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Exercise 2.1 · Q8

Q.Let A={1,2}A = \{1, 2\} and B={3,4}B = \{3, 4\}. Write A×BA \times B. How many subsets will A×BA \times B have? List them.

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The Cartesian product A×BA \times B is the set of all ordered pairs (a,b)(a,b) with a∈A,b∈Ba \in A, b \in B. Here A×B={(1,3),(1,4),(2,3),(2,4)}A \times B = \{(1,3), (1,4), (2,3), (2,4)\}. It has 44 elements, so it has 24=162^4 = 16 subsets. The subsets are listed below.

Why Cartesian product?

The Cartesian product A×BA \times B is not about multiplying numbers — it’s about pairing. Every element of AA gets paired with every element of BB, in order. So if AA has mm elements and BB has nn elements, A×BA \times B has m×nm \times n ordered pairs. That’s the “why” behind the name.

Here AA has 2 elements, BB has 2 elements, so A×BA \times B has 2×2=42 \times 2 = 4 elements.

∣A×B∣=∣A∣⋅∣B∣|A \times B| = |A| \cdot |B|

Now, the number of subsets of any set with kk elements is 2k2^k. That’s because each element can either be in or out of a subset — two choices per element, independent choices, so 2k2^k total.

So for A×BA \times B, k=4k = 4, so the number of subsets is 24=162^4 = 16.


Step-by-step

  1. Write A×BA \times B explicitly.

    Take each element of AA and pair it with each element of BB:

    • 11 with 33 gives (1,3)(1,3)
    • 11 with 44 gives (1,4)(1,4)
    • 22 with 33 gives (2,3)(2,3)
    • 22 with 44 gives (2,4)(2,4)

    So

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

  1. Count the elements.

    There are 4 ordered pairs. So ∣A×B∣=4|A \times B| = 4.

  2. Number of subsets.

    A set with nn elements has 2n2^n subsets. Here n=4n = 4, so

24=162^4 = 16

  1. List all 16 subsets.

    We list them systematically: start with the empty set, then all 1-element subsets, then 2-element, then 3-element, then the full set itself.

    • 0-element subset: ∅\varnothing
    • 1-element subsets (4 of them): {(1,3)}, {(1,4)}, {(2,3)}, {(2,4)}\{(1,3)\},\ \{(1,4)\},\ \{(2,3)\},\ \{(2,4)\}
    • 2-element subsets (6 of them): {(1,3),(1,4)}, {(1,3),(2,3)}, {(1,3),(2,4)}\{(1,3),(1,4)\},\ \{(1,3),(2,3)\},\ \{(1,3),(2,4)\} {(1,4),(2,3)}, {(1,4),(2,4)}, {(2,3),(2,4)}\{(1,4),(2,3)\},\ \{(1,4),(2,4)\},\ \{(2,3),(2,4)\} …

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