Cartesian Product of Two Sets

1

easyIf A = {1, 2} and B = {3, 4}, what is n(A × B)?

A. 2 B. 4 C. 6 D. 8

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✔ Correct: B

n(A) = 2, n(B) = 2. n(A × B) = 2 × 2 = 4.

2

easyIf A = {a, b} and B = {1}, find A × B.

A. {(a, 1), (b, 1)} B. {(1, a), (1, b)} C. {(a, b), (1, 1)} D. {(a, 1)}

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✔ Correct: A

Pairing each element of A with the only element of B gives {(a, 1), (b, 1)}.

3

mediumIf A × B = {(1, 2), (1, 3), (4, 2), (4, 3)}, find set A.

A. {1, 2} B. {2, 3} C. {1, 4} D. {1, 2, 3, 4}

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✔ Correct: C

Set A consists of the first elements of the ordered pairs: {1, 1, 4, 4} = {1, 4}.

4

mediumIf n(A) = 3 and n(B) = 2, find n(A × B) + n(B × A).

A. 6 B. 12 C. 5 D. 10

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✔ Correct: B

n(A × B) = 3 × 2 = 6. n(B × A) = 2 × 3 = 6. 6 + 6 = 12.

5

hardIf (x + 1, y - 2) = (3, 1), find x and y.

A. x = 2, y = 3 B. x = 4, y = -1 C. x = 2, y = -1 D. x = 4, y = 3

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✔ Correct: A

x + 1 = 3 ⟹ x = 2. y - 2 = 1 ⟹ y = 3.

6

hardIf A = {1, 2} and B = {3, 4}, how many subsets does A × B have?

A. 4 B. 8 C. 16 D. 32

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✔ Correct: C

n(A × B) = 4. Number of subsets = 2⁴ = 16.

7

very hardIf A = {1, 2}, B = {3, 4}, C = {5, 6}, find n(A × B × C).

A. 6 B. 8 C. 12 D. 24

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✔ Correct: B

n(A × B × C) = n(A) × n(B) × n(C) = 2 × 2 × 2 = 8.

1

Assertion (A): If A = {1, 2} and B = {3}, then A × B = {(1, 3), (2, 3)}.
Reason (R): The Cartesian product contains all possible ordered pairs (a, b) where a ∈ A and b ∈ B.

Choose the correct option:
(A) Both A and R are true and R is the correct explanation of A.
(B) Both A and R are true but R is not the correct explanation of A.
(C) A is true but R is false.
(D) A is false but R is true.

A. Option A B. Option B C. Option C D. Option D

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✔ Correct: A

Both statements are true and the reason correctly explains the definition of the Cartesian product.

2

Assertion (A): If A = {1, 2} and B = {3, 4}, then A × B = B × A.
Reason (R): Cartesian product is commutative for all sets.

Choose the correct option:
(A) Both A and R are true and R is the correct explanation of A.
(B) Both A and R are true but R is not the correct explanation of A.
(C) A is true but R is false.
(D) A is false but R is true.

A. Option A B. Option B C. Option C D. Option D

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✔ Correct: D

The assertion is false because A × B = {(1, 3), (1, 4), (2, 3), (2, 4)} while B × A = {(3, 1), (3, 2), (4, 1), (4, 2)}. The reason is also false.

1

mediumGiven A = {a, b, c} and B = {1, 2}, find: (i) A × B (ii) B × A (iii) n(A × B) + n(B × A)

Show Answer & Solution

(i) A × B = {(a, 1), (a, 2), (b, 1), (b, 2), (c, 1), (c, 2)}
(ii) B × A = {(1, a), (1, b), (1, c), (2, a), (2, b), (2, c)}
(iii) n(A × B) = 6, n(B × A) = 6. 6 + 6 = 12.

2

mediumGiven A = {1, 2}, B = {2, 3}, C = {3, 4}, find: (i) A × (B ∩ C) (ii) (A × B) ∩ (A × C) (iii) Verify if (i) = (ii)

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(i) B ∩ C = {3}. A × {3} = {(1, 3), (2, 3)}
(ii) A × B = {(1, 2), (1, 3), (2, 2), (2, 3)}. A × C = {(1, 3), (1, 4), (2, 3), (2, 4)}. Intersection = {(1, 3), (2, 3)}
(iii) Yes, both are equal.

3

hardIf A = {1, 2}, B = {3, 4}, C = {5}, find: (i) A × B × C (ii) n(A × B × C) (iii) n(A) × n(B) × n(C)

Show Answer & Solution

(i) A × B × C = {(1, 3, 5), (1, 4, 5), (2, 3, 5), (2, 4, 5)}
(ii) n(A × B × C) = 4
(iii) 2 × 2 × 1 = 4.

1

hardLet A = {1, 2, 3} and B = {3, 4, 5}. Find A × B and verify that n(A × B) = n(A) × n(B).

Show Answer & Solution

Step 1: List elements of A = {1, 2, 3} and B = {3, 4, 5}.
Step 2: Form ordered pairs (a, b): (1, 3), (1, 4), (1, 5), (2, 3), (2, 4), (2, 5), (3, 3), (3, 4), (3, 5).
Step 3: Count elements in A × B: there are 9 pairs.
Step 4: Calculate n(A) = 3 and n(B) = 3.
Step 5: Calculate n(A) × n(B) = 3 × 3 = 9. Since 9 = 9, the relation is verified.

2

very hardIf A = {x : x ∈ N, x ≤ 3} and B = {x : x ∈ N, 2 ≤ x ≤ 4}, find A × B and identify all subsets of A × B that contain exactly 2 elements.

Show Answer & Solution

Step 1: Identify A = {1, 2, 3} and B = {2, 3, 4}.
Step 2: A × B = {(1, 2), (1, 3), (1, 4), (2, 2), (2, 3), (2, 4), (3, 2), (3, 3), (3, 4)}.
Step 3: Total elements n(A × B) = 9.
Step 4: The number of subsets with 2 elements is given by ⁹C₂.
Step 5: ⁹C₂ = (9 × 8) / (2 × 1) = 36. There are 36 such subsets.

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Case Study: A school canteen offers a combo meal consisting of one drink and one snack. The available drinks are {Coffee, Tea} and the available snacks are {Samosa, Sandwich, Burger}.

Sub-question 1: (i) Represent the set of all possible combos as a Cartesian product. (ii) How many total combos are available?

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(i) Let D = {Coffee, Tea}, S = {Samosa, Sandwich, Burger}. D × S = {(Coffee, Samosa), (Coffee, Sandwich), (Coffee, Burger), (Tea, Samosa), (Tea, Sandwich), (Tea, Burger)}.
(ii) n(D) = 2, n(S) = 3. Total combos = 2 × 3 = 6.

Sub-question 2: If the canteen adds 'Juice' to the drinks and 'Pizza' to the snacks, how many total combos are now possible?

Show Solution

New D = {Coffee, Tea, Juice}, n(D) = 3. New S = {Samosa, Sandwich, Burger, Pizza}, n(S) = 4. Total combos = 3 × 4 = 12.

Definition of Cartesian Product

Cardinality of Cartesian Product

Properties of Cartesian Product

Section A: Multiple Choice Questions (MCQs)

Section B: Assertion–Reason Questions

Section C: Short Answer Questions

Section D: Long Answer Questions

Section E: Case-Based Question

CBSE Board & Sample Paper Past Year Questions

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