Permutation and Combination Notes for Competitive Exams

Permutation and Combination (P&C) is the mathematics of counting. It answers questions like "In how many ways can 5 students sit on 5 chairs?" or "How many committees of 3 can be formed from 8 people?" The chapter appears in UPSC CSAT almost every year in some form, often as a logical counting puzzle, and it is a regular part of bank PO, SSC CGL Tier 2, CDS and CAT-style aptitude sections. It also forms the base of probability. The chapter looks scary because of formulas, but at heart it rests on two simple ideas: the multiplication rule and the addition rule.

Quick Facts: P&C at a Glance

• Permutation = arrangement, where order matters.
• Combination = selection, where order does not matter.
• nPr = n! ÷ (n − r)!
• nCr = n! ÷ [r! × (n − r)!]
• Relation: nPr = nCr × r!
• Circular arrangement of n distinct objects = (n − 1)!

1. Factorial Notation

The factorial of a natural number n, written n!, is the product of all natural numbers from 1 to n.

• 1! = 1, 2! = 2, 3! = 6, 4! = 24, 5! = 120, 6! = 720, 7! = 5040
• By definition, 0! = 1.
• n! = n × (n − 1)!, which helps in simplification. Example: 10! ÷ 8! = 10 × 9 = 90.

2. Fundamental Principles of Counting

Multiplication Principle (AND)

If one task can be done in m ways and, after that, another task can be done in n ways, both tasks together can be done in m × n ways.

Example: A person has 4 shirts and 3 trousers. Number of different outfits = 4 × 3 = 12.

Addition Principle (OR)

If a task can be done in m ways or in n ways (not both together), it can be done in m + n ways.

Example: You can travel from Delhi to Jaipur by 5 trains or 3 buses. Total ways = 5 + 3 = 8.

Golden rule: "AND" means multiply; "OR" means add. Most P&C questions become easy once you split them into AND and OR steps.

3. Permutation (Arrangement)

Permutation counts the number of ways to arrange r objects out of n distinct objects, where order matters.

Formula: nPr = n! ÷ (n − r)!

• Arranging all n objects: nPn = n!
• Example: In how many ways can a president, secretary and treasurer be chosen from 10 members? Order matters (posts are different), so 10P3 = 10 × 9 × 8 = 720.
• Example: How many 3-digit numbers can be formed from 1, 2, 3, 4, 5 without repetition? 5P3 = 5 × 4 × 3 = 60.

Permutation with Repetition Allowed

If repetition is allowed, each place can be filled in n ways, so r places give n^r arrangements.

Example: How many 3-digit codes can be made from digits 0 to 9 with repetition? 10³ = 1000.

Permutation of Identical Objects

If n objects include p alike of one kind, q alike of another, and so on, the arrangements = n! ÷ (p! × q! × ...).

Example: Arrangements of the letters of "BANANA": 6 letters with A three times and N twice. Ways = 6! ÷ (3! × 2!) = 720 ÷ 12 = 60.

Example: Arrangements of "INDIA": 5 letters with I twice. Ways = 5! ÷ 2! = 60.

4. Common Arrangement Cases

Condition	Method	Example (letters of "MOTHER", 6 distinct)
No restriction	n!	6! = 720
Vowels always together	Treat vowels as one unit, then arrange inside	5! × 2! = 240
Vowels never together	Total − always together	720 − 240 = 480
Starts with M	Fix M, arrange the rest	5! = 120
Starts with M and ends with R	Fix both, arrange the rest	4! = 24

Gap method: To keep certain items apart, first arrange the others, then place the restricted items in the gaps. Example: 4 boys and 3 girls in a row with no two girls together. Arrange boys: 4! = 24. There are 5 gaps; choose and arrange 3 girls: 5P3 = 60. Total = 24 × 60 = 1440.

5. Circular Permutation

In a circle, there is no fixed starting point, so one object is fixed and the rest are arranged.

• n distinct objects in a circle: (n − 1)!
• If clockwise and anticlockwise arrangements are the same (like beads in a necklace or flowers in a garland): (n − 1)! ÷ 2

Example: 6 persons around a round table: (6 − 1)! = 120 ways.

Example: A garland of 8 different flowers: 7! ÷ 2 = 2520 ways.

6. Combination (Selection)

Combination counts the number of ways to select r objects from n, where order does not matter.

Formula: nCr = n! ÷ [r! × (n − r)!]

• nC0 = nCn = 1 and nC1 = n.
• nCr = nC(n − r). Example: 10C8 = 10C2 = 45. This saves calculation time.
• nCr + nC(r − 1) = (n + 1)Cr (Pascal's rule).
• Total number of selections from n distinct things (any number, including none) = 2ⁿ; at least one = 2ⁿ − 1.

Example: A committee of 3 from 8 people: 8C3 = (8 × 7 × 6) ÷ 6 = 56.

Example: A committee of 5 with 3 men and 2 women from 6 men and 5 women: 6C3 × 5C2 = 20 × 10 = 200.

Example ("at least"): Select 3 from 5 men and 4 women with at least 1 woman. Total 9C3 = 84; all men 5C3 = 10. Answer = 84 − 10 = 74.

7. Geometry-Based Counting

• Number of straight lines through n points (no three collinear) = nC2.
• Number of triangles from n points (no three collinear) = nC3.
• Number of diagonals of an n-sided polygon = nC2 − n.
• Handshakes among n people (each shakes hands once with every other) = nC2.
• Matches in a round-robin tournament of n teams = nC2.

Example: In a party of 12 people, each shakes hands with every other person once. Total handshakes = 12C2 = 66.

8. Permutation vs Combination: How to Decide

Ask yourself	If yes	Typical words
Does changing the order give a new result?	Permutation	Arrange, order, rank, digits, seating, posts
Is it just a group with no order?	Combination	Select, choose, committee, team, group

A useful check: choosing a team of 3 players is a combination, but choosing captain, vice-captain and wicket-keeper from them is a permutation, because the roles are different.

Common Mistakes to Avoid

1. Using permutation when the question only asks for selection.
2. Forgetting to divide by factorials of repeated letters.
3. Forgetting that a number cannot start with 0 when forming numbers from digits.
4. Using n! instead of (n − 1)! for circular arrangements.
5. Solving "at least" problems case by case when the complement method is quicker.

How to Prepare P&C

• Memorise factorials up to 8! and small nCr values (5C2 = 10, 6C2 = 15, 6C3 = 20, 8C2 = 28, 10C3 = 120).
• Solve 15 questions each on counting principles, word arrangements, circular arrangements and committees.
• For CSAT, focus on logic-based counting (numbers, codes, routes, handshakes) rather than complex formulas.
• Move on to probability only after P&C basics are clear, since probability uses the same counting.

Practice Questions with Answers

1. In how many ways can the letters of "LEADER" be arranged? Answer: 6 letters with E twice: 6! ÷ 2! = 360.
2. How many 4-digit numbers can be formed using 0, 1, 2, 3, 4 without repetition? Answer: First place cannot be 0, so 4 choices; then 4 × 3 × 2. Total = 4 × 24 = 96.
3. From 7 men and 4 women, how many ways can a team of 5 with exactly 2 women be chosen? Answer: 4C2 × 7C3 = 6 × 35 = 210.
4. In how many ways can 5 boys sit around a circular table if two particular boys must sit together? Answer: Treat the two as one unit: (4 − 1)! × 2! = 12.
5. How many triangles can be formed from 10 points, of which 4 are collinear? Answer: 10C3 − 4C3 = 120 − 4 = 116.
6. A question paper has two sections of 5 questions each. A student must answer 6 questions, taking at least 2 from each section. In how many ways? Answer: (2,4) + (3,3) + (4,2) = 10 × 5 + 10 × 10 + 5 × 10 = 200.

These six questions cover the most frequently tested patterns. Revisit any question you get wrong and identify whether the mistake was in the AND-OR split or in choosing permutation over combination.

Related Notes

• Permutation, Combination and Probability for CSAT
• Number System Notes
• Percentage Notes
• All Quantitative Aptitude Notes

Frequently Asked Questions

What is the difference between permutation and combination?

Permutation is arrangement where order matters. Combination is selection where order does not matter.

What is the formula of nPr?

nPr = n! ÷ (n − r)!. It gives the number of arrangements of r objects out of n distinct objects.

What is the formula of nCr?

nCr = n! ÷ [r! × (n − r)!]. It gives the number of selections of r objects out of n.

What is the value of 0!?

0! is defined as 1.

How many ways can n people sit around a round table?

(n − 1)! ways, because one person's position is fixed to remove identical rotations.

How do you arrange letters of a word with repeated letters?

Divide n! by the factorial of the count of each repeated letter. For BANANA, it is 6! ÷ (3! × 2!) = 60.

How many handshakes happen among n people?

nC2 handshakes, which equals n(n − 1) ÷ 2.

What is the relation between nPr and nCr?

nPr = nCr × r!, because each selection of r objects can be arranged in r! ways.

How many diagonals does a polygon of n sides have?

nC2 − n, which equals n(n − 3) ÷ 2.

Is permutation and combination asked in UPSC CSAT?

Yes. CSAT often has counting-based questions on numbers, arrangements and selections, usually framed as logical puzzles.

What does "at least one" mean in combination questions?

It means one or more. The quickest method is total selections minus the selections with none.

How should I start preparing permutation and combination?

Start with the multiplication and addition principles, then learn nPr and nCr, and practise word arrangement and committee questions.