If $A$ and $B$ are finite sets such that $A \subset B$, then $n(A \cup B) = $ ______________.

If A and B are finite sets such that A ⊂ B, then n(A ∪ B) = _______.

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Concept understanding — Subset Listing

Subset Listing: A First Look

Let's build this from the ground up — no jargon, just intuition first.

1. The Intuition: What does "subset" mean?

Imagine you have a set — a collection of distinct objects. For example:

Set A = {apple, banana, cherry}

Now, a subset is simply a selection of some (or all, or none) of these objects, taken from the original set.

You could pick all three → {apple, banana, cherry}
You could pick just two → {apple, banana}
You could pick just one → {cherry}
You could pick none → {} (the empty set)

Each of these is a subset of the original set.

2. The Precise Definition

Definition: A set $B$ is a subset of a set $A$ if every element of $B$ is also an element of $A$ .

We write this as:

B \subseteq A

If $B$ is not a subset of $A$ , we write:

B \not\subseteq A

Key points to remember:

Every set is a subset of itself.

Example: {apple, banana} ⊆ {apple, banana}
The empty set $\emptyset$ (or $\{\}$ ) is a subset of every set.

Why? Because it has no elements, so there's nothing to violate the condition.
If $B$ is a subset of $A$ but $B \neq A$ , we call $B$ a proper subset.

Notation: $B \subset A$ (some books use $\subsetneq$ )

3. How to "list" all subsets

Subset listing means writing down every possible subset of a given set.

Example: Set $S = \{a, b\}$

All subsets:

$\emptyset$ (empty set)
$\{a\}$
$\{b\}$
$\{a, b\}$ (the set itself)

So the list of all subsets is:

\{ \emptyset, \{a\}, \{b\}, \{a, b\} \}

How many subsets does a set have?

If a set has $n$ elements, it has exactly $2^n$ subsets.

$n = 0$ → $2^0 = 1$ subset (just the empty set)
$n = 1$ → $2^1 = 2$ subsets
$n = 2$ → $2^2 = 4$ subsets (as above)
$n = 3$ → $2^3 = 8$ subsets

Why $2^n$ ?

For each element, you have 2 choices: include it or exclude it. Multiply these choices: $2 \times 2 \times \dots \times 2$ ( $n$ times) = $2^n$ .

4. A systematic way to list subsets

For a set with $n$ elements, you can use a binary counting method:

Label each element with a position (1st, 2nd, 3rd, ...)
Count from $0$ to $2^n - 1$ in binary
Each binary number tells you which elements to include (1 = include, 0 = exclude)

Example: $S = \{a, b, c\}$ (3 elements)

Binary	Subset
000	$\emptyset$
001	$\{c\}$
010	$\{b\}$
011	$\{b, c\}$
100	$\{a\}$

Why this formula?

Okay, let's break down Subset Listing from the ground up. The core idea is simple: given a set, how do we systematically list all its subsets, and why does the formula $2^n$ work?

1. The Core Question

Imagine you have a set with $n$ elements, like $S = \{a, b, c\}$ (so $n=3$ ). A subset is any collection of elements from $S$ , including the empty set $\{\}$ and the set itself $\{a, b, c\}$ .

The key formula is:

Total number of subsets of a set with $n$ elements = $\mathbf{2^n}$

Let's see why this is true, not just memorize it.

2. The "Decision" or "Binary Choice" Reasoning

The most intuitive derivation comes from thinking about each element individually.

For each element in the original set, when building a subset, you have exactly two choices:

Include the element in the subset.
Exclude the element from the subset.

This is a fundamental, independent decision for every element.

Example with $S = \{a, b, c\}$

For element $a$ : Choose IN or OUT. (2 choices)
For element $b$ : Choose IN or OUT. (2 choices)
For element $c$ : Choose IN or OUT. (2 choices)

Since these choices are independent (choosing for $a$ doesn't affect the choice for $b$ ), the total number of distinct combinations of choices is the product of the number of choices for each element:

2 \times 2 \times 2 = 2^3 = 8

This directly gives the 8 subsets of $\{a, b, c\}$ :

$\{\}$ (all OUT)
$\{a\}$ (a IN, b OUT, c OUT)
$\{b\}$
$\{c\}$
$\{a, b\}$
$\{a, c\}$
$\{b, c\}$
$\{a, b, c\}$ (all IN)

3. The General Formula (Derivation)

For a set with $n$ elements, you have $n$ independent binary decisions. Therefore:

\text{Total subsets} = \underbrace{2 \times 2 \times \dots \times 2}_{n \text{ times}} = 2^n

This is the fundamental reason the formula holds. It's not a coincidence; it's a direct consequence of the counting principle for independent events.

4. Why This Matters for Exams

Don't just memorize $2^n$ . If a question asks "How many subsets does a set with 5 elements have?", you can instantly say $2^5 = 32$ . But if they ask why, you now have the reasoning. …

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When set $A$ is a subset of set $B$ , their union $A \cup B$ is simply set $B$ itself. Therefore, the number of elements in their union, $n(A \cup B)$ , is equal to the number of elements in $B$ , which is $\boxed{n(B)}$ .

Let's break down the concept of subsets and unions to understand why this result holds. The core idea here is to visualize what it means for one set to be entirely contained within another, and then consider what happens when you combine them.

Imagine you have a collection of items, say all the fruits in your basket ( $B$ ). Now, suppose you also have a smaller collection of items, specifically all the apples in your basket ( $A$ ). Since all apples are fruits, and these apples are in your basket, it means the set of apples ( $A$ ) is entirely contained within the set of fruits in your basket ( $B$ ). This is precisely what $A \subset B$ means.

When we talk about the union of these two sets, $A \cup B$ , we are asking for all items that are either apples or fruits in your basket (or both). Since every apple is already a fruit in your basket, combining the apples with all the fruits in your basket simply gives you all the fruits in your basket. You don't get any new items beyond what was already in $B$ .

Let's formalize this step-by-step.

Understanding the Subset Condition:

The notation $A \subset B$ means that every element of set $A$ is also an element of set $B$ . In simpler terms, $A$ is entirely contained within $B$ .

For example, if $B = \{1, 2, 3, 4, 5\}$ and $A = \{2, 3\}$ , then $A \subset B$ because both $2$ and $3$ are elements of $B$ .

Visualizing with a Venn Diagram:

When $A \subset B$ , a Venn diagram would show the circle representing set $A$ drawn completely inside the circle representing set $B$ . There are no elements in $A$ that are outside of $B$ .

+-----------------------+
|                       |
|       B               |
|    +-----------+      |
|    |     A     |      |
|    |           |      |
|    +-----------+      |
|                       |
+-----------------------+

Defining the Union of Sets:

The union of two sets, $A \cup B$ , is the set containing all elements that are in $A$ , or in $B$ , or in both.

Mathematically, $x \in (A \cup B)$ if and only if $x \in A$ or $x \in B$ .
Applying the Subset Condition to the Union:

Since $A \subset B$ , every element $x$ that belongs to $A$ also belongs to $B$ .

Consider an element $y \in (A \cup B)$ . By definition, this means $y \in A$ or $y \in B$ .
- If $y \in B$ , then $y$ is an element of $B$ .
- If $y \in A$ , then because $A \subset B$ , it must also be true that $y \in B$ . In both cases, any element in $A \cup B$ must necessarily be an element of $B$ . …

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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 26 on this concept.

CBSE 2026Set ANNUAL1 markMCQ
$\{1, 2, 3\}$
(a) 8
(b) 7
(c) 3
(d) 9
›Reveal solutionSolution

A set of $n$ elements has $2^n$ subsets in total; removing the empty subset gives $2^n-1$ non-empty subsets.

For the set $\{1,2,3\}$ , $n=3$ .

Total subsets $= 2^n = 2^3 = 8$ (this includes $\varnothing$ and the full set itself).

Non-empty subsets $= 2^n - 1 = 8 - 1 = 7$ .

…

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CBSE 2026Set ANNUAL1 mark
$Q. Define subset.$
›Reveal solutionSolution

$A\subseteq B$ means every element of A belongs to B.

Formally, A is a subset of B if, for every element $x$ , $x\in A \Rightarrow x\in B$ .

…

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CBSE 2026Set 1A1 mark
$C = \{x : x \text{ is an integer}, x^2 \le 4\}$
›Reveal solutionSolution

Integers with $x^2\le 4$ are $-2,-1,0,1,2$ .

We need integers $x$ with $x^2\le 4$ , i.e. $-2\le x\le 2$ . The integers in this r …

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CBSE 2025Set ANNUAL1 markMCQ
$A = \{2, 3, 4\}$
(a) $P(A) = \{2\}, \{3\}, \{4\}$
(b) $P(A) = \{\{2\}, \{3\}, \{4\}, \{2, 3, 4\}\}$
(c) $P(A) = \{\phi, \{2\}, \{3\}, \{4\}, \{2, 3\}, \{3, 4\}, \{4, 2\}, \{2, 3, 4\}\}$
(d) None of these
›Reveal solutionSolution
A set with $n$ elements has $2^n$ subsets. For $A=\{2,3,4\}$ , $n=3$ , so there are $8$ subsets total, exactly the list in option (c).

The power set $P(A)$ is the collection of ALL subsets of $A$ -- including the empty set $\phi$ and $A$ itself.

Here $n(A) = 3$ , so $n(P(A)) = 2^3 = 8$ .

Listing every subset by size:

Size 0: $\phi$

Size 1: $\{2\}, \{3\}, \{4\}$

Size 2: $\{2,3\}, \{3,4\}, \{2,4\}$

Size 3: $\{2,3,4\}$ …
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CBSE 2025Set ANNUAL1 mark
$Q. If set A has 4 elements, then number of subsets is .............$
›Reveal solutionSolution

The number of subsets of a set with $n$ elements is $2^n$ .

Set $A$ has $4$ elements, so the number of subsets is $2^4 = 16$ (this counts the empt …

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CBSE 2025Set sz1 markMCQ
$X = \{0, -1\}$
(a) 4 elements
(b) 3 elements
(c) 2 elements
(d) 6 elements
›Reveal solutionSolution

Reading the paper's $P(A)$ as the power set of the given set $X$ , a 2-element set has $2^2 = 4$ subsets.

Honest note on the printed stem: the question states "If $X = \{0, -1\}$ then $P(A)$ has :" — the set introduced is $X$ , but the power-set notation printed is $P(A)$ , not $P(X)$ . This is almost certainly a printing slip in the original paper (there is no set $A$ defined anywhere in this question), and the only way the question makes sense is if it means the power set of $X$ , i.e. $P(X)$ .

Solving for $P(X)$ : $X = \{0, -1\}$ has $n = 2$ elements.

…

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CBSE 2025Set ANNUAL1 markMCQ
$\{2, 3, 5\}$
(a) 2
(b) 8
(c) $\dfrac{2}{3}$
(d) $\dfrac{2\tan x}{1 - \tan^2 x}$
(e) $\sin 2x$
(f) 10
(g) 20
(h) $\dfrac{2\tan x}{1 + \tan^2 x}$
(i) 4
›Reveal solutionSolution

A set with $n$ elements has exactly $2^n$ subsets; here $n = 3$ gives $2^3 = 8$ .

The set $\{2, 3, 5\}$ has $n = 3$ elements. The number of subsets of a set with $n$ elements is $2^n$ (this includes the empty set $\varnothing$ and the set itself).

…

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CBSE 2024Set ANNUAL1 markMCQ
$A = \{2, 4, 6\}$
(a) 8
(b) 16
(c) 15
(d) None of these
›Reveal solutionSolution

The number of subsets of a set with $n$ elements is $2^n$ .

$A = \{2, 4, 6\}$ has $n(A) = 3$ elements. The total number of subsets of a set (i.e. the number of elements in its power set) is $2^n$ , because each element independently is either included or excluded from a subset.

…

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CBSE 2024Set ANNUAL1 markMCQ
$n$
(a) $2n$
(b) $n^2$
(c) $2^n$
(d) $2^n - 1$ .
›Reveal solutionSolution

A set with $n$ elements has exactly $2^n$ subsets.

For a set $A=\{a_1,a_2,\ldots,a_n\}$ with $n$ elements, to build any subset we decide, for each element, whether to include it or not — that is $2$ independent choices per element. By the multiplication principle, the total number of subsets is …

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CBSE 2024Set sz1 markMCQ
$\{-1, 1\}$
(a) $\phi, \{-1\}$
(b) $\phi$ only
(c) $\phi, \{-1\}, \{1\}, \{-1, 1\}$
(d) $0$
›Reveal solutionSolution
A set with $n$ elements has $2^n$ subsets; the set $\{-1,1\}$ has 2 elements so it has $2^2=4$ subsets: the empty set, the two singletons, and the set itself.

A subset of a set $S$ is any set formed by taking zero or more elements of $S$ (including none, giving the empty set $\phi$ , and all of them, giving $S$ itself).

For $S = \{-1, 1\}$ , which has $n = 2$ elements, the number of subsets is $2^n = 2^2 = 4$ .

Listing them out:

Taking 0 elements: $\phi$

Taking 1 element: $\{-1\}$ , $\{1\}$

Taking 2 elements: $\{-1, 1\}$ …
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CBSE 2024Set ANNUAL1 markMCQ
$\{1, 2\}$
(a) 1
(b) 2
(c) 3
(d) 4
›Reveal solutionSolution

A set with $n$ elements has $2^n$ subsets, so $\{1,2\}$ (2 elements) has $2^2=4$ subsets.

Step 1. The number of subsets of a set with $n$ elements is $2^n$ , since each element is independently either included or excluded.

…

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CBSE 2024Set ANNUAL1 mark
$A = \{1, 2, 3\}$
›Reveal solutionSolution

$A$ is a proper subset of $B$ because every element of $A$ lies in $B$ , but $B$ has an extra element $A$ lacks.

Step 1. Check $A \subseteq B$ : every element of $A=\{1,2,3\}$ — namely $1,2,3$ — is present in $B=\{1,2,3,4\}$ . So $A \subseteq B$ .

Step 2. Check $A\ne B$ : $B$ contains $4$ , which is not in $A$ . So $A\ne B$ .

…

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Q.If AAA and BBB are finite sets such that A⊂BA \subset BA⊂B, then n(A∪B)=n(A \cup B) = n(A∪B)= ______________.

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Q.If $A$ and $B$ are finite sets such that $A \subset B$ , then $n(A \cup B) =$ ______________.

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