Set Theory for Placement Aptitude: Formulas and Worked Problems

Set theory problems appear in the quantitative aptitude sections of TCS NQT, AMCAT, and Infosys placement tests, testing your ability to count overlapping groups using unions, intersections, and the inclusion-exclusion formula.

The core idea is simple: a set is a collection of distinct, unordered elements, and a handful of operations on sets (plus one formula) handle the bulk of aptitude questions. This article covers notation, all five standard operations, and eight fully worked problems from the types that appear in actual placement rounds.

The same placement test section that includes set problems typically also covers calendar and date problems and coding-and-decoding questions, both of which reward understanding the underlying rule over pattern-guessing.

What Is a Set?

A set is a well-defined, unordered collection of distinct objects. “Well-defined” means there’s no ambiguity about membership. “Distinct” means no duplicates: {1, 1, 2} is just {1, 2}.

The Wikipedia article on Set (mathematics) traces the formal theory back to Georg Cantor’s work in the 1870s. For placement aptitude, the applied subset of that theory is what matters.

Notation

Two styles of set notation appear in aptitude questions:

• Roster notation lists elements explicitly: {1, 2, 3} or {a, e, i, o, u}.
• Set-builder notation describes a rule: {x | x is an even integer between 1 and 10} means {2, 4, 6, 8, 10}.

Membership and Special Sets

• If element a belongs to set A, write a ∈ A.
• If element b does not belong, write b ∉ A.
• The empty set ∅ (also written {}) has no elements.
• The universal set U contains all elements under consideration for a given problem.

Sets are labelled with capital letters (A, B, C) and elements with lowercase (a, b, x). Order and repetition do not matter: {2, 4, 6}, {6, 4, 2}, and {4, 2, 6} are the same set.

Set Relationships: Subsets, Power Sets, and Cardinality

Subsets and Proper Subsets

Set A is a subset of B (written A ⊆ B) if every element of A is also in B. A ⊆ A is always true.

Set A is a proper subset of B (written A ⊂ B) if A ⊆ B and A ≠ B. So {1, 2} ⊂ {1, 2, 3}, but {1, 2, 3} is not a proper subset of itself.

Power Set

The power set P(A) is the set of all subsets of A, including ∅ and A itself.

If |A| = n elements, then |P(A)| = 2^n.

For A = {a, b, c}: the subsets are ∅, {a}, {b}, {c}, {a,b}, {a,c}, {b,c}, and {a,b,c}, giving exactly 2^3 = 8 subsets. This count mirrors the binomial coefficients in Pascal’s Triangle: 1 + 3 + 3 + 1 = 8.

Cardinality

The cardinality |A| is the count of elements in a set. If A = {10, 20, 30, 40}, then |A| = 4.

Set Operations

Five operations cover all placement-test scenarios.

Operation	Symbol	Meaning	Example (A = {1,2,3}, B = {3,4,5})
Union	A ∪ B	All elements in A or B (or both)	{1,2,3,4,5}
Intersection	A ∩ B	Only elements in both A and B	{3}
Difference	A − B	Elements in A but not in B	{1,2}
Complement	A’	Elements in U but not in A	Depends on U
Symmetric Difference	A △ B	Elements in A or B but not both	{1,2,4,5}

In set-builder form:

• Union: A ∪ B = {x | x ∈ A or x ∈ B}
• Intersection: A ∩ B = {x | x ∈ A and x ∈ B}
• Difference: A − B = {x | x ∈ A and x ∉ B}
• Complement: A’ = {x | x ∈ U and x ∉ A}
• Symmetric difference: A △ B = (A − B) ∪ (B − A)

The symmetric difference contains elements belonging to exactly one of the two sets.

Cartesian Product

A × B (read “A cross B”) is the set of all ordered pairs (a, b) where a ∈ A and b ∈ B.

If A = {1, 2} and B = {x, y}, then A × B = {(1,x), (1,y), (2,x), (2,y)}.

The size rule: |A × B| = |A| × |B|. So |A × B| = 2 × 2 = 4 in this example.

Unlike union or intersection, order matters in a Cartesian product: (1, x) and (x, 1) are different pairs.

Venn Diagrams for Aptitude Problems

A Venn diagram represents sets as overlapping circles inside a rectangle (the universal set U).

For two sets A and B:

• Left region (A only): elements in A but not B.
• Overlapping centre region: elements in A ∩ B.
• Right region (B only): elements in B but not A.
• Rectangle outside both circles: elements in neither A nor B.

When a problem states “35 like Maths, 30 like Science, 15 like both, class of 60,” the diagram fills as:

• Left (Maths only): 35 − 15 = 20
• Centre (both): 15
• Right (Science only): 30 − 15 = 15
• Outside (neither): 60 − (20 + 15 + 15) = 10

For three sets, add a third overlapping circle. Fill the centre region (all three overlap) first, then pairwise overlaps, then individual regions, then “none” last.

The Inclusion-Exclusion Principle

Two Sets

• |A ∪ B| = |A| + |B| - |A ∩ B|

The subtraction corrects for double-counting: elements in A ∩ B were counted once in |A| and once in |B|.

Neither (Complement of Union)

• |neither| = |U| - |A ∪ B|

Three Sets

• |A ∪ B ∪ C| = |A| + |B| + |C| - |A∩B| - |B∩C| - |A∩C| + |A∩B∩C|

Subtracting all three pairwise intersections over-subtracts each pair by one; adding back the triple intersection restores the correct count.

Derived Formulas

• |A - B| = |A| - |A ∩ B|
• |A’| = |U| - |A|
• If A and B are disjoint (no overlap): |A ∪ B| = |A| + |B|

Worked Aptitude Problems

Practice sets for these problem types are available at IndiaBix Venn Diagrams. The eight problems below cover every operation and question type found in typical placement rounds.

P1: Basic Inclusion-Exclusion (Two Sets)

• Problem: In a class of 60 students, 35 like Maths and 30 like Science. If 15 like both, how many like neither?
• Step 1: |M ∪ S| = 35 + 30 - 15 = 50
• Step 2: Neither = 60 - 50 = 10
• Answer: 10 students like neither subject.

P2: Finding the Intersection from the Union

• Problem: In a batch of 80 students, 50 like Cricket and 45 like Football. 25 like neither sport. How many like both?
• Step 1: Like at least one = 80 - 25 = 55
• Step 2: |C ∩ F| = 50 + 45 - 55 = 40
• Answer: 40 students like both sports.

P3: Subset Count and Proper Subsets

• Problem: How many subsets does {p, q, r, s} have? How many are proper subsets?
• Step 1: Total subsets = 2^4 = 16
• Step 2: Proper subsets exclude the set itself: 16 - 1 = 15
• Answer: 16 subsets total; 15 proper subsets.

P4: Cartesian Product

• Problem: A = {1, 2}, B = {3}. List A × B and state its size.
• Step 1: A × B = {(1,3), (2,3)}
• Step 2: |A × B| = 2 × 1 = 2
• Answer: A × B = {(1,3), (2,3)}, size 2.

P5: Symmetric Difference

• Problem: A = {1, 2, 3, 4}, B = {3, 4, 5, 6}. Find A △ B.
• Step 1: A - B = {1, 2} (in A but not B)
• Step 2: B - A = {5, 6} (in B but not A)
• Step 3: A △ B = (A - B) ∪ (B - A) = {1, 2, 5, 6}
• Answer: A △ B = {1, 2, 5, 6}.

P6: Complement

• Problem: U = {1,2,3,4,5,6,7,8}, A = {2,4,6,8}. Find A’.
• Step 1: A’ contains elements of U not in A.
• Answer: A’ = {1, 3, 5, 7}.

P7: Three-Set Inclusion-Exclusion

• Problem: 100 students; 45 study Physics, 38 study Chemistry, 27 study Biology. Pairwise overlaps: Physics ∩ Chemistry = 15, Chemistry ∩ Biology = 10, Physics ∩ Biology = 12. All three = 5. How many study at least one subject?
• Step 1: |P ∪ C ∪ B| = 45 + 38 + 27 - 15 - 10 - 12 + 5
• Step 2: Sum = 110; pairwise sum = 37; 110 - 37 + 5 = 78
• Answer: 78 students study at least one subject.

P8: Set Difference and Cardinality

• Problem: A = {1,2,3,4,5}, B = {4,5,6,7}. Find |A - B|.
• Step 1: A - B = elements in A not in B = {1, 2, 3}
• Step 2: |A - B| = 3
• Verification using formula: |A - B| = |A| - |A ∩ B| = 5 - 2 = 3 ✓
• Answer: |A - B| = 3.

Set Theory in Code and AI Pipelines

The inclusion-exclusion formula you just applied across eight problems isn’t only a placement aptitude tool. Python’s built-in set type implements every operation covered here: union(), intersection(), difference(), and symmetric_difference() are all single-line calls. The same membership-counting logic appears in AI feature pipelines whenever document corpora or knowledge bases are merged, de-duplicated, or compared.

If you want to go from placement aptitude to building with language models, TinkerLLM (₹299) is a practical starting point. The set-based filtering logic you applied in P1 to P8 shows up repeatedly in retrieval and ranking code once you get there.