Permutation and Combination Questions for Placement Exams

Comprehensive Topic Guide with 30 practice questions, shortcuts, and detailed solutions for TCS, Infosys, Wipro, Google, Microsoft, and all major placement exams.

Last Updated: March 2026

Introduction to Permutation and Combination

Permutation and Combination (P&C) is one of the most conceptual and scoring topics in quantitative aptitude. While many students find it challenging initially, mastering the fundamental principles opens up a world of problem-solving possibilities. This topic tests your logical reasoning and ability to count systematically—skills highly valued by top tech companies.

Why is Permutation and Combination Important?

Product Company Favorite: Google, Microsoft, Amazon frequently ask P&C questions
Logical Thinking: Develops systematic approach to counting problems
Real Applications: Password combinations, arrangements, probability base
Diverse Question Types: Words, numbers, seating, selection, distribution
High Differentiator: Separates top performers from the rest

Types of P&C Questions

Fundamental Principle: Multiplication and Addition rules
Arrangements: Linear, circular, with/without repetition
Selection Problems: Committee formation, team selection
Word Formation: Arranging letters of words
Number Formation: Creating numbers with given digits
Distribution Problems: Identical/distinct objects into groups
Geometric Applications: Points, lines, triangles, quadrilaterals
Derangements: Special arrangement problems

Exam Distribution

Exam Type	Questions	Difficulty Level
TCS NQT	1-2	Easy to Medium
Infosys	1-2	Medium
Google	2-3	Hard
Microsoft	2-3	Medium to Hard
Amazon	1-2	Medium
CAT	2-3	Medium to Hard
Banking	0-1	Easy

Complete Formula Sheet & Shortcuts

Fundamental Principles

Principle	When to Use	Formula
Multiplication Rule	AND situations (sequential)	m × n ways
Addition Rule	OR situations (alternatives)	m + n ways
Permutation (Order matters)	Arrangements	nPr = n!/(n-r)!
Combination (Order doesn't matter)	Selections	nCr = n!/(r!(n-r)!)

Key Formulas

Scenario	Formula	Example
nPn (all items)	n!	5! = 120
nP0	1	Arranging 0 items
nCn	1	Selecting all items
nC0	1	Selecting nothing
nCr = nC(n-r)	Symmetry	10C3 = 10C7
nPr = nCr × r!	Relation	5P3 = 5C3 × 3!

Special Cases

Situation	Formula	Description
Circular Permutation	(n-1)!	Around a table
Circular (necklace)	(n-1)!/2	Clockwise = anti-clockwise
With Repetition	n^r	Each position has n choices
Arranging with duplicates	n!/(p!×q!×r!)	p, q, r are repeats
Selecting r from n identical	1	Only one way

🚀 SHORTCUT TRICKS

Trick 1: Quick nCr Calculation

nCr = [n × (n-1) × ... × (n-r+1)] / r!

Example: 8C3 = (8×7×6)/(3×2×1) = 56

Trick 2: nCr + nC(r+1) = (n+1)C(r+1)

Example: 5C2 + 5C3 = 6C3 = 20

Trick 3: Total Selections from n items

2^n - 1 (excluding empty selection) Example: From 5 items, total non-empty selections = 31

Trick 4: Sum of all nCr

nC0 + nC1 + ... + nCn = 2^n

Trick 5: At Least One Selection

Total - None = 2^n - 1

Trick 6: Selecting from Two Groups

From m of type A and n of type B, select r: Sum over k: mCk × nC(r-k) for valid k

Trick 7: Gap Method

Arrange items, then place restricted items in gaps. Example: Arrange 5 boys, 3 girls with no two girls together: Arrange boys (5!), choose 3 gaps from 6 (6C3), arrange girls (3!) Answer: 5! × 6C3 × 3!

Trick 8: Derangement Formula

!n = n! × (1/0! - 1/1! + 1/2! - ... + (-1)^n/n!) !3 = 2, !4 = 9, !5 = 44

Practice Questions with Solutions

EASY LEVEL (Questions 1-10)

Question 1 [Easy]

How many 3-digit numbers can be formed using digits 1, 2, 3, 4, 5 without repetition?

Solution: We need to fill 3 positions: _ _ _

First digit: 5 choices
Second digit: 4 remaining choices
Third digit: 3 remaining choices

Total = 5 × 4 × 3 = 5P3 = 60 numbers

Question 2 [Easy]

In how many ways can 5 people be arranged in a row?

Solution: Number of permutations of 5 distinct objects = 5! = 5 × 4 × 3 × 2 × 1 = 120 ways

Question 3 [Easy]

How many ways can a committee of 3 be selected from 8 people?

Solution: Order doesn't matter, so we use combinations.

8C3 = (8 × 7 × 6)/(3 × 2 × 1) = 336/6 = 56 ways

Question 4 [Easy]

How many different words can be formed from the letters of the word "MATH"?

Solution: Word "MATH" has 4 distinct letters.

Number of arrangements = 4! = 24 words

Question 5 [Easy]

In how many ways can 4 people sit around a circular table?

Solution: For circular arrangements: (n-1)!

Number of ways = (4-1)! = 3! = 6 ways

Question 6 [Easy]

A bag contains 4 red and 6 blue balls. In how many ways can 3 balls be selected if at least one must be red?

Solution: Method: Total ways - All blue ways

Total ways to select 3 from 10 = 10C3 = 120 Ways to select all blue (from 6 blue) = 6C3 = 20

Ways with at least one red = 120 - 20 = 100 ways

Question 7 [Easy]

How many diagonals does a hexagon have?

Solution: Total lines from 6 vertices = 6C2 = 15 Subtract the 6 sides of hexagon

Number of diagonals = 15 - 6 = 9 diagonals

Formula: n(n-3)/2 = 6×3/2 = 9 ✓

Question 8 [Easy]

In how many ways can the letters of "BALL" be arranged?

Solution: Word "BALL" has 4 letters with L repeated twice.

Arrangements = 4!/2! = 24/2 = 12 ways

Question 9 [Easy]

How many 2-digit even numbers can be formed from digits 1, 2, 3, 4, 5 without repetition?

Solution: For even number, last digit must be 2 or 4 (2 choices).

Case 1: Last digit is 2 First digit: 4 choices (1,3,4,5) = 4 numbers

Case 2: Last digit is 4 First digit: 4 choices (1,2,3,5) = 4 numbers

Total = 4 + 4 = 8 numbers

Question 10 [Easy]

From a group of 5 men and 4 women, how many ways can 3 men and 2 women be selected?

Solution: Select 3 men from 5: 5C3 = 10 Select 2 women from 4: 4C2 = 6

Total ways = 10 × 6 = 60 ways

MEDIUM LEVEL (Questions 11-20)

Question 11 [Medium]

How many 4-letter words can be formed from "COMPUTER" with no letter repeated?

Solution: "COMPUTER" has 8 distinct letters.

We need to arrange 4 letters from 8: 8P4 = 8 × 7 × 6 × 5 = 1680 words

Question 12 [Medium]

In how many ways can 6 boys and 4 girls be seated in a row so that all girls sit together?

Solution: Treat all 4 girls as one block/unit. Now we have: 6 boys + 1 girl block = 7 units

Arrange 7 units: 7! ways Arrange girls within block: 4! ways

Total = 7! × 4! = 5040 × 24 = 120,960 ways

Question 13 [Medium]

How many triangles can be formed by joining the vertices of a decagon (10-sided polygon)?

Solution: A triangle needs 3 vertices. Number of triangles = 10C3 = (10×9×8)/(3×2×1) = 120 triangles

Question 14 [Medium]

In how many ways can 5 prizes be distributed among 4 students if each student can receive any number of prizes?

Solution: Each prize can go to any of 4 students.

Total ways = 4 × 4 × 4 × 4 × 4 = 4^5 = 1024 ways

Question 15 [Medium]

How many 4-digit numbers greater than 5000 can be formed using digits 0, 1, 3, 5, 7 without repetition?

Solution: For number > 5000, first digit must be 5 or 7 (2 choices).

Case 1: First digit is 5 Remaining 3 digits from {0,1,3,7}: 4P3 = 24 numbers

Case 2: First digit is 7 Remaining 3 digits from {0,1,3,5}: 4P3 = 24 numbers

Total = 24 + 24 = 48 numbers

Question 16 [Medium]

In how many ways can the letters of "BANANA" be arranged?

Solution: "BANANA" has 6 letters:

A appears 3 times
N appears 2 times
B appears 1 time

Arrangements = 6!/(3! × 2!) = 720/(6 × 2) = 720/12 = 60 ways

Question 17 [Medium]

A team of 5 is to be selected from 6 boys and 4 girls. In how many ways can this be done if the team must have at least 2 girls?

Solution: Cases with at least 2 girls:

2 girls + 3 boys: 4C2 × 6C3 = 6 × 20 = 120
3 girls + 2 boys: 4C3 × 6C2 = 4 × 15 = 60
4 girls + 1 boy: 4C4 × 6C1 = 1 × 6 = 6

Total = 120 + 60 + 6 = 186 ways

Alternative: Total - (0 girls + 1 girl) = 10C5 - (6C5 + 4C1×6C4) = 252 - (6 + 60) = 186 ✓

Question 18 [Medium]

In how many ways can 7 people sit around a circular table if two particular people must not sit together?

Solution: Total circular arrangements = (7-1)! = 6! = 720

Arrangements with two particular people together: Treat them as one unit: 6 units in circle = 5! ways They can swap within unit: 2! ways Total together = 5! × 2! = 120 × 2 = 240

Arrangements with them NOT together = 720 - 240 = 480 ways

Question 19 [Medium]

How many factors does the number 720 have?

Solution: Prime factorization of 720: 720 = 2^4 × 3^2 × 5^1

Number of factors = (4+1)(2+1)(1+1) = 5 × 3 × 2 = 30 factors

Question 20 [Medium]

In how many ways can 4 letters be posted in 5 letter boxes if any letter can be posted in any box?

Solution: Each letter has 5 choices.

Total ways = 5 × 5 × 5 × 5 = 5^4 = 625 ways

HARD LEVEL (Questions 21-30)

Question 21 [Hard]

In how many ways can 5 boys and 5 girls be seated in a row such that no two girls sit together?

Solution: Using the gap method:

Step 1: Arrange 5 boys first: 5! = 120 ways This creates 6 gaps (including ends): _ B _ B _ B _ B _ B _

Step 2: Choose 5 gaps from 6 for girls: 6C5 = 6 Step 3: Arrange 5 girls in these gaps: 5! = 120

Total = 120 × 6 × 120 = 86,400 ways

Question 22 [Hard]

How many 5-digit numbers can be formed from digits 0, 1, 2, 3, 4, 5 without repetition such that the number is divisible by 5?

Solution: For divisibility by 5, last digit must be 0 or 5.

Case 1: Last digit is 0 First 4 digits from {1,2,3,4,5}: 5P4 = 120 numbers

Case 2: Last digit is 5 First digit from {1,2,3,4} (4 choices, can't be 0) Next 3 digits from remaining 4: 4P3 = 24 Total = 4 × 24 = 96 numbers

Total = 120 + 96 = 216 numbers

Question 23 [Hard]

In how many ways can 12 identical chocolates be distributed among 4 children such that each child gets at least 2 chocolates?

Solution: First give 2 chocolates to each child: 8 chocolates given, 4 remaining.

Now distribute 4 identical chocolates to 4 children (can get 0 or more): Using stars and bars: (n+r-1)C(r-1) = (4+4-1)C(4-1) = 7C3 = 35 ways

Question 24 [Hard]

How many parallelograms can be formed by two sets of 5 parallel lines each, intersecting each other?

Solution: A parallelogram needs 2 lines from first set and 2 from second set.

Ways to choose 2 from first set: 5C2 = 10 Ways to choose 2 from second set: 5C2 = 10

Total parallelograms = 10 × 10 = 100

Question 25 [Hard]

In how many ways can the letters of "SUCCESS" be arranged such that no two S's are together?

Solution: Word "SUCCESS": S-S-S, U, C, C, E (7 letters, 3 S's, 2 C's)

Step 1: Arrange non-S letters (U, C, C, E): 4!/2! = 12 ways This creates 5 gaps: _ U _ C _ C _ E _

Step 2: Place 3 S's in 5 gaps: 5C3 = 10

Total = 12 × 10 = 120 ways

Question 26 [Hard]

A committee of 5 is to be formed from 6 Indians and 5 Americans. In how many ways can this be done if the committee must have a majority of Indians?

Solution: Majority Indians means:

3 Indians + 2 Americans: 6C3 × 5C2 = 20 × 10 = 200
4 Indians + 1 American: 6C4 × 5C1 = 15 × 5 = 75
5 Indians + 0 Americans: 6C5 × 5C0 = 6 × 1 = 6

Total = 200 + 75 + 6 = 281 ways

Question 27 [Hard]

In how many ways can 8 people be seated at a round table if 3 particular people must sit together?

Solution: Treat 3 particular people as one block. Now we have: 5 individuals + 1 block = 6 units

Circular arrangement of 6 units = (6-1)! = 5! = 120 Arrange 3 people within block = 3! = 6

Total = 120 × 6 = 720 ways

Question 28 [Hard]

How many 4-letter words can be formed using the letters of "PROBLEM" if each word must contain the letter P?

Solution: "PROBLEM" has 7 distinct letters.

Method: Total 4-letter words - Words without P

Total 4-letter words = 7P4 = 840 Words without P (from R,O,B,L,E,M = 6 letters) = 6P4 = 360

Words containing P = 840 - 360 = 480 words

Question 29 [Hard]

In how many ways can 10 identical balls be distributed into 4 distinct boxes such that no box is empty?

Solution: This is distributing 10 identical items to 4 distinct groups with each getting at least 1.

First give 1 ball to each box: 6 balls remaining. Distribute 6 balls to 4 boxes (can get 0 or more): Using stars and bars: (6+4-1)C(4-1) = 9C3 = 84 ways

Question 30 [Hard]

Find the number of ways to arrange 5 boys and 4 girls in a row such that: (a) All boys sit together and all girls sit together (b) No two boys sit together

Solution: (a) All boys together, all girls together: Treat boys as block B, girls as block G. Arrangements: B-G or G-B = 2 ways Arrange boys within block: 5! = 120 Arrange girls within block: 4! = 24

Total = 2 × 120 × 24 = 5,760 ways

(b) No two boys sit together: Using gap method:

Step 1: Arrange 4 girls: 4! = 24 ways Gaps created: _ G _ G _ G _ G _ (5 gaps)

Step 2: Place 5 boys in 5 gaps: 5! = 120 ways

Total = 24 × 120 = 2,880 ways

Companies and Exams That Frequently Ask P&C

🏢 Product Companies (High Weightage)

Google: 2-3 questions (Hard level, conceptual)
Microsoft: 2-3 questions (Medium to Hard)
Amazon: 1-2 questions (Medium)
Adobe: 2-3 questions (Hard)
Oracle: 1-2 questions (Medium)
SAP: 1-2 questions (Medium)

🏢 Service Companies

TCS: 1-2 questions (Easy to Medium)
Infosys: 1-2 questions (Medium)
Wipro: 1-2 questions (Easy)
Cognizant: 1-2 questions (Medium)
Accenture: 1-2 questions (Easy)

🏛️ Government & Competitive Exams

CAT: 2-3 questions (Medium to Hard)
GATE: 1-2 questions (Medium)
Banking: 0-1 questions (Easy)
SSC: 1-2 questions (Easy to Medium)
UPSC CSAT: 1 question (Medium)

Preparation Tips for Permutation and Combination

🎯 Before the Exam

Master Fundamental Principles: The multiplication rule (AND) and addition rule (OR) are the foundation. Every problem reduces to these principles.
Understand the Difference: Permutation = Order matters (arrangements). Combination = Order doesn't matter (selections). This distinction is crucial.
Memorize Key nCr Values: Know values like 5C2=10, 6C3=20, 8C3=56, 10C3=120. Saves calculation time.
Practice Pattern Recognition: Questions follow patterns: word formation, seating, distribution, selection. Recognize the pattern to apply the right formula.
Learn the Gap Method: Essential for "not together" problems. Arrange unrestricted items first, then place restricted items in gaps.

📝 During the Exam

Draw Diagrams: For circular arrangements, geometric problems, visualize with a quick sketch.
Use Complementary Counting: "At least" problems are often easier as: Total - (Unwanted cases)
Check for Repetitions: Always check if items are identical or distinct before applying formulas.

📚 Study Resources

RS Aggarwal Quantitative Aptitude: Chapters on Permutation and Combination
Arun Sharma CAT Quant: Excellent for hard-level problems
NMAT Official Guide: Good for medium-level practice
Previous Year CAT Papers: Must-solve for serious aspirants

Frequently Asked Questions

Q1: How do I know whether to use permutation or combination?

Example: Selecting a team of 3 from 10 → Combination. Arranging 3 people in 3 seats → Permutation.

Q2: What's the most common mistake in P&C problems?

Q3: How do I handle "at least one" type questions?

Q4: Are circular arrangements always (n-1)!?

Q5: How important is P&C for product companies?

Quick Reference Card

PERMUTATION & COMBINATION - KEY POINTS

FUNDAMENTAL RULES:
• AND → Multiply (sequential)
• OR → Add (alternatives)

FORMULAS:
• nPr = n!/(n-r)! (arrangements)
• nCr = n!/(r!(n-r)!) (selections)
• nCr = nC(n-r) (symmetry)

SPECIAL CASES:
• nPn = n!
• nC0 = nCn = 1
• Circular: (n-1)!
• Necklace: (n-1)!/2
• With repetition: n^r
• With duplicates: n!/(p!q!r!)

SHORTCUTS:
• nCr = n×(n-1)×...×(n-r+1)/r!
• Total selections from n: 2^n - 1
• nCr + nC(r+1) = (n+1)C(r+1)

COMMON RATES:
• Google/Microsoft: 2-3 questions
• TCS/Infosys: 1-2 questions
• CAT: 2-3 questions

Best of luck with your placement preparation! 🎯

Permutation and Combination requires practice, but once you master the patterns, it becomes one of the most scoring topics. Focus on understanding the logic rather than memorizing formulas!