Class 11 Maths Chapter 6: Permutations and Combinations

Permutations and Combinations is Chapter 6 of NCERT Class 11 Maths. It teaches smart ways of counting arrangements and selections without listing every possibility, using the fundamental principle of counting, factorials, permutations and combinations. Watch the 10 video lessons below, then use the notes, formulas and solved examples to practise.

What you will learn

The fundamental principle of counting and how to apply it step by step
Factorial notation and how to simplify factorial expressions
Permutations of distinct objects, with repetition, and with some objects alike
Combinations and the link between nPr and nCr
How to decide whether a question needs an arrangement or a selection

Video Lessons

Permutations and Combinations - Part 1

Permutations and Combinations - Part 2

Permutations and Combinations - Part 3

Permutations and Combinations - Part 4

Permutations and Combinations - Part 5

Permutations and Combinations - Part 6

Permutations and Combinations - Part 7

Permutations and Combinations - Part 8

Permutations and Combinations - Part 9

Permutations and Combinations - Chapter Revision

Key Concepts and Formulas

Fundamental Principle of Counting

If one task can be done in m ways and, after it, a second task can be done in n ways, then both tasks together can be done in m × n ways. This multiplication rule extends to any number of tasks done one after another.

Multiplication principle (A and then B): m × n
Addition idea (A or B, not both): m + n

Use the addition idea only when the methods cannot happen together
Draw blanks or boxes for each position and fill the number of choices in each before multiplying

Factorial Notation

The product of the first n natural numbers is written n! and read "n factorial". Factorials appear in every formula of this chapter.

n! = 1 × 2 × 3 × ... × n
0! = 1
n! = n × (n − 1)!
5! = 120, 6! = 720
7! = 5040, 8! = 40320

The result n! = n × (n − 1)! helps in cancelling large factorials

Permutations

A permutation is an arrangement of objects in a definite order. Changing the order gives a new permutation, so ABC and BAC are different.

nPr = n! / (n − r)!,  0 ≤ r ≤ n
nPn = n!
With repetition allowed: nʳ
With like objects: n! / (p! × q! × ...)
where p, q, ... = counts of each repeated kind
nP0 = 1,  nP1 = n

nPr counts arrangements of n different objects taken r at a time, without repetition

Combinations

A combination is a selection where order does not matter. Choosing Riya and Aman for a team is the same as choosing Aman and Riya.

nCr = n! / (r! (n − r)!)
nPr = nCr × r!
nCr = nC(n − r)
nC0 = nCn = 1,  nC1 = n
nCr + nC(r − 1) = (n + 1)Cr
If nCa = nCb, then a = b or a + b = n

nPr = nCr × r! because each selection of r objects can be arranged in r! ways
nCr = nC(n − r) because choosing r is the same as rejecting n − r
The identity nCr + nC(r − 1) = (n + 1)Cr is called Pascal's rule

Arrangement or Selection

The biggest skill in this chapter is reading the question correctly. Ask yourself whether swapping two chosen items gives a new outcome.

Permutation vs combination diagram listing ordered pairs and selections of A, B, C for Class 11 Maths Chapter 6 — Six arrangements but only three selections; P(3, 2) is 3P2

Words like arrange, order, rank, digits of a number, seating or words from letters usually need permutations
Words like choose, select, committee, team, group or handshakes usually need combinations
Many problems need both: first select with nCr, then arrange with r!
When some items must stay together, tie them into one unit, arrange the units, then arrange inside the unit

Solved Examples

Example 1: A café offers 4 starters, 6 main courses and 3 desserts. In how many ways can a customer choose one of each?

Ways = 4 × 6 × 3
∴ Ways = 72

Example 2: How many 3-digit even numbers with distinct digits can be formed from 1, 2, 4, 6, 7, 9?

Units place (2, 4 or 6): 3 ways
Tens place: 5 ways
Hundreds place: 4 ways
Total = 3 × 5 × 4
∴ 60 numbers

Fundamental principle of counting boxes for 3-digit even numbers, Class 11 Permutations and Combinations example — Fill the units place first, then multiply

Example 3: In how many ways can the letters of the word JANAKPURI be arranged?

Total letters = 9, A repeated twice
Arrangements = 9! / 2!
= 362880 / 2
∴ 181440 ways

Example 4: Five different books are placed on a shelf so that two particular books are always together. Find the number of arrangements.

Tie the 2 books into one unit: 4 units
Arrange units: 4! = 24
Arrange inside the unit: 2! = 2
Total = 24 × 2
∴ 48 ways

Example 5: A committee of 4 is to be formed from 7 boys and 5 girls with exactly 2 girls. In how many ways can this be done?

Choose 2 girls: 5C2 = 10
Choose 2 boys: 7C2 = 21
Total = 10 × 21
∴ 210 committees

Example 6: If nC10 = nC6, find nC3.

10 ≠ 6, so n = 10 + 6 = 16
16C3 = (16 × 15 × 14)/(3 × 2 × 1)
= 3360/6
∴ nC3 = 560

Example 7: How many diagonals does a decagon (10 sides) have?

Segments joining 2 vertices = 10C2 = 45
Diagonals = 45 − 10 sides
∴ 35 diagonals

Common Mistakes to Avoid

Using nCr when order matters, or nPr when it does not. Check by swapping two chosen items and asking if the outcome changes.
Forgetting that 0 cannot be in the first place of a number when 0 is one of the given digits.
Ignoring repeated letters in word problems. Always divide by the factorial of each repeated count.
In "together" questions, forgetting to multiply by the internal arrangements of the grouped items.

Quick Revision

Multiplication principle: m × n
n! = n × (n − 1)!,  0! = 1
nPr = n! / (n − r)!
Repetition allowed: nʳ
Alike objects: n! / (p! q! r!)
nCr = n! / (r! (n − r)!)
nPr = nCr × r!
nCr = nC(n − r)
nCr + nC(r − 1) = (n + 1)Cr