Chapter 6 of 14

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CBSE Class 11 Mathematics★ 6-8 marks · CBSE Class 11 Mathematics Chapter 6

Permutations and Combinations

How many ways can you arrange 5 books on a shelf? How many ways can you choose a team of 11 from 15 players? PERMUTATIONS count ARRANGEMENTS (order matters). COMBINATIONS count SELECTIONS (order doesn't matter). CBSE Class 11 Mathematics Chapter 6.

⏱ 22 min readMedium difficulty✓ Free · NCERT-aligned

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By the end of this chapter you'll be able to…

1Apply the Fundamental Principle of Counting (multiplication principle) to multi-stage counting problems
2Compute factorials and use them to evaluate nPr and nCr for given values
3Solve permutation problems involving arrangements of distinct and identical objects
4Solve combination problems involving selection of objects from a group
5Distinguish correctly between problems that require permutations and those that require combinations

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Why this chapter matters

Permutations and Combinations are the foundation of probability theory and directly prerequisite the Binomial Theorem. Counting questions appear regularly in JEE Main and are the basis for combinatorial proofs in mathematics.

Before you start — revise these

A 5-minute refresher here will save you 30 minutes of confusion below.

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Sets

Set notation used in defining sample spaces; combinations count subsets

Permutations and Combinations

"Counting is the most basic mathematical operation. But counting POSSIBILITIES requires the tools in this chapter."

1. Chapter Overview

How many ways can things be arranged or selected? This chapter provides the systematic tools: the Fundamental Principle of Counting, Factorial notation (n!), Permutations (arrangements — order matters: nPr), and Combinations (selections — order doesn't matter: nCr). These form the basis of PROBABILITY (next chapter).

2. Fundamental Principle of Counting (Multiplication Principle)

If one event can occur in m ways, and (for each of these m ways) another event can occur in n ways → the two events TOGETHER can occur in m × n ways
This extends to any number of events: m₁ × m₂ × m₃ × ...

Example

A restaurant has 3 starters and 4 main courses.
A customer wants 1 starter AND 1 main course.
Total combinations: 3 × 4 = 12 possible meals.

3. Factorial Notation — n!

n! = n × (n-1) × (n-2) × ... × 3 × 2 × 1
0! = 1 (by convention — makes the formulas work)
5! = 5 × 4 × 3 × 2 × 1 = 120
n! = n × (n-1)!

4. Permutations (nPr) — Order MATTERS

Definition

The number of ways to ARRANGE r objects selected from n distinct objects, where ORDER MATTERS
nPr = n! / (n — r)!

Special Cases

Arranging ALL n objects: nPn = n! (no restriction)
Arranging n objects where some are IDENTICAL: n! / (p₁! p₂! ... pₖ!) where p₁, p₂ are the counts of each type of identical object

Key Principle: Multiply choices

Arranging 3 letters from {A, B, C, D, E}: 5 × 4 × 3 = 60 = ⁵P₃
First position: 5 choices. Second: 4 remaining. Third: 3 remaining.

5. Combinations (nCr) — Order DOES NOT MATTER

Definition

The number of ways to SELECT r objects from n distinct objects, where ORDER DOES NOT MATTER
nCr = n! / [r! (n — r)!]
nCr = nPr / r! (take the permutations and divide by the number of ways to rearrange the r selected objects)
Also written as C(n, r) or (ⁿᵣ)

Key Properties

nCr = nC(n-r). Choosing r objects is THE SAME as choosing the n-r objects to LEAVE OUT.
nC₀ = nCn = 1
nC₁ = n
nCr + nC(r+1) = (n+1)C(r+1)

6. Difference Between Permutations and Combinations

	Permutations (nPr)	Combinations (nCr)
Order?	ORDER MATTERS (ABC ≠ CBA)	Order DOES NOT MATTER (ABC = CBA)
What it counts	ARRANGEMENTS	SELECTIONS
Formula	n!/(n-r)!	n!/[r!(n-r)!]
Value	nPr ≥ nCr (larger or equal)	nCr ≤ nPr
When to use	Forming numbers, words. Seating arrangements. President/VP/Secretary.	Choosing a team. Selecting a committee.

7. Exam Focus

Fundamental Principle of Counting (multiplication principle)
n! — factorial definition, 0! = 1
nPr — formula, when to use (order matters)
nCr — formula, properties (nCr = nC(n-r))
Distinguishing permutations from combinations — KEY SKILL

8. Key Formulas

nPr = n! / (n-r)!
nCr = n! / [r! (n-r)!]
nCr = nC(n-r)
nCr = nPr / r!
Arranging n objects with p identical of one type, q of another: n!/(p! q!)

9. Conclusion

Counting possibilities is systematic — not guesswork:

PRINCIPLE OF COUNTING: Multiply the number of choices at each step
PERMUTATIONS (nPr): Arrangements. Order matters. President, VP, Secretary — who gets which office MATTERS.
COMBINATIONS (nCr): Selections. Order doesn't matter. Choosing a committee — the COMMITTEE is the same regardless of the order in which members were chosen.

The difference between permutations and combinations is one word: ORDER. If you're making a queue → permutations. If you're picking a team → combinations.

Key formulas & results

Everything you need to memorise, in one card. Screenshot this for revision.

Factorial

n! = n×(n−1)×(n−2)×...×2×1; 0! = 1

0! = 1 is a convention that makes the formulas for nP0 and nC0 work correctly

Permutations (nPr)

nPr = n! / (n−r)!

Number of ways to ARRANGE r objects from n distinct objects — order matters

Combinations (nCr)

nCr = n! / [r!(n−r)!] = nPr / r!

Number of ways to SELECT r objects from n — order does NOT matter

Symmetry of Combinations

nCr = nC(n−r)

Choosing r objects is equivalent to choosing the (n−r) objects to leave behind

Pascal's Identity

nCr + nC(r−1) = (n+1)Cr

Used in the Binomial Theorem; every entry in Pascal's Triangle follows this rule

Arrangements with Identical Objects

Arrangements of n objects with p identical of one type, q of another = n! / (p! × q!)

Divide by factorial of each group of identical objects

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Common mistakes & fixes

These are the exact errors that cost students marks in board exams. Read them once, save yourself the trouble.

WATCH OUT

✗ Using nPr instead of nCr for selection problems (e.g., choosing a team)

✓ Ask: does the ORDER of selection matter? Choosing a committee of 3 from 10 — the committee is the same regardless of order → use nCr. Choosing President, VP, Secretary from 10 → different roles mean order matters → use nPr.

WATCH OUT

✗ Treating 0! as 0

✓ 0! = 1 by definition. This ensures nC0 = n!/[0!×n!] = 1, which is correct (there is exactly one way to choose nothing).

WATCH OUT

✗ Applying the multiplication principle when events are mutually exclusive (not sequential)

✓ The multiplication principle (m×n) applies when both events happen TOGETHER (one AND the other). Use addition (m+n) when only one happens at a time (one OR the other, mutually exclusive).

WATCH OUT

✗ Forgetting to divide by repeated arrangements when objects are identical

✓ For the word MISSISSIPPI, not all 11! arrangements are distinct. Divide by 4! (four Is), 4! (four Ss), and 2! (two Ps): 11!/(4!×4!×2!) arrangements.

NCERT exercises (with solutions)

Every NCERT exercise from this chapter — what it covers and how many questions to expect.

Ex 6.1

Exercise 6.1

Fundamental Principle of Counting — multiplication principle with multi-stage problems

Questions

Ex 6.2

Exercise 6.2

Permutations — factorial notation, nPr formula, special cases

Questions

Ex 6.3

Exercise 6.3

Permutations of objects with restrictions, circular arrangements

Questions

Ex 6.4

Exercise 6.4

Combinations — nCr formula, properties, selection problems

Questions

Practice problems

Try each one yourself before tapping "Show solution". Active recall > rereading.

Q1EASY· Combinations

In how many ways can a team of 3 students be selected from a group of 8?

▶ Show solution

Order does not matter (a team is a team regardless of selection order). Use ⁸C₃ = 8!/(3!×5!) = (8×7×6)/(3×2×1) = 336/6 = 56.

Q2MEDIUM· Permutations

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

▶ Show solution

We need to arrange 4 distinct digits selected from 5 distinct digits. Order matters (4321 ≠ 1234). Answer = ⁵P₄ = 5!/(5−4)! = 5!/1! = 120. Alternatively: 5×4×3×2 = 120 choices by multiplication principle.

Q3HARD· Restricted Combinations

A committee of 5 is to be formed from 6 men and 4 women such that there are at least 2 women. Find the number of ways.

▶ Show solution

Cases: (i) Exactly 2 women: ⁴C₂×⁶C₃ = 6×20 = 120. (ii) Exactly 3 women: ⁴C₃×⁶C₂ = 4×15 = 60. (iii) Exactly 4 women: ⁴C₄×⁶C₁ = 1×6 = 6. Total = 120+60+6 = 186.

5-minute revision

The whole chapter, distilled. Read this the night before the exam.

•n! = n×(n−1)×...×1 and 0! = 1 (by convention)
•nPr = n!/(n−r)! counts ARRANGEMENTS (order matters); nCr = n!/[r!(n−r)!] counts SELECTIONS (order irrelevant)
•Distinguishing key: does swapping the order of selection give a different result? If yes → permutation; if no → combination
•nCr = nC(n−r): choosing 3 from 10 is the same count as choosing 7 from 10
•nC₀ = nCn = 1; nC₁ = n; nCr + nC(r−1) = (n+1)Cr (Pascal's identity)
•Identical objects: arrangements = n!/(p!×q!×r!...) dividing by factorial of each repeated group
•Multiplication principle: if task 1 has m ways and task 2 has n ways (independent), total = m×n
•Addition principle: if event A can happen in m ways and event B in n ways (mutually exclusive), total = m+n

CBSE marks blueprint

Where the marks come from in this chapter — so you can plan your prep.

Typical chapter weightage: 6-8 marks

Question type	Marks each	Typical count	What it tests
Short Answer	2	1	Direct application of nPr or nCr formula
Long Answer	4-6	1	Multi-case problems (at least, at most), arrangements with restrictions

Prep strategy

The most important skill is identifying whether a problem needs permutation or combination — practise this distinction on 20 varied problems
For 'at least' and 'at most' problems, list all valid cases separately and add — never try to use a single formula
Memorise small factorials (1!=1, 2!=2, 3!=6, 4!=24, 5!=120, 6!=720, 7!=5040) to speed up computations

Where this shows up in the real world

This chapter isn't just an exam topic — it lives in the world around you.

Lottery and Gambling Odds

A lottery requiring you to choose 6 numbers from 49 has ⁴⁹C₆ possible tickets — a direct application of combinations to calculate the probability of winning.

Password and PIN Generation

The number of 4-digit PINs using digits 0–9 without repetition is ¹⁰P₄ = 5040 — a permutation problem because the order of digits determines the PIN.

Tournament Scheduling

To schedule round-robin matches between n teams (each pair plays once), the number of matches = nC₂ — a combination because Team A vs Team B is the same match as Team B vs Team A.

Exam strategy

Battle-tested tips from teachers and toppers for this chapter.

Always write down which formula you are using (nPr or nCr) before computing — this earns method marks even if arithmetic errors follow
Break multi-condition problems into separate cases, compute each, then sum — do NOT try to account for all conditions in one formula
Check your answer for reasonableness: nCr ≤ nPr always, and nCr should decrease as r approaches n/2 moves away from the extremes
For 'at least one' problems, use the complement: total arrangements − arrangements with NONE of the required condition

Going beyond the textbook

For olympiad aspirants and curious learners — topics that build on this chapter.

Inclusion-exclusion principle for counting: |A∪B∪C| = |A|+|B|+|C|−|A∩B|−|B∩C|−|A∩C|+|A∩B∩C| — generalises the set cardinality formula
Pigeonhole principle: if n+1 objects are placed in n boxes, at least one box contains ≥ 2 objects — a powerful olympiad tool

Where else this chapter is tested

CBSE board isn't the only one — other exams test this chapter too.

CBSE Class 11 BoardHigh

JEE MainHigh

JEE AdvancedHigh

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Questions students ask

The real ones — pulled from the Q&A community and tutor sessions.

Ask one question: does the order of selection matter? If YES (e.g., electing President then VP, forming a PIN number) → Permutation. If NO (e.g., choosing a team, selecting books to read) → Combination.

nCr counts the number of subsets of size r from a set of n elements. The 'r-element subsets' of an n-element set number exactly nCr.

Every time you select r objects from n, you simultaneously leave out n−r objects. The number of ways to choose which r to INCLUDE equals the number of ways to choose which n−r to EXCLUDE.

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