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Mathematics Grade XI

Grade 11 Mathematics

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Chapter 6: Permutations and Combinations

Class 11 • Mathematics • Chapter 6

Permutations and Combinations

Counting arrangements and selections without writing out every possibility.

CBSE / NCERT Aligned   School Revise

Chapter Roadmap

Counting Principle • Factorials • Permutations • Arranging Alike Objects • Combinations • Permutation vs Combination • Key Formulae • Applications
1

Why Permutations and Combinations Matter

Suppose a school has to pick a captain and a vice-captain from 30 students, or a quiz team of 4 from a class of 40. Writing out every possibility by hand would take hours, and you could easily miss some or repeat others. Counting is one of the oldest and most useful parts of mathematics, and this chapter gives you a fast, reliable way to count without listing.

The whole chapter rests on one simple rule, the Fundamental Principle of Counting, and from it we build two big ideas. A permutation is an arrangement, where the order matters, such as the order of runners finishing a race. A combination is a selection, where the order does not matter, such as the players chosen for a team. Telling these two apart is the single most important skill in this chapter, and once it becomes clear, the formulae fall into place naturally.

Key idea
If the order of the chosen items matters, you are counting permutations. If only which items are chosen matters and not their order, you are counting combinations. Captain and vice-captain is a permutation; a team of players is a combination.
2

Key Terms You Must Know
Term Meaning Example
Factorial The product of all positive integers up to n, written n!. 5! = 5 × 4 × 3 × 2 × 1 = 120
Permutation An arrangement of objects in a definite order. ABC and ACB are different permutations
Combination A selection of objects where order does not matter. {A, B} is the same as {B, A}
nPr The number of permutations of n different objects taken r at a time. ⁿPᵣ = n!/(n − r)!
nCr The number of combinations of n different objects taken r at a time. ⁿCᵣ = n!/(r!(n − r)!)
Fundamental Principle If one job can be done in m ways and another in n ways, both can be done in m × n ways. 2 shirts, 3 caps give 6 outfits
Zero factorial By definition 0! = 1, which keeps the formulae working. ⁿC₀ = n!/(0! · n!) = 1
3

Core Concepts, Step by Step

1. The Fundamental Principle of Counting

This is the rule everything else is built on. If one task can be carried out in m different ways, and a second task can then be carried out in n different ways, then the two tasks together can be carried out in m × n ways. The principle extends to any number of tasks: just multiply the number of choices at every stage. Suppose you own 2 shirts and 3 caps. For each shirt you may pick any of the 3 caps, so the total number of shirt-and-cap outfits is 2 × 3 = 6, as the grid below shows.

Counting outfits: 2 shirts and 3 caps

Shirt / Cap Cap 1 Cap 2 Cap 3
Shirt A A1 A2 A3
Shirt B B1 B2 B3

Each box is one possible outcome. With 2 choices for the first slot and 3 for the second, the grid holds 2 × 3 = 6 outcomes in total. This is the Fundamental Principle of Counting: multiply the number of choices at each stage.

2. Factorials

When we arrange several different objects, products like 5 × 4 × 3 × 2 × 1 appear again and again, so mathematics gives them a short name. The factorial of a positive integer n, written n!, is the product of all the whole numbers from n down to 1. So 4! = 4 × 3 × 2 × 1 = 24. A neat feature is that n! = n × (n − 1)!, which lets you build each factorial from the one before. We also define 0! = 1, which may look strange but keeps every counting formula correct.

The first few factorials

n n! written out Value
0 (by definition) 1
1 1 1
2 2 × 1 2
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

3. Permutations: When Order Matters

A permutation is an arrangement of objects in a definite order. The number of ways to arrange r objects chosen from n different objects, where order matters, is written ⁿPᵣ. Think of filling r empty places one by one. The first place has n choices, the second has (n − 1) left, the third has (n − 2), and so on. The figure below fills three places from seven different books, giving 7 × 6 × 5 = 210 arrangements. In factorial form this is ⁿPᵣ = n!/(n − r)!.

Filling 3 places from 7 different books

7
1st place
×
6
2nd place
×
5
3rd place

Seven books can fill the first place, then six remain for the second, then five for the third. By the counting principle the total is 7 × 6 × 5 = 210, which is ₇P₃.

4. Arranging Objects When Some Are Alike

If all the objects are different, n of them arrange in n! ways. But when some objects are identical, swapping the identical ones does not create a new arrangement, so we have counted too many. To correct this we divide by the factorial of the number of each repeated kind. If there are n objects in which one kind repeats p times and another repeats q times, the number of distinct arrangements is n!/(p! × q!). For example, the word LEVEL has 5 letters with L twice and E twice, so the number of distinct arrangements is 5!/(2! × 2!) = 120/4 = 30.

5. Combinations: When Order Does Not Matter

A combination is a selection in which order does not matter. Choosing players A and B for a team is the same as choosing B and A. The number of ways to select r objects from n different objects is written ⁿCᵣ and equals n!/(r!(n − r)!). There is a clean link between the two ideas: every combination of r objects can itself be arranged in r! orders, so ⁿPᵣ = r! × ⁿCᵣ. In words, a permutation is a combination that has then been put in order. So you first choose, then arrange.

6. Telling Permutations and Combinations Apart

The hardest part is deciding which one a problem needs, so always ask: does the order matter? Ranks, positions, seats in a row, and digits in a number all depend on order, so they are permutations. Teams, committees, groups, and handshakes do not depend on order, so they are combinations. The table below sets the two side by side.

Permutation or combination?

Question to ask Permutation Combination
Does order matter? Yes No
What is it? Arrangement Selection
Everyday example Captain and vice-captain A team of players
Formula ⁿPᵣ = n!/(n − r)! ⁿCᵣ = n!/(r!(n − r)!)
Link ⁿPᵣ = r! × ⁿCᵣ Choose first, then arrange
4

Key Results & Proofs

Three results carry this entire chapter, and each one follows directly from the Fundamental Principle of Counting. Seeing why they are true makes them far easier to recall than learning them as bare formulae.

Result 1: The Permutation Formula

Statement. The number of permutations of n different objects taken r at a time is ⁿPᵣ = n!/(n − r)!.

Proof

We fill the places one at a time and multiply the choices.

Imagine r empty places to be filled in order.  –  set up the problem
The 1st place can be filled in n ways.  –  all n objects are available
The 2nd place can then be filled in (n − 1) ways.  –  one object is already used
Continuing, the r-th place can be filled in (n − r + 1) ways.  –  r − 1 objects already used
ⁿPᵣ = n(n − 1)(n − 2) … (n − r + 1) multiply, by the counting principle
ⁿPᵣ = n!/(n − r)! multiply top and bottom by (n − r)!

Putting r = n gives ⁿPₙ = n!/0! = n!, the number of ways to arrange all n objects.
Result 2: The Combination Formula

Statement. The number of combinations of n different objects taken r at a time is ⁿCᵣ = n!/(r!(n − r)!).

Proof

We connect combinations to permutations, which we have already counted.

Each selection of r objects can be arranged among itself in r! orders.  –  order does not matter in a combination
So every combination produces r! different permutations.  –  one selection, r! arrangements
ⁿPᵣ = r! × ⁿCᵣ total permutations = (combinations) × (arrangements of each)
ⁿCᵣ = ⁿPᵣ / r! divide both sides by r!
ⁿCᵣ = n!/(r!(n − r)!) substitute ⁿPᵣ = n!/(n − r)!

Putting r = 0 gives ⁿC₀ = n!/(0! · n!) = 1: there is exactly one way to choose nothing.
Result 3: The Symmetry Rule

Statement. For all whole numbers r from 0 to n, ⁿCᵣ = ⁿCₙ₋ᵣ.

Proof

Choosing r objects to keep is the same as choosing (n − r) objects to leave out.

ⁿCₙ₋ᵣ = n!/((n − r)!(n − (n − r))!) replace r by (n − r) in the formula
ⁿCₙ₋ᵣ = n!/((n − r)! · r!) since n − (n − r) = r
ⁿCₙ₋ᵣ = n!/(r!(n − r)!) rearrange the two factors
ⁿCₙ₋ᵣ = ⁿCᵣ this is exactly the formula for ⁿCᵣ

This is why ⁹C₇ is far quicker as ⁹C₂ = 36: choosing 7 to keep equals choosing 2 to drop.
5

Worked Examples
Example 1

Question: How many 3-digit numbers can be formed from the digits 1 to 9 if no digit is repeated?

▶ Show full working

We are filling three places (hundreds, tens, units) in order, so this is a permutation. Each used digit cannot appear again.

hundreds place = 9 choices any of the nine digits
tens place = 8 choices one digit already used
units place = 7 choices two digits already used
total = 9 × 8 × 7 Fundamental Principle of Counting
total = 504 multiply

Answer: 504 such 3-digit numbers.
Example 2

Question: Evaluate 7! / 5!.

▶ Show full working

Write the larger factorial in terms of the smaller one, then cancel. There is no need to compute both fully.

7! = 7 × 6 × 5! since n! = n × (n − 1)!
7!/5! = (7 × 6 × 5!) / 5! substitute
7!/5! = 7 × 6 cancel the common 5!
7!/5! = 42 multiply

Answer: 42.
Example 3

Question: Find ₇P₃, the number of permutations of 7 objects taken 3 at a time.

▶ Show full working

Use ⁿPᵣ = n!/(n − r)! with n = 7 and r = 3, or simply fill three places.

₇P₃ = 7! / (7 − 3)! permutation formula
₇P₃ = 7! / 4! simplify the bracket
₇P₃ = 7 × 6 × 5 cancel 4!
₇P₃ = 210 multiply

Answer: 210.
Example 4

Question: In how many ways can the letters of the word DELHI be arranged?

▶ Show full working

DELHI has 5 different letters, so we are arranging all of them. Order matters, so it is a permutation of 5 distinct objects.

ways = 5! arrange 5 different letters
ways = 5 × 4 × 3 × 2 × 1 write out the factorial
ways = 120 multiply

Answer: 120 arrangements.
Example 5

Question: How many distinct arrangements are there of the letters of the word LEVEL?

▶ Show full working

LEVEL has 5 letters, but L appears twice and E appears twice. Swapping the two L’s, or the two E’s, gives no new word, so we divide.

arrangements = 5! / (2! × 2!) divide by the repeats of L and E
arrangements = 120 / (2 × 2) evaluate the factorials
arrangements = 120 / 4 multiply the denominator
arrangements = 30 divide

Answer: 30 distinct arrangements.
Example 6

Question: Find ₈C₂, the number of ways to choose 2 objects from 8.

▶ Show full working

Order does not matter when we simply choose 2 objects, so this is a combination. Use ⁿCᵣ = n!/(r!(n − r)!).

₈C₂ = 8! / (2! × 6!) combination formula
₈C₂ = (8 × 7) / (2 × 1) cancel 6! and expand
₈C₂ = 56 / 2 multiply top and bottom
₈C₂ = 28 divide

Answer: 28 ways.
Example 7

Question: A committee of 3 is to be chosen from 10 people. In how many ways can this be done?

▶ Show full working

A committee has no ranks, so order does not matter. This is a combination of 3 from 10.

ways = ₁₀C₃ choose 3 from 10
ways = 10! / (3! × 7!) combination formula
ways = (10 × 9 × 8) / (3 × 2 × 1) cancel 7!
ways = 720 / 6 evaluate
ways = 120 divide

Answer: 120 ways.
Example 8

Question: From 5 men and 4 women, a group of 2 men and 1 woman is to be chosen. In how many ways?

▶ Show full working

Choose the men and the women separately (each a combination), then multiply by the counting principle.

choosing 2 men = ₅C₂ = 10 10 ways from 5 men
choosing 1 woman = ₄C₁ = 4 4 ways from 4 women
total = 10 × 4 Fundamental Principle of Counting
total = 40 multiply

Answer: 40 ways.
Example 9

Question: How many ways can the letters of EQUATION be arranged?

▶ Show full working

EQUATION has 8 letters and all of them are different, so we arrange 8 distinct objects.

ways = 8! arrange 8 different letters
ways = 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 write out the factorial
ways = 40320 multiply

Answer: 40320 arrangements.
Example 10

Question: How many diagonals does a hexagon (6 sides) have?

▶ Show full working

Joining any 2 of the 6 vertices gives a line. Choosing 2 vertices is a combination. But 6 of those lines are the sides themselves, not diagonals, so we subtract them.

lines from 2 vertices = ₆C₂ choose 2 of 6 vertices
₆C₂ = (6 × 5)/(2 × 1) = 15 evaluate the combination
Of these 15 lines, 6 are the sides of the hexagon.  –  sides are not diagonals
diagonals = 15 − 6 subtract the 6 sides
diagonals = 9 simplify

Answer: 9 diagonals.
Example 11

Question: In how many ways can the first, second and third prizes be given to 5 students?

▶ Show full working

The prizes are ranked, so order matters. This is a permutation of 3 from 5.

ways = ₅P₃ arrange 3 of 5 in order
ways = 5! / (5 − 3)! permutation formula
ways = 5 × 4 × 3 cancel 2!
ways = 60 multiply

Answer: 60 ways.
Example 12

Question: If ⁿC₂ = 10, find n.

▶ Show full working

Write the combination as a formula, set it equal to 10, and solve the resulting equation.

ⁿC₂ = n(n − 1)/2 since ⁿC₂ = n!/(2!(n − 2)!)
n(n − 1)/2 = 10 given value
n(n − 1) = 20 multiply both sides by 2
n = 5 because 5 × 4 = 20

Answer: n = 5.
6

Where You Meet This in Real Life

Passwords and PIN codes

Counting how many codes are possible tells you how safe a PIN or password is. A 4-digit PIN with repeats allowed has 10 × 10 × 10 × 10 = 10,000 possibilities, which is why longer codes are far harder to guess.

Teams and committees

Selecting a debate team, a project group or a school committee from a larger set is a pure combination problem, since the chosen members have no internal order.

Seating and queues

Arranging guests around a table, students in a row, or books on a shelf is a permutation, because changing the order produces a genuinely different arrangement.

Tournaments and fixtures

Working out how many matches are needed in a round-robin tournament, where every team plays every other once, is the combination ⁿC₂.

Number plates and tickets

Designing vehicle number plates or lottery tickets relies on the counting principle to make sure there are enough unique labels for everyone.
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Practice Sets A to D

Practice Set A – Basics

A1. In how many ways can 6 different books be arranged on a shelf?

▶ Reveal full working

All 6 books are different, so arrange all of them: that is 6!.

ways = 6! arrange 6 different books
ways = 720 6 × 5 × 4 × 3 × 2 × 1

Answer: 720.

A2. Find ₅P₂.

▶ Reveal full working

Fill two places in order from 5 objects, or use the formula.

₅P₂ = 5 × 4 first place 5 ways, second 4 ways
₅P₂ = 20 multiply

Answer: 20.

A3. Find ₆C₂.

▶ Reveal full working

Choose 2 from 6, order not mattering.

₆C₂ = (6 × 5)/(2 × 1) combination formula
₆C₂ = 15 simplify

Answer: 15.

A4. How many 2-digit numbers can be formed from the digits 1, 2, 3 with no repetition?

▶ Reveal full working

Two places to fill in order, so it is a permutation.

tens place = 3 choices any of the three digits
units place = 2 choices one digit already used
total = 3 × 2 = 6 counting principle

Answer: 6 numbers.

Practice Set B – Conceptual

B1. Explain the difference between a permutation and a combination with one example each.

▶ Reveal full working

The deciding question is whether order matters.

A permutation is an arrangement where order matters; example: the order in which runners finish a race.  –  order counts
A combination is a selection where order does not matter; example: choosing 3 students for a committee.  –  order does not count

Answer: Permutation = arrangement (order matters); combination = selection (order does not).

B2. Why is 0! defined to be 1?

▶ Reveal full working

The definition is chosen so that the formulae keep working.

The rule n! = n × (n − 1)! with n = 1 gives 1! = 1 × 0!.  –  apply the rule at n = 1
Since 1! = 1, we need 0! = 1 for this to hold.  –  solve for 0!
It also makes ⁿC₀ = 1, since there is one way to choose nothing.  –  keeps combinations correct

Answer: 0! = 1 keeps n! = n × (n − 1)! and the counting formulae consistent.

B3. Show that ₇C₃ = ₇C₄.

▶ Reveal full working

Use the symmetry rule ⁿCᵣ = ⁿCₙ₋ᵣ with n = 7.

₇C₃ = ₇C₇₋₃ symmetry rule
₇C₃ = ₇C₄ since 7 − 3 = 4

Answer: Both equal 35, so ₇C₃ = ₇C₄.

B4. Verify that ₅P₃ = 3! × ₅C₃.

▶ Reveal full working

Compute each side using the formulae and compare.

₅P₃ = 60 5 × 4 × 3
₅C₃ = 10 (5 × 4 × 3)/(3 × 2 × 1)
3! × ₅C₃ = 6 × 10 = 60 matches ₅P₃

Answer: Both sides equal 60, confirming ⁿPᵣ = r! × ⁿCᵣ.

Practice Set C – Application / Numerical

C1. How many distinct arrangements are there of the letters of the word BANANA?

▶ Reveal full working

BANANA has 6 letters: A appears 3 times and N appears 2 times, so divide by their repeats.

arrangements = 6! / (3! × 2!) divide by repeats of A and N
arrangements = 720 / (6 × 2) evaluate factorials
arrangements = 720 / 12 simplify denominator
arrangements = 60 divide

Answer: 60 arrangements.

C2. From 7 boys and 5 girls, a team of 2 boys and 2 girls is chosen. In how many ways?

▶ Reveal full working

Choose boys and girls separately (each a combination), then multiply.

choosing 2 boys = ₇C₂ = 21 from 7 boys
choosing 2 girls = ₅C₂ = 10 from 5 girls
total = 21 × 10 counting principle
total = 210 multiply

Answer: 210 ways.

C3. How many 4-letter codes can be formed from the 26 English letters if repetition is allowed?

▶ Reveal full working

Each of the 4 places can be any of 26 letters, so multiply 26 four times.

codes = 26 × 26 × 26 × 26 repetition allowed at each place
codes = 26⁴ write as a power
codes = 456976 evaluate

Answer: 456976 codes.

C4. If ⁿP₂ = 20, find n.

▶ Reveal full working

Write ⁿP₂ as a product and solve the equation.

ⁿP₂ = n(n − 1) first two places
n(n − 1) = 20 given
n = 5 because 5 × 4 = 20

Answer: n = 5.

Practice Set D – HOTS / Word Problems

D1. How many diagonals does a decagon (10 sides) have?

▶ Reveal full working

Lines join any 2 of the 10 vertices; subtract the 10 sides.

lines = ₁₀C₂ choose 2 of 10 vertices
₁₀C₂ = (10 × 9)/(2 × 1) = 45 evaluate
Subtract the 10 sides, which are not diagonals.  –  sides are not diagonals
diagonals = 45 − 10 = 35 subtract

Answer: 35 diagonals.

D2. In how many arrangements of the letters of PENCIL do the two vowels E and I stay together?

▶ Reveal full working

Tie the vowels into one block, arrange the block with the other letters, then arrange the vowels inside the block.

Treat (EI) as a single block. Then PNCL plus the block make 5 units.  –  glue the vowels together
arranging 5 units = 5! = 120 arrange the 5 units in a row
inside the block = 2! = 2 E and I can swap
total = 120 × 2 = 240 counting principle

Answer: 240 arrangements.

D3. A cricket squad of 9 is to be chosen from 13 players. In how many ways?

▶ Reveal full working

A squad has no internal order, so it is a combination; use the symmetry rule to keep numbers small.

ways = ₁₃C₉ choose 9 of 13
₁₃C₉ = ₁₃C₄ symmetry rule, 13 − 9 = 4
₁₃C₄ = (13 × 12 × 11 × 10)/(4 × 3 × 2 × 1) combination formula
₁₃C₄ = 17160 / 24 = 715 evaluate

Answer: 715 ways.

D4. How many 3-digit even numbers can be formed from the digits 1, 2, 3, 4, 5 with no repetition?

▶ Reveal full working

An even number must end in an even digit, so fix the units place first, then fill the rest.

The units digit must be even: 2 or 4, giving 2 choices.  –  even number ends in 2 or 4
hundreds place = 4 choices any of the remaining four digits
tens place = 3 choices two digits now used
total = 2 × 4 × 3 = 24 counting principle

Answer: 24 even numbers.

Chapter Summary

Everything in One Glance

Counting Principle

If one task has m ways and the next has n ways, together they have m × n ways. Multiply the choices at each stage.

  

Factorials

n! = n × (n − 1) × … × 1, with 0! = 1. Each factorial is n times the one before.
 

Permutations

Arrangements where order matters. ⁿPᵣ = n!/(n − r)!, and all of n arrange in n! ways.

  

Alike Objects

With repeats, divide: n objects with repeats p and q arrange in n!/(p! q!) ways.
 

Combinations

Selections where order does not matter. ⁿCᵣ = n!/(r!(n − r)!).

  

Two Key Links

ⁿPᵣ = r! × ⁿCᵣ, and ⁿCᵣ = ⁿCₙ₋ᵣ (choosing what to keep equals choosing what to drop).
 
8

Are You Exam-Ready?

8-Point Exam Quick-Check

1 State the Fundamental Principle of Counting in one sentence.
 
2 Evaluate 6!/4! without a calculator.
 
3 Find ₈P₃ and ₈C₃.
 
4 How many distinct arrangements has the word BALLOON?
 
5 Decide: choosing a 4-member committee from 12, is it a permutation or a combination?
 
6 If ⁿC₂ = 15, find n.
 
7 How many diagonals does an octagon (8 sides) have?
 
8 In how many ways can 5 people be seated in a row?
 

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Class 11 Mathematics Chapter 6: Permutations and Combinations, Complete Notes and Practice

These free Class 11 Maths Chapter 6 Permutations and Combinations notes follow the latest NCERT 2026-27 syllabus and give complete, exam-ready coverage for board exams and competitive foundations. Includes worked examples, step-by-step proofs and graded practice, free on SchoolRevise.com.
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