Let X be a non-empty set containing n elements. Th – Class 11 Mathematics | ConceptFirst

Question of 100

X

(a)

2^{n^2}

(b)

2^n

(c)

2^{2n}

(d)

n^2

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Concept understanding — Cartesian Product

Cartesian Product: From Intuition to Definition

Imagine you're ordering a pizza. You have two choices to make: the size (Small, Medium, Large) and the topping (Cheese, Pepperoni, Veggie). How many different pizzas can you order?

You can pair each size with each topping:

Small + Cheese, Small + Pepperoni, Small + Veggie
Medium + Cheese, Medium + Pepperoni, Medium + Veggie
Large + Cheese, Large + Pepperoni, Large + Veggie

That's $3 \times 3 = 9$ possible pizzas. What you just did — systematically pairing every element of one set with every element of another — is the Cartesian product in action.

The Intuition

The Cartesian product is a way to combine two sets to create a new set of ordered pairs. The order matters: (Small, Cheese) is different from (Cheese, Small) — one is a pizza order, the other is nonsense.

Think of it like a multiplication table for sets. If set $A$ has $m$ items and set $B$ has $n$ items, their Cartesian product has $m \times n$ items.

Note

The name comes from René Descartes, who used this idea to create the coordinate plane — every point $(x, y)$ on a graph is an element of the Cartesian product of the x-axis and y-axis.

The Precise Definition

Let $A$ and $B$ be two sets. The Cartesian product of $A$ and $B$ , written $A \times B$ , is the set of all ordered pairs $(a, b)$ where $a$ is from $A$ and $b$ is from $B$ .

A \times B = \{(a, b) \mid a \in A \text{ and } b \in B\}

The vertical bar means "such that." So read it as: "The set of all ordered pairs (a, b) such that a belongs to A and b belongs to B."

Key Properties to Remember

Order matters: $A \times B$ is generally not the same as $B \times A$ . For example, if $A = \{1, 2\}$ and $B = \{x, y\}$ :
- $A \times B = \{(1, x), (1, y), (2, x), (2, y)\}$
- $B \times A = \{(x, 1), (x, 2), (y, 1), (y, 2)\}$
These are different sets because the pairs are ordered differently.
Size formula: If $|A| = m$ and $|B| = n$ , then $|A \times B| = m \times n$ . This holds even if one set is empty — then the product is empty.
Empty set: $A \times \emptyset = \emptyset$ and $\emptyset \times B = \emptyset$ . You can't form any pairs if one set has nothing to contribute.

Watch out

A common mistake: thinking $A \times B$ contains all possible combinations of elements from $A$ and $B$ without caring about order. But $(a, b)$ and $(b, a)$ are different pairs unless $a = b$ . Always treat ordered pairs as distinct based on position.

Examples to Cement the Idea

Example 1: $A = \{1, 2\}$ , $B = \{3, 4\}$

A \times B = \{(1, 3), (1, 4), (2, 3), (2, 4)\}

Four pairs, as expected ( $2 \times 2 = 4$ ).

Example 2: $A = \{a\}$ , $B = \{1, 2, 3\}$

A \times B = \{(a, 1), (a, 2), (a, 3)\}

Three pairs — every element of $B$ gets paired with the single element of $A$ .

Example 3: $A = \{0, 1\}$ , $B = \{0, 1\}$

A \times B = \{(0, 0), (0, 1), (1, 0), (1, 1)\}

This is the set of all possible 2-bit binary strings — a foundation for computer science.

Why This Matters

The Cartesian product is the mathematical backbone of: …

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Number of relations on $X$ $=2^{n^2}$ ; option (a).

…

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 27 on this concept.

CBSE 2026Set ANNUAL1 markMCQ
$A = \{1, 2\}$
(a) 6
(b) 36
(c) 32
(d) 64
›Reveal solutionSolution

A relation from $A$ to $B$ is any subset of $A \times B$ ; with $|A \times B| = 6$ , there are $2^6 = 64$ subsets, hence 64 relations.

A relation from $A$ to $B$ is defined as any subset of the Cartesian product $A \times B$ . Here $|A| = 2$ and $|B| = 3$ , so $|A \times B| = 2 \times 3 = 6$ . The total number of subsets of a set with 6 elements is $2^6 = 64$ . Since every subset of …

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CBSE 2026Set ANNUAL1 markMCQ
$A = \{1, 2\}$
(a) 4
(b) $2^4$
(c) 2
(d) 1
›Reveal solutionSolution

The number of relations from a set A to a set B is $2^{|A|\cdot|B|}$ , since every relation is a subset of $A \times B$ .

Given $A=\{1,2\}$ , $B=\{3,4\}$ , so $|A|=2$ , $|B|=2$ .

$A \times B$ has $|A|\times|B| = 2 \times 2 = 4$ ordered pairs: $(1,3),(1,4),(2,3),(2,4)$ .

…

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CBSE 2026Set ANNUAL1 mark
$A \times B = \{(p, q), (p, r)\}$
›Reveal solutionSolution

In $A\times B=\{(p,q),(p,r)\}$ , A is the set of first coordinates and B is the set of second coordinates; here $A=\{p\}$ , not $\{p,q,r\}$ .

Every ordered pair $(x,y)\in A\times B$ has $x\in A$ and $y\in B$ .

Here both ordered pairs $(p,q)$ and $(p,r)$ have first component $p$ , so the set of first components — which is exactly A — is $A=\{p\}$ .

…

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CBSE 2025Set ANNUAL1 markMCQ
$A = \{1, 2\}, B = \{3\} \Rightarrow A \times B =$
(a) $\{1, 2, 3\}$
(b) $\{(1,3), (2,3), (1,2)\}$
(c) $\{(1,3), (2,3)\}$
(d) $\{(1,2), (3,1)\}$
›Reveal solutionSolution

$A \times B = \{(1,3), (2,3)\}$ .

For sets $A$ and $B$ , $A \times B = \{(a, b) : a \in A, b \in B\}$ — every element of $A$ is paired, in order, with every element of $B$ .

…

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CBSE 2025Set ANNUAL1 markMCQ
$A = \{a, b\}, B = \{c, d\}$
(a) 8
(b) 16
(c) 32
(d) 64
›Reveal solutionSolution

The number of relations from $A$ to $B$ is $2^{|A \times B|} = 2^4 = 16$ .

A relation from $A$ to $B$ is defined as any subset of the Cartesian product $A \times B$ . If $A$ has $m$ elements and $B$ has $n$ elements, $A \times B$ has $mn$ elements, and a set with $mn$ elements has $2^{mn}$ subsets.

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. If A = {1, 2} and B = {3, 4}, then A \times B is:$
(a) {3, 4, 6, 8}
(b) {3, 8}
(c) {(1, 3), (1, 4), (2, 3), (2, 4)}
(d) None of these
›Reveal solutionSolution

$A \times B$ is the set of all ordered pairs $(a,b)$ with $a \in A$ , $b \in B$ .

Given $A=\{1,2\}$ and $B=\{3,4\}$ .

Pair each element of $A$ with every element of $B$ :

$1 \to (1,3),(1,4)$

$2 \to (2,3),(2,4)$

…

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CBSE 2025Set ANNUAL1 mark
$A = \{x, y, z\}$
›Reveal solutionSolution

A relation from $A$ to $B$ is any subset of $A \times B$ ; with $|A|=3$ and $|B|=2$ , $|A \times B| = 6$ , so there are $2^6 = 64$ relations.

$A = \{x, y, z\}$ has 3 elements, $B = \{1, 2\}$ has 2 elements.

…

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CBSE 2024Set ANNUAL1 markMCQ
$A$
(a) 8
(b) 9
(c) 10
(d) 6
›Reveal solutionSolution

If $A$ has $m$ elements and $B$ has $n$ elements, then $A\times B$ has $mn$ elements.

The Cartesian product $A\times B$ consists of all ordered pairs $(a,b)$ with $a\in A$ and $b\in B$ . Since $A$ has 3 elements and $B=\{3,4,5\}$ has 3 elements, each of the 3 choices for the first coordinate can be paire …

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CBSE 2024Set ANNUAL1 markMCQ
$Q. Let A = {x, y, z} and B = {1, 2}, then number of relations from A into B will be:$
(a) 6
(b) 9
(c) 24
(d) 64
›Reveal solutionSolution

A relation from A to B is any subset of $A\times B$ ; with $|A|=3,|B|=2$ , there are $2^{3\times2}=64$ subsets.

A relation from set $A$ to set $B$ is defined as any subset of the Cartesian product $A \times B$ .

$|A| = 3$ (elements $x,y,z$ ), $|B| = 2$ (elements $1,2$ ).

So $|A \times B| = 3 \times 2 = 6$ .

…

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CBSE 2024Set ANNUAL1 markMCQ
$n((A\times B)\cap(A\times C))=8$
(a) $8$
(b) $6$
(c) $16$
(d) $4$
›Reveal solutionSolution

Using $(A\times B)\cap(A\times C)=A\times(B\cap C)$ , we get $n(A)=4$ .

For sets, $(A\times B)\cap(A\times C)=A\times(B\cap C)$ , so

$n\big((A\times B)\cap(A\times C)\big)=n(A)\times n(B\cap C).$ …

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CBSE 2024Set ANNUAL1 markMCQ
$n$
(a) $2^n$
(b) $2^{n^2}$
(c) $n^2$
(d) $n^n$ .
›Reveal solutionSolution

The number of relations on a set of $n$ elements is $2^{n^2}$ .

A relation $R$ on set $A$ is defined as any subset of the Cartesian product $A\times A$ . Since $A$ has $n$ elements, $A\times A$ has $n\times n = n^2$ ordered pairs. The number of subsets of a set with $n^2$ elements is $2^{n^2}$ (by the subset-counting rule). Each such subset is a valid relation on $A$ (including the …

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CBSE 2023Set ANNUAL1 markMCQ
$A = \{a, b, c\}$
(a) 8
(b) 5
(c) $2^9$
(d) 9
›Reveal solutionSolution

The Cartesian product's size is the product of the two sets' sizes: $3 \times 3 = 9$ .

The Cartesian product $A \times B$ consists of all ordered pairs $(a, b)$ where $a \in A$ and $b \in B$ . For each of the $n(A)$ choices of $a$ , there are $n(B)$ choices of $b$ , giving:

$n(A \times B) = n(A) \cdot n(B)$ …

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