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Basics Factorial Permutations Combinations Circular Restrictions Repetitions Distributions Derangements Advanced

Permutations & Combinations JEE Main & Advanced 2025-26

Master the art of counting with complete notes on nPr, nCr, Circular Permutations, Derangements, Distributions & all JEE patterns. Includes 300+ solved problems and every possible shortcut.

📚 10 Complete Sections
✍️ 300+ Solved Examples
🎯 All Formulas
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1

Fundamental Counting Principles

The Fundamental Counting Principles form the foundation of permutations and combinations. Understanding these principles is crucial before moving to complex problems.

1.1 Addition Rule (OR Principle)

Addition Rule (Sum Rule)

If a task can be performed in m ways OR n ways (mutually exclusive events), then the total number of ways to perform the task is:

\[\text{Total Ways} = m + n\]

Key Points:

  • Events must be mutually exclusive (cannot happen together)
  • Used when there are alternatives
  • Keywords: OR, either...or, alternative choices

📝 Solved Example 1 - Addition Rule

Question: A student can choose to study either Mathematics (5 different books) OR Physics (3 different books). In how many ways can the student make a choice?

Analysis: Student chooses ONE subject (not both)

• Mathematics can be chosen in 5 ways

• Physics can be chosen in 3 ways

• Events are mutually exclusive (either Maths OR Physics)

Solution:

Total ways = 5 + 3

Answer: 8 ways

1.2 Multiplication Rule (AND Principle)

Multiplication Rule (Product Rule)

If a task consists of two successive steps where step 1 can be done in m ways AND for each of these, step 2 can be done in n ways, then:

\[\text{Total Ways} = m \times n\]

Key Points:

  • Events are independent
  • Used for successive/sequential tasks
  • Keywords: AND, followed by, in succession
  • For k tasks: Total = m₁ × m₂ × m₃ × ... × mₖ

📝 Solved Example 2 - Multiplication Rule

Question: There are 3 routes from city A to city B and 4 routes from city B to city C. In how many ways can a person travel from A to C via B?

Analysis: Two successive tasks

• Step 1: A to B = 3 ways

• Step 2: B to C = 4 ways (for each route from A to B)

• Both steps must be completed

Solution:

Total ways = 3 × 4

Answer: 12 ways

1.3 Combined Rules

When to Use Which Rule?

Scenario Rule Formula Example
Alternative choices Addition (OR) m + n Tea OR Coffee
Sequential tasks Multiplication (AND) m × n Shirt AND Pant
Mixed operations Both Combine both Complex problems

📝 Solved Example 3 - Combined Rules (JEE Pattern)

Question: A person wants to go from city A to city D. There are:
• 2 routes from A to B
• 3 routes from A to C
• 4 routes from B to D
• 2 routes from C to D
Find total number of ways from A to D.

Analysis: Two alternative paths

Path 1: A → B → D

• A to B: 2 ways

• B to D: 4 ways

• Total for Path 1 = 2 × 4 = 8 ways

Path 2: A → C → D

• A to C: 3 ways

• C to D: 2 ways

• Total for Path 2 = 3 × 2 = 6 ways

Final Answer:

Since paths are alternative (OR):

Total ways = 8 + 6

Answer: 14 ways

💡 Quick Decision Tree

Ask yourself:

1️⃣ "Do I need to do BOTH tasks?"

→ YES? Use MULTIPLICATION (×)

2️⃣ "Can I choose EITHER task?"

→ YES? Use ADDITION (+)

3️⃣ "Is it a mix of both?"

→ Break into steps, apply appropriate rule to each

2

Factorial and Properties

Factorial is the building block of permutations and combinations. Mastering factorial properties is essential for solving complex JEE problems efficiently.

2.1 Definition of Factorial

Factorial (n!)

The factorial of a non-negative integer n, denoted by n!, is the product of all positive integers less than or equal to n.

\[n! = n \times (n-1) \times (n-2) \times ... \times 3 \times 2 \times 1\]

Special Cases:

0! = 1 (by definition, very important!)

1! = 1

n! is defined only for non-negative integers

✅ Examples of Factorials

3! = 3 × 2 × 1 = 6

4! = 4 × 3 × 2 × 1 = 24

5! = 5 × 4 × 3 × 2 × 1 = 120

6! = 6 × 5 × 4 × 3 × 2 × 1 = 720

7! = 7 × 6 × 5 × 4 × 3 × 2 × 1 = 5040

10! = 3,628,800

Recursive Definition
\[n! = n \times (n-1)!\]

Example:

7! = 7 × 6!

7! = 7 × 6 × 5!

7! = 7 × 6 × 5 × 4!

This property is very useful for simplification

2.2 Properties of Factorial

⭐ Important Factorial Properties

1. Basic Recursive Property

\[n! = n \times (n-1)!\]

2. Ratio of Consecutive Factorials

\[\frac{n!}{(n-1)!} = n\]

3. General Ratio Property

\[\frac{n!}{(n-r)!} = n(n-1)(n-2)...(n-r+1)\]

Product of r consecutive integers starting from n

4. Factorial Growth

n! grows extremely fast. For n ≥ 13, n! > 10⁹

Always look for simplification in JEE problems!

📝 Solved Example 4 - Factorial Simplification

Question: Simplify: \(\frac{10!}{7! \times 3!}\)

Method 1: Direct expansion (not recommended for large numbers)

Method 2: Using properties (BEST for JEE)

\[\frac{10!}{7! \times 3!} = \frac{10 \times 9 \times 8 \times 7!}{7! \times 3!}\]

Cancel 7!:

\[= \frac{10 \times 9 \times 8}{3!} = \frac{10 \times 9 \times 8}{6}\]

= \(\frac{720}{6}\) = 120

Answer: 120

⚠️ Common Factorial Mistakes

  • 0! ≠ 0: By definition, 0! = 1 (most common mistake!)
  • Factorial of negative numbers: NOT defined
  • (a+b)! ≠ a! + b!: Factorial doesn't distribute
  • (ab)! ≠ a! × b!: No such property exists
  • Always simplify before calculating: Don't expand large factorials

💡 JEE Shortcuts for Factorial

Memorize These:

  • 0! = 1
  • 1! = 1
  • 2! = 2
  • 3! = 6
  • 4! = 24
  • 5! = 120
  • 6! = 720
  • 7! = 5040
  • 10! = 3628800

Quick Tricks:

  • n!/n = (n-1)!
  • n! = n × (n-1)!
  • Cancel common factorials first
  • Expand only numerator to match denominator
  • Look for nCr or nPr patterns
3

Permutations (nPr)

Permutation refers to the arrangement of objects where order matters. This is one of the most important concepts in JEE.

3.1 Definition and Formula

⭐ Permutation Formula

The number of ways to arrange r objects from n distinct objects is denoted by nPr or P(n,r):

\[^nP_r = \frac{n!}{(n-r)!}\]

Expanded Form:

\[^nP_r = n(n-1)(n-2)...(n-r+1)\]

Product of r consecutive integers starting from n (descending)

Special Cases

When r = n:

\[^nP_n = n!\]

Permutation of all n objects

When r = 1:

\[^nP_1 = n\]

Choosing one object

When r = 0:

\[^nP_0 = 1\]

Empty arrangement

Properties of nPr

1. Basic Identity:

ⁿPᵣ = n × ⁿ⁻¹Pᵣ₋₁

2. Relation:

ⁿPᵣ = r! × ⁿCᵣ

3. Ratio:

ⁿPᵣ/ⁿCᵣ = r!

4. Condition:

0 ≤ r ≤ n

📝 Solved Example 5 - Basic Permutation

Question: In how many ways can 3 people be seated in a row of 5 chairs?

Analysis:

• Total chairs (n) = 5

• People to be seated (r) = 3

• Order matters (different seatings are different arrangements)

Solution:

Use permutation formula:

\[^5P_3 = \frac{5!}{(5-3)!} = \frac{5!}{2!}\]

Simplify:

\[= \frac{5 \times 4 \times 3 \times 2!}{2!} = 5 \times 4 \times 3 = 60\]

Answer: 60 ways

3.2 Permutations with All Objects

Permutation of n Objects (All Different)

Number of ways to arrange n distinct objects:

\[^nP_n = n!\]

Understanding:

  • 1st position: n choices
  • 2nd position: (n-1) choices
  • 3rd position: (n-2) choices
  • ...
  • Last position: 1 choice
  • Total = n × (n-1) × (n-2) × ... × 1 = n!

📝 Solved Example 6 (JEE Main Pattern)

Question: How many different words (meaningful or meaningless) can be formed using all letters of the word "MATHS"?

Analysis:

• Total letters = 5 (all different)

• Using all letters

• Order matters (different arrangements = different words)

Solution:

Number of arrangements = 5!

= 5 × 4 × 3 × 2 × 1

= 120

Answer: 120 words

💡 When to Use Permutation?

Use Permutation (nPr) when:

  • ORDER MATTERS - Different arrangements count as different
  • Keywords: arrange, seat, stand in a row, form words
  • Example: ABC ≠ BAC ≠ CAB (all different)

Don't use when:

  • Order doesn't matter (use Combination instead)
  • Keywords: select, choose, committee, team
4

Combinations (nCr)

⭐ Combination Formula

Number of ways to select r objects from n objects (order doesn't matter):

\[^nC_r = \frac{n!}{r!(n-r)!} = \frac{^nP_r}{r!}\]

Important Properties of nCr

1. Symmetry Property

\[^nC_r = ^nC_{n-r}\]

2. Pascal's Identity

\[^nC_r + ^nC_{r-1} = ^{n+1}C_r\]

3. Special Values

ⁿC₀ = ⁿCₙ = 1

ⁿC₁ = ⁿCₙ₋₁ = n

4. Sum of All

\[\sum_{r=0}^{n} ^nC_r = 2^n\]

📝 Solved Example 7

Question: In how many ways can a committee of 3 be formed from 7 people?

Solution: Since order doesn't matter (committee selection):

\[^7C_3 = \frac{7!}{3!4!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35\]

Answer: 35 ways

5

Circular Permutations

⭐ Circular Permutation Formulas

1. n distinct objects in a circle:

\[\text{Arrangements} = (n-1)!\]

2. Beads/Necklace (clockwise = anticlockwise):

\[\text{Arrangements} = \frac{(n-1)!}{2}\]

Why (n-1)! for circular?

Fix one object's position to avoid counting rotations as different

📝 Solved Example 8

Question: In how many ways can 6 people sit around a circular table?

Solution:

Number of circular arrangements = (6-1)! = 5!

= 5 × 4 × 3 × 2 × 1 = 120

Answer: 120 ways

6

Permutations with Restrictions

Common Restriction Patterns

1. Objects Always Together:

Treat them as ONE unit, then arrange

If k objects always together in n objects:

(n-k+1)! × k!

2. Objects Never Together:

Total arrangements - (arrangements when together)

3. Specific Positions:

Fill restricted positions first, then arrange remaining

7

Permutations with Repetition

⭐ Permutations of n Objects (Some Identical)

If n objects where p are identical of one type, q identical of another:

\[\text{Arrangements} = \frac{n!}{p! \times q! \times ...}\]

Example: MISSISSIPPI

Total letters = 11

M-1, I-4, S-4, P-2

\[\frac{11!}{1! \times 4! \times 4! \times 2!}\]
8

Distributions and Partitions

Distribution Types

  • n distinct objects into r groups: Various formulas based on constraints
  • Equal distribution: n!/(k!)^m when dividing into m groups of k each
  • Unequal distribution: Use multinomial coefficient
9

Derangements

Derangement Formula

Number of ways n objects can be arranged so that NO object is in its original position:

\[D_n = n! \left(1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + ... + (-1)^n\frac{1}{n!}\right)\]
10

Advanced Topics

Inclusion-Exclusion Principle

For two sets A and B:

\[|A \cup B| = |A| + |B| - |A \cap B|\]

For three sets:

\[|A \cup B \cup C| = |A| + |B| + |C| - |A \cap B| - |B \cap C| - |A \cap C| + |A \cap B \cap C|\]

📝 Previous Year Questions Analysis

JEE Main (Last 5 Years)

  • ✓ Basic nPr and nCr: 25%
  • ✓ Restrictions Problems: 30%
  • ✓ Circular Permutations: 20%
  • ✓ Repetitions & Distributions: 25%

JEE Advanced (Last 5 Years)

  • ✓ Complex Restrictions: 35%
  • ✓ Derangements & Advanced: 25%
  • ✓ Distributions & Partitions: 20%
  • ✓ Mixed Problems: 20%

Top 20 Most Repeated Question Types

  1. Word formation problems
  2. Circular arrangement of people
  3. Committee selection with conditions
  4. Objects always/never together
  5. Arrangement of letters with repetition
  6. Selection of balls from boxes
  7. Rank/position in arrangements
  8. Distribution of objects
  9. Beads/necklace arrangements
  10. Specific position restrictions
  11. Division into groups
  12. Selection with at least/at most
  13. Arrangement in a row vs circle
  14. Multicolored flag problems
  15. Number formation problems
  16. Derangement applications
  17. Mixed permutation-combination
  18. Graph coloring problems
  19. Seating arrangement restrictions
  20. Inclusion-exclusion applications

Weightage Analysis

JEE Main: 20-24 marks (5-6 questions)
JEE Advanced: 15-18 marks (3-4 questions)
Difficulty Level: Easy to Very Hard
Time Required for Prep: 6-7 hours

🎯 Practice Problem Set

Level 1: Foundation

  1. Find 8P3
  2. Find 10C4
  3. How many 3-digit numbers can be formed from digits 1-9 without repetition?
  4. In how many ways can 5 boys sit in a row?
  5. Select 4 cards from 52 cards
  6. Prove: nCr = nC(n-r)
  7. How many words from PENCIL?
  8. Find number of diagonals in a polygon with 8 sides

Level 2: Intermediate (JEE Main)

  1. Arrange letters of MATHEMATICS
  2. 5 men and 5 women in a row, women always together
  3. Circular arrangement of 6 people
  4. Form committee of 5 from 6 men and 4 women (at least 2 women)
  5. Words from DAUGHTER starting with D
  6. Number of triangles from 10 points (no 3 collinear)
  7. Distribute 10 identical balls in 4 boxes
  8. Arrangement where vowels occupy even positions

Level 3: Advanced (JEE Advanced)

  1. Number of ways to distribute n distinct objects into r identical groups
  2. Find derangements of 5 objects
  3. Prove using inclusion-exclusion: Surjective functions
  4. Words from ASSASSINATION with no two S together
  5. Circular arrangement: men and women alternate, specific constraints
  6. Number of ways to select r objects from n when at least k are chosen
  7. Coefficient of x^20 in (1+x)^50
  8. Complex distribution with restrictions
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Permutations and Combinations - Complete Guide for JEE 2025-26

Why P&C is MOST IMPORTANT for JEE?

Permutations and Combinations has the HIGHEST weightage (7-8%) in JEE Mathematics:

  • Direct Questions: 5-6 questions every year
  • Probability: Foundation for all probability
  • Binomial Theorem: Uses nCr extensively
  • Graph Theory: Counting problems
  • Statistics: Sample selection

Critical Formulas

Permutation:

ⁿPᵣ = n!/(n-r)!

Combination:

ⁿCᵣ = n!/[r!(n-r)!]

Circular:

(n-1)!

Repetition:

n!/(p!×q!×...)

Relation:

ⁿPᵣ = r! × ⁿCᵣ