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Sets revision notes

Every idea in the Sets chapter for JEE Main, with the usual mistakes and the fast methods. Topic 1 is open to everyone; sign in free to read all 13 and download the PDF.

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How we count: 7× asked means 7 past JEE Main questions in the PrepWiser bank needed that concept, and "last" is the latest year one appeared. ★ marks ideas asked 8 or more times.

Sets, types and subsets

Roster and set-builder form
7× asked, last 2025
{x∈Z:x2<9}={−2,−1,0,1,2}\{x\in\mathbb Z: x^2<9\}=\{-2,-1,0,1,2\}
Roster lists each element once; order and repeats do not change a set. In JEE, N={1,2,3,… }\mathbb N=\{1,2,3,\dots\}.
  • Not one element per pair (m,n)(m,n). Equal values are one element. {m/n:m<n; m,n∈{1,2,3,4}}\{m/n: m<n;\ m,n\in\{1,2,3,4\}\} has 6 pairs but 5 elements, since 2/4=1/22/4=1/2. Count distinct values.
  • Not starting {3k+1:k∈N}\{3k+1: k\in\mathbb N\} at 1. k≥1k\ge1, so the set starts at 4. One stray first term changes a sum of elements by exactly that term.
  • Quick method. Elements of A∩BA\cap B given by formulas like {ak+b}\{ak+b\} in a range: write the common elements as one AP (step = lcm of the steps), then sum as (number of terms) ×\times (first + last)/2.
Types of sets
n(∅)=0,n({∅})=1n(\varnothing)=0,\qquad n(\{\varnothing\})=1
Empty, singleton, finite, infinite. Equal: same elements. Equivalent: same number of elements. UU holds every set in the problem.
Subsets and power set
4× asked, last 2024
n(P(A))=2n,proper subsets=2n−1,non-empty proper=2n−2n(P(A))=2^{n},\qquad \text{proper subsets}=2^{n}-1,\qquad \text{non-empty proper}=2^{n}-2
n=n(A)n=n(A). A⊆BA\subseteq B when every element of AA is in BB; ∅\varnothing and AA itself are always subsets of AA.
∅\varnothing{a}\{a\}{b}\{b\}{c}\{c\}{a,b}\{a,b\}{a,c}\{a,c\}{b,c}\{b,c\}{a,b,c}\{a,b,c\}
P({a,b,c})P(\{a,b,c\}): 23=82^3=8 subsets; drop the top one for the 7 proper subsets, and ∅\varnothing too for the 6 non-empty proper ones
  • Nested power sets: n(P(P(A)))=22nn(P(P(A)))=2^{2^n}. P(∅)={∅}P(\varnothing)=\{\varnothing\} has 1 element and P(P(∅))P(P(\varnothing)) has 2. Elements of P(A)P(A) are sets: a∈Aa\in A gives {a}∈P(A)\{a\}\in P(A).
  • Quick method. 2m−2n=k2^m-2^n=k: write k=2n(2m−n−1)k=2^n(2^{m-n}-1). The power of 2 in kk gives nn, the odd part gives m−nm-n. E.g. 48=16×348=16\times3 gives n=4n=4, m=6m=6.
Counting subsets with a condition
5× asked, last 2026
#{C⊆A:C∩B≠∅}=2n(A)−2n(A−B),#{C⊆A:x∈C}=2n(A)−1  (x∈A)\#\{C\subseteq A: C\cap B\ne\varnothing\}=2^{n(A)}-2^{n(A-B)},\qquad \#\{C\subseteq A: x\in C\}=2^{n(A)-1}\ \ (x\in A)
Count the complement: subsets that avoid BB are the subsets of A−BA-B. With kk fixed elements of AA forced in: 2n−k2^{n-k}.
  • 'Either p∉Tp\notin T or q∈Tq\in T': the failures have p∈Tp\in T, q∉Tq\notin T, which is 2n−22^{n-2} subsets, so 2n−2n−22^n-2^{n-2} pass. For B∪CB\cup C, add only the members of CC that fail BB.
  • Not stopping after the complement when a size rule is added. 'product even' is all minus the all-odd subsets (∅\varnothing goes with them); 'at least 2 elements' then drops single evens. {1,2,3,4}\{1,2,3,4\}: 16−4−2=1016-4-2=10.
  • Quick method. Sum not a multiple of 3: sort elements by remainder. Multiples of 3 are free (factor 2 each); count remainder patterns of the rest. {1,2,3}\{1,2,3\}: {1},{2}\{1\},\{2\}, times 2 for the 3, gives 4.

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Practise Sets

9 more concepts in 3 topics

Intervals and set operations, Cardinality and Venn diagrams and Cartesian product and point sets. Read the first part free. Sign in free with Google to read the rest and download the PDF.

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Intervals and set operations

  • Intervals and inequality sets10×
  • Union, intersection, difference, complement6×
  • Algebraic laws and De Morgan's laws3×

Cardinality and Venn diagrams

  • Two sets: inclusion–exclusion7×
  • Three-set inclusion–exclusion4×
  • Exactly one, exactly two, at least one3×
  • Largest and smallest overlap4×

Cartesian product and point sets

  • Cartesian product1×
  • Sets of points in the plane3×

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About 40 past JEE Main questions from this chapter, with instant answer checking and step-by-step solutions where available. Free, and you can start without signing up.

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