JEE Corner — Sets for JEE Main and Advanced

Why This Section Exists

The rationalised textbook chapter stops at complements — but the official JEE Main syllabus for Unit 1 explicitly lists "Sets and their representation; union, intersection and complement of sets and their algebraic properties; power set", and JEE question setters freely use subset-counting and the cardinality formulas that older textbook editions carried. This section restores every JEE-required tool:

1. Power set $P(A)$ and counting subsets;
2. Laws of the algebra of sets — stated as a usable toolkit with proofs;
3. Cardinality formulas — $n(A \cup B)$, its three-set version, and survey problems;
4. Symmetric difference $A \triangle B$ — a JEE favourite for one-liner MCQs.

The worked examples and practice questions here are modelled on the JEE Main and Advanced pattern and difficulty; they are practice questions in the exam style, not reproductions of specific past papers.

[Board Note] CBSE students: power set and $n(A \cup B)$ counting are no longer in your syllabus. Treat this section as enrichment — the ideas are short and they deepen everything you learned in Sections 1-4.

The Power Set and Counting Subsets

Key Point (Definition): The power set of A, written $P(A)$, is the collection of all subsets of A — including $\phi$ and A itself. In $P(A)$, every element is itself a set.

For A = {1, 2}: $P(A) = \{\phi, \{1\}, \{2\}, \{1, 2\}\}$.

The master count

Building a subset of an n-element set means making n independent in-or-out decisions — one per element:

$$n(A) = m \quad \Rightarrow \quad n(P(A)) = 2^m$$

The counting toolkit that follows from it:

• Subsets of an m-element set: $2^m$
• Proper subsets: $2^m - 1$ (exclude the set itself)
• Non-empty proper subsets: $2^m - 2$
• Subsets containing a fixed element: $2^{m-1}$ (fix it in; the rest choose freely)
• Subsets of even size / odd size: $2^{m-1}$ each (a neat symmetry)
• $n(P(P(A)))$ when $A = \phi$: $P(\phi) = \{\phi\}$ has 1 element, so $P(P(\phi))$ has $2^1 = 2$ elements.

Key Point: Distinguish $\in$ and $\subset$ inside power sets: $\{1\} \in P(A)$ but $\{\{1\}\} \subset P(A)$. Membership in $P(A)$ means being a subset of A: $X \in P(A) \iff X \subset A$.

[JEE Tip] "If a set has 240 subsets more than another…" — such problems reduce to solving $2^m - 2^k = 240$. Factor out the smaller power: $2^k(2^{m-k} - 1) = 240 = 16 \times 15$, forcing $2^k = 16$ and $2^{m-k} - 1 = 15$, so $k = 4$, $m = 8$. The even-times-odd factorisation cracks every problem of this family.

The Laws of the Algebra of Sets — the Full Toolkit

Collected in one place, these are the rules that let you manipulate set expressions as fluently as algebraic ones:

• Idempotent: $A \cup A = A$; $A \cap A = A$
• Identity: $A \cup \phi = A$; $A \cap U = A$
• Domination: $A \cup U = U$; $A \cap \phi = \phi$
• Commutative: $A \cup B = B \cup A$; $A \cap B = B \cap A$
• Associative: $(A \cup B) \cup C = A \cup (B \cup C)$; likewise for $\cap$
• Distributive: $A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$ and $A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$
• Absorption: $A \cup (A \cap B) = A$; $A \cap (A \cup B) = A$
• Complement: $A \cup A' = U$; $A \cap A' = \phi$; $(A')' = A$; $\phi\,' = U$; $U' = \phi$
• De Morgan: $(A \cup B)' = A' \cap B'$; $(A \cap B)' = A' \cup B'$

Difference identities (constant JEE fodder)

• $A - B = A \cap B'$ — the most useful rewrite in the topic
• $A - B = A - (A \cap B) = (A \cup B) - B$
• $A - (B \cup C) = (A - B) \cap (A - C)$ and $A - (B \cap C) = (A - B) \cup (A - C)$ (De Morgan for differences)

Proof of $A - B = A \cap B'$: $x \in A - B \iff x \in A$ and $x \notin B \iff x \in A$ and $x \in B' \iff x \in A \cap B'$. ∎

[JEE Tip] When an MCQ shows a jungle like $(A' \cup B')' \cup (A \cap B')$, mechanically apply De Morgan and distributivity: $(A' \cup B')' = A \cap B$, so the expression is $(A \cap B) \cup (A \cap B') = A \cap (B \cup B') = A \cap U = A$. Two laws, done. Converting every difference to $\cap$-with-complement form makes such simplifications automatic.

Cardinality Formulas and Survey Problems

For finite sets, counting elements interacts beautifully with the operations. The workhorse — the inclusion-exclusion principle for two sets:

$$n(A \cup B) = n(A) + n(B) - n(A \cap B)$$

Adding $n(A)$ and $n(B)$ counts the overlap twice; subtracting $n(A \cap B)$ repairs it. If A and B are disjoint, this collapses to $n(A \cup B) = n(A) + n(B)$.

Companion formulas, all readable off a Venn diagram:

• $n(A - B) = n(A) - n(A \cap B)$ — "only A" count
• $n(A \triangle B) = n(A) + n(B) - 2\,n(A \cap B)$ — "exactly one of the two"
• $n(A' ) = n(U) - n(A)$ and $n(A' \cap B') = n(U) - n(A \cup B)$ — "neither"

Three sets

$$n(A \cup B \cup C) = n(A) + n(B) + n(C) - n(A \cap B) - n(B \cap C) - n(A \cap C) + n(A \cap B \cap C)$$

Survey-problem protocol ("In a group of 400 people, 250 speak Hindi, 200 speak English…"):

1. Name the sets and write every given number as an $n(\cdot)$ value.
2. Decide what is asked: "at least one" = union, "both" = intersection, "only A" = $n(A) - n(A \cap B)$, "neither" = $n(U) - n(A \cup B)$.
3. Apply inclusion-exclusion; for three sets, filling the 8-region Venn diagram from the innermost region outward is faster and safer.

[JEE Tip] Bounds questions — "what is the minimum possible $n(A \cap B)$?" — use $n(A \cap B) \geq n(A) + n(B) - n(U)$ (from $n(A \cup B) \leq n(U)$) and $n(A \cap B) \leq \min(n(A), n(B))$. These two inequalities settle every max/min survey MCQ.

Symmetric Difference

Key Point (Definition): The symmetric difference of A and B is $A \triangle B = (A - B) \cup (B - A)$ — the set of elements in exactly one of A and B.

Equivalent forms, each useful:

$$A \triangle B = (A \cup B) - (A \cap B) = (A \cap B') \cup (A' \cap B)$$

Example: A = {1, 2, 3, 4}, B = {3, 4, 5, 6}: $A - B$ = {1, 2}, $B - A$ = {5, 6}, so $A \triangle B$ = {1, 2, 5, 6} — the shared 3 and 4 vanish.

Properties worth knowing cold:

• $A \triangle B = B \triangle A$ (commutative — unlike plain difference!)
• $A \triangle A = \phi$ and $A \triangle \phi = A$
• $A \triangle B = \phi \iff A = B$
• Associative: $(A \triangle B) \triangle C = A \triangle (B \triangle C)$

[JEE Tip] "$A \triangle B = A \triangle C$ implies B = C" is TRUE — symmetric difference cancels (unlike $\cup$ and $\cap$ individually). Reason: $\triangle$ B on both sides of $A \triangle B = A \triangle C$ using associativity and $A \triangle A = \phi$. A slick fact that has appeared as a direct MCQ.

Solved Examples

Example 1. Write $P(A)$ for A = {a, b, c} and verify $n(P(A)) = 2^3$.

Solution.

Step 1 — list subsets by size to miss nothing. Size 0: $\phi$. Size 1: {a}, {b}, {c}. Size 2: {a, b}, {a, c}, {b, c}. Size 3: {a, b, c}.

Step 2 — collect into the power set. $P(A) = \{\phi, \{a\}, \{b\}, \{c\}, \{a,b\}, \{a,c\}, \{b,c\}, \{a,b,c\}\}$.

Step 3 — verify the count. $1 + 3 + 3 + 1 = 8 = 2^3$ ✓ — three in/out decisions, one per element.

Takeaway: listing by size is systematic and self-checking; the size-counts 1, 3, 3, 1 are binomial coefficients — a preview of the Binomial Theorem chapter.

Example 2. Find $P(\phi)$, $P(P(\phi))$ and their cardinalities.

Solution.

Step 1 — subsets of $\phi$. The empty set has exactly one subset: itself. So $P(\phi) = \{\phi\}$, and $n(P(\phi)) = 2^0 = 1$.

Step 2 — go one level up. $P(P(\phi))$ is the power set of $\{\phi\}$, a ONE-element set: its subsets are $\phi$ and $\{\phi\}$. So $P(P(\phi)) = \{\phi, \{\phi\}\}$ with $2^1 = 2$ elements.

Step 3 — keep the three objects straight. $\phi$ (size 0), $\{\phi\}$ (size 1), $\{\phi, \{\phi\}\}$ (size 2) — three DIFFERENT sets.

Takeaway: iterating power sets from $\phi$ climbs sizes $0, 1, 2, 4, 16, \ldots$ — each application doubles exponentially; the first two levels are standard trick-question territory.

Example 3. Set A has m elements and set B has k elements. If A has 240 more subsets than B, find m and k.

Solution.

Step 1 — translate to an equation. Subset counts are $2^m$ and $2^k$: $2^m - 2^k = 240$.

Step 2 — factor out the smaller power. $2^k(2^{m-k} - 1) = 240$.

Step 3 — split 240 into even × odd. $240 = 16 \times 15 = 2^4 \times 15$. Since $2^{m-k} - 1$ is ODD, it must equal the odd part 15, and $2^k$ must absorb all the 2's: $2^k = 16$.

Step 4 — solve. $k = 4$; and $2^{m-k} = 16$ gives $m - k = 4$, so $m = 8$. Check: $2^8 - 2^4 = 256 - 16 = 240$ ✓.

Takeaway: every "$2^m - 2^k = N$" problem cracks the same way — split N into (power of 2) × (odd part); the two factors identify $2^k$ and $2^{m-k} - 1$ uniquely.

Example 4. Let A = {1, 2}. Which are true? (i) $\{1\} \in P(A)$ (ii) $1 \in P(A)$ (iii) $\{\{1\}\} \subset P(A)$ (iv) $\phi \in P(A)$ (v) $\phi \subset P(A)$.

Solution.

Step 1 — write out the power set. $P(A) = \{\phi, \{1\}, \{2\}, \{1, 2\}\}$; membership in $P(A)$ means being a SUBSET of A.

(i) Step 2. Is {1} a subset of A? Yes → $\{1\} \in P(A)$. True.

(ii) Step 3. Is the number 1 a subset of A? No — 1 is an element of A, not a set of elements. False.

(iii) Step 4. Unpack {{1}}: its only element is {1}, which IS an element of $P(A)$ (by (i)). So $\{\{1\}\} \subset P(A)$. True.

(iv) Step 5. $\phi \subset A$ holds, so $\phi$ qualifies for membership: $\phi \in P(A)$. True.

(v) Step 6. $\phi$ is a subset of EVERY set, including $P(A)$ itself. True.

Takeaway: (iv) and (v) are both true for different reasons — $\phi$ is simultaneously a member of every power set (because it is a subset of A) and a subset of it (because it is a subset of everything).

Example 5. Simplify: (i) $(A \cap B) \cup (A \cap B')$ (ii) $(A' \cup B')' \cup (A \cap B')$ (iii) $A \cap (A' \cup B)$.

Solution.

(i) Step 1 — factor A out by distributivity. $(A \cap B) \cup (A \cap B') = A \cap (B \cup B')$.

Step 2 — finish with the complement law. $B \cup B' = U$, so the expression is $A \cap U = A$.

(ii) Step 3 — kill the primed bracket with De Morgan. $(A' \cup B')' = (A')' \cap (B')' = A \cap B$.

Step 4 — reuse part (i). The expression becomes $(A \cap B) \cup (A \cap B') = A$.

(iii) Step 5 — distribute. $A \cap (A' \cup B) = (A \cap A') \cup (A \cap B) = \phi \cup (A \cap B) = A \cap B$.

Takeaway: two moves — De Morgan to strip primes off brackets, distributivity to factor — reduce almost any expression; $B \cup B' = U$ and $A \cap A' = \phi$ are the standard finishers.

Example 6. Prove that $A - (B \cup C) = (A - B) \cap (A - C)$.

Solution.

Step 1 — rewrite every difference in intersection form. Using $A - X = A \cap X'$: the left side is $A \cap (B \cup C)'$.

Step 2 — apply De Morgan inside. $(B \cup C)' = B' \cap C'$, so the left side $= A \cap B' \cap C'$.

Step 3 — expand the right side. $(A - B) \cap (A - C) = (A \cap B') \cap (A \cap C') = A \cap A \cap B' \cap C' = A \cap B' \cap C'$ (idempotent and commutative laws).

Step 4 — match. Both sides equal $A \cap B' \cap C'$. ∎

Takeaway: the rewrite $A - X = A \cap X'$ turns difference identities into three-line De Morgan exercises — never element-chase what algebra does mechanically.

Example 7. In a group of 400 people, 250 can speak Hindi and 200 can speak English. Each person speaks at least one of the two languages. How many can speak both?

Solution.

Step 1 — name the sets and translate the data. $n(H) = 250$, $n(E) = 200$. "Each person speaks atleast one" means the union is everyone: $n(H \cup E) = 400$.

Step 2 — write inclusion-exclusion. $n(H \cup E) = n(H) + n(E) - n(H \cap E)$.

Step 3 — substitute and solve. $400 = 250 + 200 - n(H \cap E)$, so $n(H \cap E) = 450 - 400 = 50$.

Step 4 — interpret. 50 people speak both languages — exactly the amount by which the two headcounts overshoot the group size.

Takeaway: "atleast one" pins down the union; the overlap is whatever the totals exceed it by.

Example 8. In a class of 60 students, 35 like cricket, 30 like football, and 10 like neither. Find how many like (i) both games (ii) only cricket (iii) exactly one game.

Solution.

Step 1 — convert "neither" into the union. Students liking atleast one game: $60 - 10 = 50 = n(C \cup F)$.

(i) Step 2 — inclusion-exclusion for the overlap. $n(C \cap F) = n(C) + n(F) - n(C \cup F) = 35 + 30 - 50 = 15$.

(ii) Step 3 — strip the overlap from cricket. Only cricket $= n(C) - n(C \cap F) = 35 - 15 = 20$.

(iii) Step 4 — exactly one. Either by the formula $n(C) + n(F) - 2n(C \cap F) = 35 + 30 - 30 = 35$, or by regions: only-cricket 20 + only-football $(30 - 15) = 15$ → 35 ✓.

Step 5 — global check. Regions: 20 + 15 + 15 + 10 = 60 = class size ✓.

Takeaway: handle "neither" first to get the union, then cascade; the ten-second region check catches most slips.

Example 9. In a survey of 100 students, 72 read newspaper A and 45 read newspaper B. What are the minimum and maximum possible numbers reading both?

Solution.

Step 1 — maximum. The overlap sits inside each set, so it cannot exceed the smaller: $n(A \cap B) \leq \min(72, 45) = 45$ (achieved when B $\subset$ A).

Step 2 — minimum. The union cannot exceed the population: $n(A) + n(B) - n(A \cap B) = n(A \cup B) \leq 100$, so $n(A \cap B) \geq 72 + 45 - 100 = 17$.

Step 3 — state the range. $17 \leq n(A \cap B) \leq 45$, both ends achievable.

Takeaway: two one-line inequalities — union capped by $n(U)$, intersection capped by the smaller set — settle every min/max survey question.

Example 10. In a survey of 60 people, 25 read magazine H, 26 read T, 26 read I; 9 read both H and I, 11 read both H and T, 8 read both T and I; 3 read all three. Find how many read (i) at least one magazine (ii) exactly one magazine.

Solution.

(i) Step 1 — three-set inclusion-exclusion. $n(H \cup T \cup I) = 25 + 26 + 26 - 11 - 8 - 9 + 3$. Compute: singles $77$, pairs $28$, so $77 - 28 + 3 = 52$.

(ii) Step 2 — fill the Venn diagram from the CENTRE outward. All three: 3. Then the pairwise-only regions: H∩T only $= 11 - 3 = 8$; T∩I only $= 8 - 3 = 5$; H∩I only $= 9 - 3 = 6$.

Step 3 — the only-one regions. H only $= 25 - 8 - 6 - 3 = 8$; T only $= 26 - 8 - 5 - 3 = 10$; I only $= 26 - 5 - 6 - 3 = 12$.

Step 4 — answer and check. Exactly one: $8 + 10 + 12 = 30$. Check all regions: $8 + 10 + 12 + 8 + 5 + 6 + 3 = 52$ ✓ matches (i), and $60 - 52 = 8$ read none.

Takeaway: for three sets, the centre-outward fill is faster than formula-juggling and self-checks — all eight regions must sum to $n(U)$.

Example 11. A = {1, 2, 3, 4, 5}, B = {4, 5, 6, 7}. Compute $A \triangle B$ two ways: as $(A - B) \cup (B - A)$ and as $(A \cup B) - (A \cap B)$.

Solution.

Step 1 — way 1: private parts. $A - B = \{1, 2, 3\}$ (strike the shared 4, 5); $B - A = \{6, 7\}$. Union: $\{1, 2, 3, 6, 7\}$.

Step 2 — way 2: union minus intersection. $A \cup B = \{1, \ldots, 7\}$; $A \cap B = \{4, 5\}$; difference: $\{1, 2, 3, 6, 7\}$.

Step 3 — match and count-check. Both routes agree ✓, and $n(A \triangle B) = n(A) + n(B) - 2n(A \cap B) = 5 + 4 - 4 = 5$ ✓.

Takeaway: two routes, one answer — use whichever the data makes cheaper, and the other as verification.

Example 12. Let $X = \{4^n - 3n - 1 : n \in \mathbb{N}\}$ and $Y = \{9(n - 1) : n \in \mathbb{N}\}$. Show that $X \subset Y$ (and hence $X \cup Y = Y$).

Solution.

Step 1 — expand $4^n$ binomially. $4^n = (1 + 3)^n = 1 + 3n + \binom{n}{2}3^2 + \binom{n}{3}3^3 + \cdots$

Step 2 — subtract the linear part. $4^n - 3n - 1 = 3^2\binom{n}{2} + 3^3\binom{n}{3} + \cdots = 9\left[\binom{n}{2} + 3\binom{n}{3} + \cdots\right]$ — a multiple of 9 for every $n \geq 2$; for $n = 1$ the value is $4 - 3 - 1 = 0 = 9 \times 0$, also a multiple of 9.

Step 3 — identify Y. $Y = \{9(n-1) : n \in \mathbb{N}\} = \{0, 9, 18, 27, \ldots\}$ — ALL non-negative multiples of 9.

Step 4 — conclude. Every element of X is a non-negative multiple of 9, so $X \subset Y$, and therefore $X \cup Y = Y$. (The inclusion is proper: $18 \in Y$ but X's elements run $0, 9, 54, 243, \ldots$ — 18 is skipped.)

Takeaway: the binomial expansion of $(1 + 3)^n$ is the standard tool for "multiple of 9" claims about $4^n - 3n - 1$ — an evergreen exam pair.

Example 13. For A = {1, 2, 3, 4, 5, 6}, count the subsets that (i) contain both 1 and 2 (ii) contain 1 but not 6 (iii) contain at least one of 1, 2.

Solution.

(i) Step 1. Fix 1 IN and 2 IN; elements 3, 4, 5, 6 choose freely: $2^4 = 16$.

(ii) Step 2. Fix 1 IN and 6 OUT; elements 2, 3, 4, 5 free: $2^4 = 16$.

(iii) Step 3 — complement count. Subsets avoiding BOTH 1 and 2: the other four elements choose freely, $2^4 = 16$. Total subsets: $2^6 = 64$. Subsets with atleast one of 1, 2: $64 - 16 = 48$.

Takeaway: each fixed element halves the count ($2^{n-k}$ survives); and "atleast one" is fastest by counting the avoiders and subtracting.

Example 14. Given $A \triangle B = A \triangle C$, prove B = C.

Solution.

Step 1 — apply $A \,\triangle$ to both sides. $A \triangle (A \triangle B) = A \triangle (A \triangle C)$.

Step 2 — regroup by associativity. $(A \triangle A) \triangle B = (A \triangle A) \triangle C$.

Step 3 — simplify. $A \triangle A = \phi$ (a set differs from itself nowhere), and $\phi \triangle B = B$, $\phi \triangle C = C$. Hence B = C. ∎

Takeaway: under $\triangle$, $\phi$ is the identity and every set is its own inverse — so $\triangle$ cancels like addition does. It is the ONLY basic set operation with a genuine cancellation law ($\cup$ and $\cap$ individually have none).