Q.Decide, among the following sets, which sets are subsets of one and another: A = { x : x ∈ R and x satisfy x2 – 8x + 12 = 0 }, B = { 2, 4, 6 }, C = { 2, 4, 6, 8, . . . }, D = { 6 }.
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⊆A
If B is not a subset of A, we write:
B⊆A
Key points to remember:
-
Every set is a subset of itself.
Example: {apple, banana} ⊆ {apple, banana}
-
The empty set ∅ (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=A, we call B a proper subset.
Notation: B⊂A (some books use ⊊)
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:
- ∅ (empty set)
- {a}
- {b}
- {a,b} (the set itself)
So the list of all subsets is:
{∅,{a},{b},{a,b}}
How many subsets does a set have?
If a set has n elements, it has exactly 2n subsets.
- n=0 → 20=1 subset (just the empty set)
- n=1 → 21=2 subsets
- n=2 → 22=4 subsets (as above)
- n=3 → 23=8 subsets
Why 2n?
For each element, you have 2 choices: include it or exclude it. Multiply these choices: 2×2×⋯×2 (n times) = 2n.
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 2n−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 | ∅ |
| 001 | {c} |
| 010 | {b} |
| 011 | {b,c} |
| 100 | {a} |
| 101 | {a,c} |
| 110 | {a,b} |
| 111 | {a,b,c} |
That's all 8 subsets.
5. Why does this matter?
Subset listing is the foundation for:
- Probability (sample spaces and events)
- Combinatorics (counting possibilities)
- Set theory (understanding relationships between sets)
- Computer science (power sets, Boolean algebra)
Quick Check: Test Yourself
Q: List all subsets of T={x,y}.
Answer:
∅, {x}, {y}, {x,y}
Q: How many subsets does a set with 5 elements have?
Answer: 25=32
Remember: Subset listing is just systematically writing down every possible selection from a set — from picking nothing to picking everything. That's all there is to it.
Searches like "how to list all subsets of a set" and "number of subsets formula 2 to the power n" point straight to the Sets chapter of the NCERT/CBSE Class 11 Mathematics syllabus, where subset listing is introduced. The binary-counting method for listing subsets is also a handy shortcut for JEE Main set-theory and probability questions.
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 2n 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 = 2n
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:
2×2×2=23=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:
Total subsets=n times2×2×⋯×2=2n
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 2n. If a question asks "How many subsets does a set with 5 elements have?", you can instantly say 25=32. But if they ask why, you now have the reasoning.
- Watch out for "proper subsets". A proper subset is any subset except the original set itself. So the number of proper subsets is 2n−1.
- Watch out for "non-empty subsets". That's 2n−1 as well (excluding the empty set).
- Watch out for "non-empty proper subsets". That's 2n−2 (excluding both the empty set and the original set).
5. Quick Summary Table
| Type of Subset | Formula | Reasoning |
|---|---|---|
| All subsets | 2n | n independent binary choices (include/exclude) |
| Proper subsets | 2n−1 | All subsets minus the set itself |
| Non-empty subsets | 2n−1 | All subsets minus the empty set |
| Non-empty proper subsets | 2n−2 | All subsets minus the set itself and the empty set |
Final takeaway: The formula 2n is not magic. It's the product of n independent "yes/no" decisions. Always trace back to that binary choice when you need to derive or explain it.
Concept: Set Membership and Subset Relations
First, identify each set explicitly.
Set A contains real solutions to x2−8x+12=0. Factoring: (x−2)(x−6)=0, so x=2 or x=6. Thus A={2,6}.
Set B={2,4,6} is given.
Set C={2,4,6,8,…} is the set of all positive even integers.
Set D={6} is a singleton.
Now check subset relations (X⊆Y means every element of X is in Y):
- D={6}⊆A={2,6} ✓
- D={6}⊆B={2,4,6} ✓
- D={6}⊆C (since 6 is even) ✓
- A={2,6}⊆B={2,4,6} ✓
- A={2,6}⊆C (both are even) ✓
- B={2,4,6}⊆C (all are even) ✓
The subset relations are: D⊆A⊆B⊆C, D⊆B, and D⊆C.
Solve the quadratic to find A={2,6}, then check every pair: D⊂A⊂B⊂C forms a chain, with several other subset relations holding as well.
The question asks us to identify all subset relationships among four sets. A set X is a subset of Y (written X⊆Y) when every element of X also belongs to Y. The strategy is straightforward: first determine what each set actually contains, then systematically compare them.
Finding set A
Set A is defined by a condition: x∈R satisfying x2−8x+12=0.
Factoring the quadratic:
x2−8x+12=(x−2)(x−6)=0
So x=2 or x=6, giving us A={2,6}.
Identifying the other sets
- B={2,4,6} is explicitly listed
- C={2,4,6,8,…} is the set of all positive even integers
- D={6} is a singleton set
Checking all subset relationships
Now we compare each pair. There are (24)=6 pairs to check, plus we should verify if any set is a subset of itself (which is always true, but trivial).
1. Is A⊆B?
A={2,6} and B={2,4,6}. Both 2 and 6 are in B, so yes, A⊆B.
2. Is A⊆C?
C contains all positive even integers. Since 2 and 6 are both positive and even, yes, A⊆C.
3. Is A⊆D?
D={6} contains only 6, but A contains 2 as well. So no, A⊆D.
4. Is B⊆A?
B contains 4, which is not in A={2,6}. So no, B⊆A.
5. Is B⊆C?
B={2,4,6} and all three elements are positive even integers, so they're all in C. Yes, B⊆C.
6. Is B⊆D?
B has three elements but D has only one. No, B⊆D.
7. Is C⊆A?
C is infinite while A has only two elements. No, C⊆A.
8. Is C⊆B?
C contains 8,10,12,… which are not in B. No, C⊆B.
9. Is C⊆D?
C is much larger than the singleton D. No, C⊆D.
10. Is D⊆A?
D={6} and 6∈A. Yes, D⊆A.
11. Is D⊆B?
6∈B, so yes, D⊆B.
12. Is D⊆C?
6 is a positive even integer, so yes, D⊆C.
Notice the chain: D⊂A⊂B⊂C. Each set in this sequence is properly contained in the next, which automatically gives us many of the subset relations.
Summary of all subset relationships
| Subset relation | Valid? |
|---|---|
| D⊆A | ✓ |
| D⊆B | ✓ |
| D⊆C | ✓ |
| A⊆B | ✓ |
| A⊆C | ✓ |
| B⊆C | ✓ |
All other potential subset relations (like B⊆A, C⊆B, etc.) are false.
The subset relationships are: D⊆A⊆B⊆C, along with D⊆B and D⊆C (which follow from transitivity).
Method: Direct Set Listing and Subset Testing
Step 1: List each set explicitly
Set A — Solve x2−8x+12=0
Factor: (x−2)(x−6)=0
So x=2 or x=6
Thus A = { 2, 6 }
Set B = { 2, 4, 6 }
Set C = { 2, 4, 6, 8, … } (all positive even integers)
Set D = { 6 }
Step 2: Test subset relationships
A set X is a subset of Y (X⊆Y) if every element of X is also in Y.
-
D ⊆ A?
D = {6}, A = {2, 6} → 6 ∈ A → Yes
-
D ⊆ B?
6 ∈ B → Yes
-
D ⊆ C?
6 ∈ C → Yes
-
A ⊆ B?
A = {2, 6}, B = {2, 4, 6} → both 2 and 6 are in B → Yes
-
A ⊆ C?
2 and 6 are both even positive integers → Yes
-
B ⊆ C?
B = {2, 4, 6}, C = {2, 4, 6, 8, …} → all three are in C → Yes
-
B ⊆ A?
4 ∉ A → No
-
C ⊆ B?
8 ∉ B → No
Step 3: Summarise all subset relations
D ⊆ A ⊆ B ⊆ C
Also:
- D ⊆ B, D ⊆ C
- A ⊆ B, A ⊆ C
- B ⊆ C
Final Answer:
The chain of subsets is:
D⊆A⊆B⊆C
Common Mistakes on Set Membership & Subset Questions
Mistake 1: Solving the Quadratic Incorrectly
The error: Students often factor x2−8x+12=0 as (x−6)(x−2)=0 but then write the solution set as {6,−2} or {2,−6}.
Why it happens: Rushing through factorization or sign errors.
How to avoid: Always double-check:
- x2−8x+12=0
- Factors: (x−2)(x−6)=0
- Solutions: x=2 or x=6
- So A = {2, 6}
Pro tip: Expand your factors mentally to verify: (x−2)(x−6)=x2−8x+12 ✓
Mistake 2: Confusing Set C's Definition
The error: Students think C = {2, 4, 6, 8, ...} means "all even numbers" and incorrectly include 0 or negative numbers.
Why it happens: The "..." notation is ambiguous — students assume it starts from 2 and continues indefinitely.
How to avoid: Read the pattern carefully:
- C starts at 2 and increases by 2 each step
- So C = {2, 4, 6, 8, 10, ...} — only positive even numbers
- C does NOT contain 0, -2, -4, etc.
Mistake 3: Forgetting the Empty Set & Improper Subsets
The error: Students list subsets but forget that every set is a subset of itself.
Why it happens: Focusing only on "proper subsets" when the question asks for all subsets.
How to avoid: Remember the definition:
- Set X is a subset of set Y if every element of X is also in Y
- This is always true when X = Y
For this problem:
- A ⊆ A ✓ (always)
- B ⊆ B ✓
- C ⊆ C ✓
- D ⊆ D ✓
Mistake 4: Missing the D ⊆ A Relationship
The error: Students see D = {6} and A = {2, 6} but say "D is not a subset of A because A has an extra element."
Why it happens: Misunderstanding that subsets can be smaller — a subset doesn't need to contain all elements of the superset.
How to avoid: Check element-by-element:
- D = {6}
- Is 6 ∈ A? Yes (A = {2, 6})
- Therefore D ⊆ A ✓
Key insight: A subset can be "smaller" — it just needs every element to belong to the larger set.
Mistake 5: Incorrectly Claiming A ⊆ B or B ⊆ A
The error: Students say A ⊆ B because both contain 2 and 6, forgetting that A also contains 2.
Why it happens: Not checking every element systematically.
How to avoid: Use the element test:
- A = {2, 6}, B = {2, 4, 6}
- Is every element of A in B? 2 ∈ B ✓, 6 ∈ B ✓ → A ⊆ B ✓
- Is every element of B in A? 4 ∈ A? No → B ⊈ A ✗
Mistake 6: Confusing "∈" with "⊆"
The error: Writing "6 ⊆ A" instead of "6 ∈ A" or "D ⊆ A".
Why it happens: Mixing up element membership vs. subset notation.
How to avoid: Remember:
- ∈ = "is an element of" (for individual elements)
- ⊆ = "is a subset of" (for sets)
Correct usage:
- 6 ∈ A ✓ (6 is an element of set A)
- D ⊆ A ✓ (set D is a subset of set A)
- {6} ⊆ A ✓ (the set containing 6 is a subset of A)
Final Correct Answer
A = {2, 6}, B = {2, 4, 6}, C = {2, 4, 6, 8, ...}, D = {6}
Subset relationships:
- A ⊆ B ✓ (2 and 6 are both in B)
- D ⊆ A ✓ (6 is in A)
- D ⊆ B ✓ (6 is in B)
- D ⊆ C ✓ (6 is in C)
- A ⊆ C ✓ (2 and 6 are both in C)
- B ⊆ C ✓ (2, 4, 6 are all in C)
- Every set is a subset of itself
Showing the 12 most recent of 26 on this concept.
- CBSE 2026Set ANNUAL1 markMCQQ.The number of all non-empty subsets of the set {1,2,3} is:(a) 8(b) 7(c) 3(d) 9
›Reveal solutionSolution
A set of n elements has 2n subsets in total; removing the empty subset gives 2n−1 non-empty subsets.
For the set {1,2,3}, n=3.
Total subsets =2n=23=8 (this includes ∅ and the full set itself).
Non-empty subsets =2n−1=8−1=7.
Listing them confirms it: {1},{2},{3},{1,2},{1,3},{2,3},{1,2,3} — exactly 7 subsets.
✓Final answerThe correct option is (b) 7.
- CBSE 2026Set ANNUAL1 markQ.Define subset.
›Reveal solutionSolution
A⊆B means every element of A belongs to B.
Formally, A is a subset of B if, for every element x, x∈A⇒x∈B.
For example, {1,2} is a subset of {1,2,3}, since both 1 and 2 are also elements of {1,2,3}.
✓Final answerA set A is a subset of set B (A⊆B) if every element of A is also an element of B.
- CBSE 2026Set 1A1 markQ.List all the elements of the set C={x:x is an integer,x2≤4}.
›Reveal solutionSolution
Integers with x2≤4 are −2,−1,0,1,2.
We need integers x with x2≤4, i.e. −2≤x≤2. The integers in this range are −2,−1,0,1,2.
✓Final answerC={−2,−1,0,1,2}.
- CBSE 2025Set ANNUAL1 markMCQQ.All possible subsets of set A={2,3,4} are(a) P(A)={2},{3},{4}(b) P(A)={{2},{3},{4},{2,3,4}}(c) P(A)={ϕ,{2},{3},{4},{2,3},{3,4},{4,2},{2,3,4}}(d) None of these
›Reveal solutionSolution
A set with n elements has 2n 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 ϕ and A itself.
Here n(A)=3, so n(P(A))=23=8.
Listing every subset by size:
- Size 0: ϕ
- Size 1: {2},{3},{4}
- Size 2: {2,3},{3,4},{2,4}
- Size 3: {2,3,4}
That totals 1+3+3+1=8 subsets, matching option (c) exactly ({4,2} is the same set as {2,4}, just written with elements swapped).
✓Final answer(c) P(A)={ϕ,{2},{3},{4},{2,3},{3,4},{4,2},{2,3,4}}
- CBSE 2025Set ANNUAL1 markQ.If set A has 4 elements, then number of subsets is .............
›Reveal solutionSolution
The number of subsets of a set with n elements is 2n.
Set A has 4 elements, so the number of subsets is 24=16 (this counts the empty set and A itself too).
✓Final answerNumber of subsets =16.
- CBSE 2025Set sz1 markMCQQ.If X={0,−1} then P(A) has :(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 22=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.
The power set of a set with n elements always has 2n elements, because each element can independently be included or excluded from a subset.
So ∣P(X)∣=22=4.
Listing them explicitly confirms this: P(X)={∅,{0},{−1},{0,−1}} — exactly 4 subsets.
✓Final answerThe correct option is (a) 4 elements.
- CBSE 2025Set ANNUAL1 markMCQQ.Match the column: Column A entry 'Number of subsets of the set {2,3,5}' — find the matching value from Column B.(a) 2(b) 8(c) 32(d) 1−tan2x2tanx(e) sin2x(f) 10(g) 20(h) 1+tan2x2tanx(i) 4
›Reveal solutionSolution
A set with n elements has exactly 2n subsets; here n=3 gives 23=8.
The set {2,3,5} has n=3 elements. The number of subsets of a set with n elements is 2n (this includes the empty set ∅ and the set itself).
So the number of subsets =23=8.
✓Final answerThe number of subsets of {2,3,5} matches option (b) 8.
- CBSE 2024Set ANNUAL1 markMCQQ.If A={2,4,6}, then number of total subsets of A is(a) 8(b) 16(c) 15(d) None of these
›Reveal solutionSolution
The number of subsets of a set with n elements is 2n.
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 2n, because each element independently is either included or excluded from a subset.
Here n=3, so the number of subsets =23=8.
These 8 subsets are: ∅,{2},{4},{6},{2,4},{2,6},{4,6},{2,4,6}.
✓Final answer(a) 8.
- CBSE 2024Set ANNUAL1 markMCQQ.The number of subsets of a set having n elements is:(a) 2n(b) n2(c) 2n(d) 2n−1.
›Reveal solutionSolution
A set with n elements has exactly 2n subsets.
For a set A={a1,a2,…,an} 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
2×2×⋯×2 (n times)=2n.
This count includes the empty set ∅ and the full set A itself.
✓Final answerThe correct option is (c) 2n.
- CBSE 2024Set sz1 markMCQQ.Subsets of set {−1,1} are:(a) ϕ,{−1}(b) ϕ only(c) ϕ,{−1},{1},{−1,1}(d) 0
›Reveal solutionSolution
A set with n elements has 2n subsets; the set {−1,1} has 2 elements so it has 22=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 ϕ, and all of them, giving S itself).
For S={−1,1}, which has n=2 elements, the number of subsets is 2n=22=4.
Listing them out:
- Taking 0 elements: ϕ
- Taking 1 element: {−1}, {1}
- Taking 2 elements: {−1,1}
So the complete list of subsets is ϕ,{−1},{1},{−1,1}, which matches option (C) exactly — option (A) lists only 2 of the 4 subsets, and option (D) is not a set of subsets at all.
✓Final answerThe correct option is (C) ϕ,{−1},{1},{−1,1}.
- CBSE 2024Set ANNUAL1 markMCQQ.The number of subsets of the set {1,2} is:(a) 1(b) 2(c) 3(d) 4
›Reveal solutionSolution
A set with n elements has 2n subsets, so {1,2} (2 elements) has 22=4 subsets.
Step 1. The number of subsets of a set with n elements is 2n, since each element is independently either included or excluded.
Step 2. Here n=2 (elements 1,2), so the number of subsets =22=4.
Step 3. Listing them confirms this: ϕ,{1},{2},{1,2} — exactly 4 subsets.
✓Final answerThe correct option is (D) 4.
- CBSE 2024Set ANNUAL1 markQ.Write true or false: A={1,2,3} is a proper subset of B={1,2,3,4}.
›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⊆B: every element of A={1,2,3} — namely 1,2,3 — is present in B={1,2,3,4}. So A⊆B.
Step 2. Check A=B: B contains 4, which is not in A. So A=B.
Step 3. A subset that is contained in B but not equal to B is called a proper subset.
✓Final answerTrue — A is a proper subset of B.
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