On this page
Most students can recite "a set with n elements has 2ⁿ subsets". Fewer can say why, and that is what the harder questions test: subsets with conditions, power sets of sets that contain sets, and equations in 2ⁿ. This post explains the reason in one line, then uses it on six question types, with a check for each.
Is the power set still in your syllabus?
This is a topic where the documents disagree, so read them carefully.
The JEE (Main) 2026 syllabus lists "Power set" in Unit 1, Sets, Relations and Functions.
The rationalised NCERT Class 11 Sets chapter (2026-27 reprint) runs from 1.1 Introduction to 1.10 Complement of a Set and covers subsets, but has no section on the power set.
The CBSE Mathematics curriculum for 2026-27 lists sets, subsets, Venn diagrams and operations for Class 11, without the power set.
So a CBSE student may never meet the power set in school, while a JEE Main aspirant is expected to know it. If that is you, this post covers what you need. Check the syllabus for your own exam year.
Why does a set with n elements have 2ⁿ subsets?
Build a subset by walking through the elements one at a time and deciding, for each, "in" or "out". That is 2 choices for each of n elements, so by the multiplication principle there are 2 × 2 × … × 2 = 2ⁿ subsets.
For A = {a, b, c}, the eight choices give ∅, {a}, {b}, {c}, {a, b}, {a, c}, {b, c} and {a, b, c}. "Out, out, out" gives the empty set; "in, in, in" gives A itself.
The power set P(A) is the set of all these subsets, so n(P(A)) = 2ⁿ. Two related counts:
Proper subsets (all except A itself): 2ⁿ − 1.
Non-empty proper subsets: 2ⁿ − 2.
Worked example 1: conditions on the subsets
How many subsets of {1, 2, 3, 4, 5, 6} contain 1 but not 2?
The choices for 1 and 2 are fixed. The other four elements are free: 2⁴ = 16.
How many subsets of a 5-element set have at least two elements? Total 2⁵ = 32. Remove the empty set and the 5 single-element sets: 32 − 1 − 5 = 26.
How many have an even number of elements? Count by size: C(5, 0) + C(5, 2) + C(5, 4) = 1 + 10 + 5 = 16, which is exactly half of 32. This half-and-half split holds for every n ≥ 1, and our post on the binomial theorem for JEE shows the identity behind it.
The sizes also give a check: C(5, 0) + C(5, 1) + … + C(5, 5) = 1 + 5 + 10 + 10 + 5 + 1 = 32 = 2⁵.
Worked example 2: sets that contain sets
Let A = {1, {2}, ∅}. How many elements does P(A) have?
A has three elements: the number 1, the set {2} and the empty set ∅. So n(P(A)) = 2³ = 8. Among those subsets are ∅ (choose nothing) and {∅} (choose only the element ∅). These are different: the first has no elements, the second has one.
Now the nested chain:
P(∅) = {∅}, which has 1 element.
P(P(∅)) = {∅, {∅}}, which has 2 elements.
P(P(P(∅))) has 2² = 4 elements.
Each step applies 2ⁿ to the size of the previous set. Students who write n(P(∅)) = 0 are confusing "has no elements" with "has no subsets".
Worked example 3: solving an equation in 2ⁿ
The power sets of A and B differ in size by 112, with A the larger set. Find n(A) and n(B).
We need 2ᵃ − 2ᵇ = 112. Factor out the smaller power: 2ᵇ(2ᵃ⁻ᵇ − 1) = 112 = 16 × 7. The bracket is odd, so it must be 7, and 2ᵇ = 16. That gives b = 4 and 2ᵃ⁻ᵇ = 8, so a = 7.
Check: 2⁷ − 2⁴ = 128 − 16 = 112. Factoring is faster and safer than guessing.
Worked example 4: pairs of subsets
Let S = {1, 2, 3}. How many ordered pairs (A, B) are there with A ⊆ B ⊆ S?
Use the row-of-choices idea, but now each element has three options: in neither set, in B only, or in both. So the count is 3³ = 27. A computer check that lists all 8 × 8 pairs and keeps those with A ⊆ B also finds 27.
A similar argument shows that the number of ordered pairs of disjoint subsets of S is also 3³: each element is in A only, B only, or neither.
Worked example 5: power sets and operations
Is P(A) ∪ P(B) the same as P(A ∪ B)? Try A = {1} and B = {2}.
P(A) ∪ P(B) = {∅, {1}, {2}}, with 3 elements. P(A ∪ B) has 2² = 4 elements, including {1, 2}, which is in neither P(A) nor P(B). So in general P(A) ∪ P(B) ⊆ P(A ∪ B), with equality only in special cases.
Intersection behaves better: P(A) ∩ P(B) = P(A ∩ B) always, because a set is a subset of both A and B exactly when it is a subset of A ∩ B.
Which mistakes cost the most marks?
Forgetting ∅ or the set itself when listing subsets.
Confusing ∅ with {∅}, and ∈ with ⊆. For A above, ∅ ∈ A and also ∅ ⊆ A; {2} ∈ A but {2} ⊄ A, since 2 is not an element of A.
Writing 2ⁿ for proper subsets. It is 2ⁿ − 1.
Counting inner elements. {1, {2}} has 2 elements, the number 1 and the set {2}. Count top-level elements only.
Assuming P(A ∪ B) = P(A) ∪ P(B). Test a small case.
A practice routine
Day 1: For three small sets, list every subset by the in/out method and count.
Day 2: Ten condition questions ("contains", "does not contain", "at least k").
Day 3: Five nested-brace questions. Count top-level elements aloud first.
Day 4: Five equations in 2ⁿ and two pair-counting questions using 3ⁿ.
For Venn diagram counting and relations, which come next in the same unit, read our post on sets and relations for Class 11. If you want a guided start on a weak area, Fermi's AI tutor can work out what you already know and build a plan for the session.
Try this now: how many subsets of {1, 2, …, 8} contain at least one even number?
