Class 11 Maths Chapter 6 Permutations and Combinations NCERT Solutions | Full Guide with Examples

Introduction

Permutations and Combinations is one of the most important chapters in Class 11 Maths. It teaches how to count arrangements and selections without actually listing them.

This chapter builds the foundation for probability and helps in solving real-life counting problems like seating arrangements, passwords, and selections.

Basic Concepts

1. Factorial Notation

n! = n × (n−1) × (n−2) × ... × 1
0! = 1

Important Formulas

1. Permutations (Arrangement matters)

$$^{n}P_{r} = \frac{n!}{(n-r)!}$$

2. Combinations (Order does not matter)

$$^{n}C_{r} = \frac{n!}{r!(n-r)!}$$

3. Relation between P and C

nPr = nCr × r!

4. Special Results

nC0 = nCn = 1
nCr = nC(n−r)

NCERT SOLVED EXAMPLES (STEP BY STEP)

 Example 1

Find 5!

Solution:
5! = 5 × 4 × 3 × 2 × 1 = 120

 Example 2

Find 6P2

Solution:
6P2 = 6! / (6−2)!
= 6! / 4!
= (6 × 5 × 4!) / 4!
= 6 × 5 = 30

Example 3

Find 6C2

Solution:
6C2 = 6! / (2! × 4!)
= (6 × 5 × 4!) / (2 × 1 × 4!)
= (6 × 5) / 2 = 15

Example 4

In how many ways can 3 students be selected from 7?

Solution:
7C3 = 7! / (3! × 4!)
= (7 × 6 × 5) / (3 × 2 × 1)
= 35

 Example 5

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

Solution:
= 4P3
= 4! / (4−3)!
= 4! / 1! = 24

 Example 6

Arrange letters of the word “CAT”

Solution:
Total letters = 3
Arrangements = 3! = 6

 Example 7

Arrange letters of “APPLE”

Solution:
Total letters = 5
Repeated letter = P (2 times)

Number of arrangements = 5! / 2!
= 120 / 2 = 60

 Example 8

In how many ways can 5 people sit in a row?

Solution:
= 5! = 120

 Example 9

How many ways to choose 2 boys from 5 boys?

Solution:
= 5C2 = 10

 Example 10

How many ways to arrange 4 books out of 6?

Solution:
= 6P4
= 6! / 2!
= 360

ALL EXERCISE QUESTIONS (IMPORTANT TYPES COVERED)

Instead of copying entire NCERT text, here are all key question types with full solutions:

Type 1: Factorial Simplification

Evaluate: 8! / 6!

Solution:
= (8 × 7 × 6!) / 6!
= 8 × 7 = 56

Type 2: Find nPr

Find 7P3

Solution:
= 7 × 6 × 5 = 210

Type 3: Find nCr

Find 8C3

Solution:
= (8 × 7 × 6) / (3 × 2 × 1) = 56

 Type 4: Word Arrangement

Arrange “LEVEL”

Solution:
Letters = 5
L = 2 times, E = 2 times

= 5! / (2! × 2!)
= 120 / 4 = 30

Type 5: Selection Problems

Choose 3 students from 10

= 10C3 = 120

 Type 6: Mixed Problems

Number of ways to arrange 3 letters from 5 letters

= 5P3 = 60

15 FAQs 

1. What is permutation?

Permutation means arrangement where order matters.

2. What is combination?

Combination means selection where order does not matter.

3. What is nPr formula?

nPr = n! / (n-r)!

4. What is nCr formula?

nCr = n! / (r!(n-r)!)

5. What is factorial?

n! = n × (n−1) × ... × 1

6. What is 0! value?

0! = 1

7. When to use permutation?

When order matters.

8. When to use combination?

When order does not matter.

9. What is nC0?

nC0 = 1

10. What is relation between nPr and nCr?

nPr = nCr × r!

11. How to arrange repeated letters?

Use formula n! / (repetition factorials)

12. What is 5C2?

5C2 = 10

13. What is 6P2?

6P2 = 30

14. What is symmetric property?

nCr = nC(n-r)

15. Why is this chapter important?

It is used in probability and competitive exams.