,(a_2,b_2)\iff b_1=b_2$. Statement-I: $R$ is an equivalence relation. Statement-II: For some $(a,b)\in X$, the set $S=\{(x,y)\in X:(x,y)\,R\,(a,b)\}$ represents a line parallel to $y=x$.

▸ SHOW SOLUTION

Step 1. Statement I

The relation compares the second entries: $f(x,y)=y$ takes equal values. So $R$ is an equivalence relation.

Step 2. Statement II

$S=\{(x,y):y=b\}$, the horizontal line $y=b$. It is parallel to the $x$ axis for every $b$, never to $y=x$.

Answer: option (2)

Example 53 (JEE Main 2025, 22 January Shift 1). The number of non-empty equivalence relations on the set $\{1,2,3\}$ is:

▸ SHOW SOLUTION

Step 1. Partitions of {1, 2, 3}

$$\{1,2,3\}$$$$\{1,2\}\{3\},\quad\{1,3\}\{2\},\quad\{2,3\}\{1\}$$$$\{1\}\{2\}\{3\}$$

Step 2. Count

$5$ partitions, so $5$ equivalence relations. Each contains the diagonal, so each is non-empty.

Answer: option (3)

Example 54 (IITianForum). Find the number of equivalence relations on $A=\{1,2,3,4\}$, and the number of them that contain $(1,2)$.

▸ SHOW SOLUTION

Step 1. Partitions by block sizes

Block sizes	Number of partitions
$4$	$1$
$3,1$	${}^4C_3=4$
$2,2$	$\frac12\,{}^4C_2=3$
$2,1,1$	${}^4C_2=6$
$1,1,1,1$	$1$

swipe for more →

$$1+4+3+6+1=15$$

Step 2. Containing (1, 2)

$1$ and $2$ must be in one block. Treat them as a single object: partitions of the three objects $\{1,2\}$, $3$, $4$ number $5$.

Answer: $15$ and $5$

Example 55 (JEE Advanced 2026, Paper 1). Let $S=\{1,2,3,\dots,10\}$. Consider the set $X$ of all equivalence relations $R$ on $S$ such that $R$ has exactly $42$ elements. Then the number of elements in $X$ is ______.

▸ SHOW SOLUTION

Step 1. Size of R from the classes

If the classes have sizes $n_1,n_2,\dots,n_k$, then

$$n_1+\cdots+n_k=10$$$$n_1^2+\cdots+n_k^2=42$$

Step 2. Find the block sizes

A block of $7$ or more gives at least $49$, too many. With largest block $6$, the other blocks add to $4$ and their squares to $6$: only $2,1,1$. With largest block $5$, the rest add to $5$ with squares adding to $17$: only $4,1$. With largest block $4$ or less, the squares add to at most $16+16+4=36$.

So the block sizes are $6,2,1,1$ or $5,4,1$.

Step 3. Count each shape

$$6,2,1,1:\ {}^{10}C_6\cdot{}^4C_2=210\cdot6=1260$$$$5,4,1:\ {}^{10}C_5\cdot{}^5C_4=252\cdot5=1260$$

The two singletons in the first shape need no ordering; choosing the block of $2$ fixes them.

Answer: $2520$

Which method to use

A decision chart: from a box asking what is asked, eight branches lead to the method; simplify a set expression leads to writing minus as intersection with a complement; survey numbers to filling the Venn regions from the centre; count subsets to in or out for each element; count a relation to fixing the first entry and counting the second; is it reflexive, symmetric, transitive to checking each and finding one counterexample; count reflexive or symmetric relations to the diagonal then partner pairs; pairs to add to mirrors, diagonal, then closure; equivalence relations to counting partitions into classes

Quick recall

Practise these methods on real exam questions in the doMath Sets and Relations chapters. A function is a relation with one output for each input: the Functions note continues from here.