Permutations and Combinations Class 11 Notes for IIT JEE

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Fundamental Principles of Counting

Before permutations and combinations, two counting principles form the foundation of every problem in this chapter.

Multiplication Principle (AND Rule)

If task A can be done in m ways AND task B can be done in n ways, then both tasks together can be done in m × n ways.

Example: A person has 4 shirts and 3 trousers. Total outfits = 4 × 3 = 12 ways

The multiplication principle applies whenever tasks are performed in sequence — completing one AND then the other.

Addition Principle (OR Rule)

If task A can be done in m ways OR task B can be done in n ways (and tasks are mutually exclusive), then either task can be done in m + n ways.

Example: A student can travel by 3 bus routes OR 2 train routes. Total ways = 3 + 2 = 5 ways

The addition principle applies whenever only one of multiple alternatives is chosen.

How to identify which principle to use:

💡 Expert Tip by Saransh Gupta, IIT Bombay AIR-41: "Every permutation and combination problem, no matter how complex, is built entirely on these two principles. Before writing nPr or nCr, identify the steps and whether each step uses AND (multiply) or OR (add). Students who skip this structure jump to formulas and get wrong answers on JEE problems with constraints."

India's Best Exam Preparation for Class 11th - Download Now

Permutations and Combinations Class 11 Notes 

India's Best Exam Preparation for Class 11th - Download Now

India's Best Exam Preparation for Class 11th - Download Now

India's Best Exam Preparation for Class 11th - Download Now

India's Best Exam Preparation for Class 11th - Download Now

India's Best Exam Preparation for Class 11th - Download Now

India's Best Exam Preparation for Class 11th - Download Now

Factorial Notation — Definition and Key Results 

Definition

For any positive integer n, n! (n factorial) is defined as:

n! = n × (n−1) × (n−2) × ... × 3 × 2 × 1

Special case: 0! = 1 (by convention — essential for the nCr formula to work when r = 0 or r = n)

Key Factorial Values to Memorise

Useful Factorial Relations

• n! = n × (n−1)!
• (n+1)! = (n+1) × n!
• n! = n × (n−1) × (n−2)! — useful for simplifying expressions

Permutations — Formula, Types and Solved Examples 

Definition

A permutation is an arrangement of objects in a definite order. The number of ways of choosing and arranging r objects from n distinct objects is:

nPr = n! / (n−r)! where 0 ≤ r ≤ n

Key Results for Permutations

Permutations With Repetition Allowed

When repetition is allowed, each of the r positions can be filled by any of the n objects:

Number of arrangements = nʳ

Example: Number of 3-digit numbers formed from digits {1, 2, 3, 4} with repetition = 4³ = 64

Permutations of Objects Not All Distinct (With Identical Elements)

When n objects include identical groups — p objects of one type, q of another, r of another:

Number of arrangements = n! / (p! × q! × r!)

Example: Number of arrangements of letters in the word MISSISSIPPI: Total letters = 11; M=1, I=4, S=4, P=2 Arrangements = 11! / (1! × 4! × 4! × 2!) = 34650

Circular Permutations

When objects are arranged in a circle (not a line), one position is fixed to remove rotational duplicates:

Circular arrangements of n distinct objects = (n−1)!

When clockwise and anticlockwise are the same (e.g., necklaces, garlands): = (n−1)! / 2

Example: 6 people sitting around a circular table = (6−1)! = 5! = 120 ways

Combinations — Formula, Properties and Solved Examples 

Definition

A combination is a selection of objects without regard to order. The number of ways of choosing r objects from n distinct objects is:

nCr = n! / [r! × (n−r)!] where 0 ≤ r ≤ n

nCr is also written as C(n,r) or (nr)\binom{n}{r} (rn​) (binomial coefficient).

Key Combination Properties

The Symmetry Property — Most Used in JEE

nCr = nC(n−r)

This means nC3 = nC7 implies n = 10 (since 3 + 7 = 10 = n). Questions of the type "if nCx = nCy, find n" are solved directly using this property.

💡 Expert Tip: "The symmetry property nCr = nC(n−r) and Pascal's identity nCr + nC(r−1) = (n+1)Cr are the two combination identities that appear most in JEE. Know them as tools, not as statements to quote. When a JEE problem gives you a sum like nC0 + nC1 + nC2 + ... + nCn = 2ⁿ, it is using the binomial theorem with x = 1. Understanding the connection between combinations and the binomial theorem multiplies your problem-solving range significantly."

Solved Example

In how many ways can a committee of 3 men and 2 women be formed from 6 men and 5 women?

Selecting 3 men from 6: ⁶C₃ = 6!/(3!×3!) = 20 Selecting 2 women from 5: ⁵C₂ = 5!/(2!×3!) = 10

Both selections must happen (AND principle): Total = 20 × 10 = 200 ways

Key Difference: Permutation vs Combination 

This is the most important conceptual distinction in the chapter — and the most common source of errors in JEE problems.

How to Identify Permutation vs Combination

Ask: Does changing the order of selection produce a different outcome?

• 3 people sitting in 3 specific seats: permutation (swapping seats gives a different arrangement)
• 3 people forming a committee: combination (swapping names gives the same committee)
• PIN code 1234 vs 4321: permutation (different codes — order matters)
• Ingredients in a recipe: combination (same ingredients regardless of order listed)

Important Results and Identities for JEE 

These results go beyond NCERT and are tested regularly in JEE Main:

Division into Groups

Dividing 2n objects into two equal groups of n: = (2n)! / (n! × n!) — if the groups are distinguishable (labelled) = (2n)! / (n! × n! × 2!) — if the groups are indistinguishable (unlabelled)

Dividing mn objects into m groups of n each: = (mn)! / [(n!)^m × m!] — if groups are indistinguishable

Distribution of Distinct Objects into Distinct Groups

n distinct objects into r distinct groups (any number per group, including empty): = rⁿ (each object independently chooses one of r groups)

Number of Selections from Groups

Selecting at least one object from n distinct objects: = 2ⁿ − 1 (total subsets minus the empty set)

Rank of a Word

To find the rank (position in dictionary order) of a word:

1. Count letters smaller than the first letter — say k letters
2. Arrangements starting with each smaller letter = (n−1)!
3. Fix first letter; repeat for remaining positions
4. Sum all counts + 1 = rank

Example: Rank of the word "RICE" among all arrangements of R, I, C, E:

Letters in alphabetical order: C, E, I, R

Words starting with C: 3! = 6 Words starting with E: 3! = 6 Words starting with I: 3! = 6 (all before RICE)

Words starting with R: — RCxx: 2! = 2 — RExx: 2! = 2 — RICx: next is RICE itself (1 more)

Rank = 6 + 6 + 6 + 2 + 2 + 1 = 23

Most Important Topics for JEE Main and JEE Advanced 

For JEE Main (1–2 questions per paper)

1. Committee/group formation — combinations with constraints (at least one from a group, exactly k from a subset)
2. Arrangement with identical elements — n!/p!q!r! type
3. Circular permutations — people around tables, beads on a necklace
4. nCr properties and identities — symmetry, Pascal's identity, sum = 2ⁿ
5. Rank of a word in dictionary order
6. Distribution problems — objects into groups, boxes

For JEE Advanced (multi-step problems)

1. Constrained selection — at least/at most conditions requiring inclusion-exclusion
2. Gaps method — arrangements where certain objects must not be adjacent
3. Multinomial arrangements — objects split across multiple categories
4. Combinatorics + probability hybrid — selecting then computing probability
5. Geometric combinations — number of triangles, diagonals from n points

The Gaps Method — Critical for JEE

When objects must not be adjacent, use the gaps method:

1. First arrange the other objects (without the restricted ones) — say in P ways
2. Count the gaps created (n objects create n+1 gaps, including ends)
3. Place the restricted objects in the gaps using combinations

Example: Arrange 4 boys and 3 girls in a row such that no two girls are adjacent.

Step 1: Arrange 4 boys in a row: 4! = 24 ways Step 2: Gaps created: _ B _ B _ B _ B _ = 5 gaps. Step 3: Choose 3 gaps from 5 and arrange 3 girls: ⁵P₃ = 60

Total = 24 × 60 = 1440 ways

How to Study Permutations and Combinations for JEE Class 11 

Step 1: Understand AND vs OR Before Touching Formulas (half day)

Solve 10 counting problems using only the multiplication and addition principles — no nPr, no nCr. This builds the structural intuition that makes all later formula applications correct.

Step 2: Derive nPr and nCr once, Then Memorise (half day)

Derive the permutation formula from the multiplication principle (n choices for position 1, n−1 for position 2, etc.) and the combination formula from nCr = nPr/r!. Understanding the derivation means you never confuse when to apply each formula.

Step 3: Build Your Formula Card (half day)

Write every formula from these notes on one A4 sheet — factorial results, nPr, nCr, circular permutations, identical elements formula, and the identities table. This card becomes your revision anchor throughout the year.

Step 4: Categorise Every Practice Problem Before Solving

Before writing a single calculation, write:

• Type: permutation/combination / both
• Constraint: none / with repetition / with identical elements / circular / gaps method
• Principle: multiplication/addition / both

This categorisation step takes 10 seconds and prevents the most common error in this chapter — applying the right formula to the wrong type of problem.

Step 5: Solve NCERT Completely, Then Progress to Module Problems

NCERT Chapter 7 covers the foundational question types. After solving all NCERT examples and exercises, work through eSaral's module problems, which introduce constrained selection, gaps method, and distribution problems — the types that appear in JEE Main but not in NCERT.