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Class 11 Maths Notes

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Permutations and Combinations Class 11 Notes

Complete, exam-ready notes on permutations and combinations: the fundamental principle of counting, factorials, arrangements (permutations) and selections (combinations), the relation between nPr and nCr, circular permutations, and arrangements of identical objects — written for CBSE boards and JEE revision.

Class:11Subject:MathematicsCovers:CBSE · JEE
7 Key Formulas4 Practice MCQs
DWritten byDeep Narayan
Updated
Key Concept Summary

What is the core idea of this chapter?

Permutations count ordered arrangements and combinations count unordered selections — both built on the fundamental principle of counting and factorials.

01

Fundamental Principle of Counting

Definition

Multiplication principle

If one task can be done in m ways and a second independent task can be done in n ways, the two tasks together can be done in m × n ways. This extends to any number of sequential tasks.

Definition

Addition principle

If one task can be done in m ways and a different, mutually exclusive task can be done in n ways, exactly one of the two tasks can be done in m + n ways.

Multiplication vs addition

Use multiplication when tasks happen in sequence (AND). Use addition when tasks are alternatives (OR). Choosing between them is the first decision in every counting problem.
02

Factorials and Permutations

Factorial
Definition

Permutation

A permutation is an arrangement of r objects chosen from n distinct objects where order matters. The number of such arrangements is .

  • Permutations with repetition allowed: each of the r positions has n choices, giving n^r arrangements.
  • All n objects arranged in a row: n! permutations.
  • Arranging n objects in a circle: (n − 1)! circular permutations (fixing one object eliminates rotational duplicates).
03

Permutations of Objects Not All Distinct

Permutations with identical objects

When n objects include p of one kind, q of another, r of a third, and so on, the number of distinct arrangements is . For example, the letters of MISSISSIPPI (M=1, I=4, S=4, P=2) give distinct arrangements.

04

Circular Permutations

Definition

Circular permutation

In a circular arrangement, rotations of the same order are considered identical. Fixing one object's position eliminates rotational symmetry, leaving (n − 1)! arrangements of the remaining n − 1 objects.

  • If clockwise and anticlockwise order are both the same (e.g., a necklace), the count is (n − 1)!/2.
  • If direction matters (e.g., people seated around a table with a distinguished side), the count is (n − 1)!.
  • Fixing one object is the standard technique — it removes all rotational duplicates.
05

Combinations

Definition

Combination

A combination is a selection of r objects from n distinct objects where order does not matter. The number of such selections is .

  • Relation between permutations and combinations:
  • .
  • Symmetry property:
  • .
  • Pascal's identity:
  • .
  • Sum of all combinations:
  • .
06

Key Properties and Relations

  • nPr = r! · nCr — permutations count order, combinations do not.
  • nCr = nC(n − r) — choosing r to include equals choosing n − r to exclude.
  • nCr + nC(r − 1) = (n + 1)Cr — Pascal's recurrence, the basis of Pascal's triangle.
  • For r objects taken one at a time with repetition from n objects: n^r ways.
  • When all n objects are taken (r = n): nPn = nCn = 1 arrangement/selection.

Order matters or not?

The word 'arrange' or 'order' signals a permutation; the word 'select', 'choose' or 'group' signals a combination. Misreading this is the most common mistake.
07

Solved Examples

Example: In how many ways can a committee of 4 men and 3 women be chosen from 7 men and 5 women?

Solution: Choose 4 men from 7: . Choose 3 women from 5: . By the multiplication principle, total = 35 × 10 = 350 ways.

Example: Find the number of permutations of the letters of the word ARRANGE.

Solution: ARRANGE has 7 letters with A repeated 2 times and R repeated 2 times. Distinct arrangements = .

08

NCERT Exercise Questions

From the textbook

All NCERT questions for this chapter — NCERT Maths Part I, Chapter 6 (Permutations and Combinations). Questions follow the textbook order so you can match them to the book.

Exercises (NCERT Q1–Q31)

  • 1) How many 3-digit numbers can be formed from the digits 1, 2, 3, 4 and 5 assuming that (i) repetition of the digits is allowed? (ii) repetition is not allowed?
  • 2) How many 3-digit even numbers can be formed from the digits 1, 2, 3, 4, 5, 6 if the digits can be repeated?
  • 3) How many 4-letter codes can be formed using the first 10 letters of the English alphabet, if no letter can be repeated?
  • 4) How many 5-digit telephone numbers can be constructed using the digits 0 to 9 if each number starts with 67 and no digit appears more than once?
  • 5) A coin is tossed 3 times and the outcomes are recorded. How many possible outcomes are there?
  • 6) Given 5 flags of different colours, how many different signals can be generated if each signal requires the use of 2 flags, one below the other?
  • 7) Evaluate: (i) 8! (ii) 4! − 3!.
  • 8) Is 3! + 4! = 7! ?
  • 9) Compute 8! / (6! × 2!).
  • 10) If 1/6! + 1/7! = x/8!, find x.
  • 11) Evaluate n!/(n − r)! when (i) n = 6, r = 2; (ii) n = 9, r = 5.
  • 12) How many 3-digit numbers can be formed by using the digits 1 to 9 if no digit is repeated?
  • 13) How many 4-digit numbers are there with no digit repeated?
  • 14) How many 3-digit even numbers can be made using the digits 1, 2, 3, 4, 6, 7, if no digit is repeated?
  • 15) Find the number of 4-digit numbers that can be formed using the digits 1, 2, 3, 4, 5 if no digit is repeated. How many of these will be even?
  • 16) From a committee of 8 persons, in how many ways can we choose a chairman and a vice chairman assuming one person cannot hold more than one position?
  • 17) Find n if ⁿ⁻¹P₃ : ⁿP₄ = 1 : 9.
  • 18) Find r if (i) ⁵Pᵣ = 2 · ⁶Pᵣ₋₁, (ii) ⁵Pᵣ = ⁶Pᵣ₋₁.
  • 19) How many words, with or without meaning, can be formed using all the letters of the word EQUATION, using each letter exactly once?
  • 20) How many words, with or without meaning, can be made from the letters of the word MONDAY, assuming that no letter is repeated, if (i) 4 letters are used at a time, (ii) all letters are used at a time, (iii) all letters are used but the first letter is a vowel?
  • 21) In how many of the distinct permutations of the letters in MISSISSIPPI do the four I's not come together?
  • 22) In how many ways can the letters of the word PERMUTATIONS be arranged if (i) words start with P and end with S, (ii) vowels are all together, (iii) there are always 4 letters between P and S?
  • 23) If ⁿC₈ = ⁿC₂, find ⁿC₂.
  • 24) Determine n if (i) ²ⁿC₃ : ⁿC₃ = 12 : 1, (ii) ²ⁿC₃ : ⁿC₃ = 11 : 1.
  • 25) How many chords can be drawn through 21 points on a circle?
  • 26) In how many ways can a team of 3 boys and 3 girls be selected from 5 boys and 4 girls?
  • 27) Find the number of ways of selecting 9 balls from 6 red, 5 white and 5 blue balls if each selection consists of 3 balls of each colour.
  • 28) Determine the number of 5-card combinations out of a deck of 52 cards if there is exactly one ace in each combination.
  • 29) In how many ways can one select a cricket team of eleven from 17 players in which only 5 players can bowl, if each cricket team of 11 must include exactly 4 bowlers?
  • 30) A bag contains 5 black and 6 red balls. Determine the number of ways in which 2 black and 3 red balls can be selected.
  • 31) In how many ways can a student choose a programme of 5 courses if 9 courses are available and 2 specific courses are compulsory for every student?

Answers at a glance

  1. 1
    Q1 → (i) 125, (ii) 60.
  2. 2
    Q2 → 108.
  3. 3
    Q3 → 5040.
  4. 4
    Q4 → 336.
  5. 5
    Q5 → 8.
  6. 6
    Q6 → 20.
  7. 7
    Q7 → (i) 40320, (ii) 18.
  8. 8
    Q8 → No (30 ≠ 5040).
  9. 9
    Q9 → 28.
  10. 10
    Q10 → x = 64.
  11. 11
    Q11 → (i) 30, (ii) 15120.
  12. 12
    Q12 → 504.
  13. 13
    Q13 → 4536.
  14. 14
    Q14 → 60.
  15. 15
    Q15 → 120 total, 48 even.
  16. 16
    Q16 → 56.
  17. 17
    Q17 → n = 9.
  18. 18
    Q18 → (i) r = 3, (ii) r = 4.
  19. 19
    Q19 → 40320.
  20. 20
    Q20 → (i) 360, (ii) 720, (iii) 240.
  21. 21
    Q21 → 33810.
  22. 22
    Q22 → (i) 1814400, (ii) 2419200, (iii) 25401600.
  23. 23
    Q23 → 45.
  24. 24
    Q24 → (i) n = 5, (ii) n = 6.
  25. 25
    Q25 → 210.
  26. 26
    Q26 → 40.
  27. 27
    Q27 → 2000.
  28. 28
    Q28 → 778320.
  29. 29
    Q29 → 3960.
  30. 30
    Q30 → 200.
  31. 31
    Q31 → 35.
09

CBSE Pattern Questions

How CBSE asks this chapter

CBSE Class 11 Mathematics board paper: Section A (20 MCQs × 1M), Section B (5 × 2M), Section C (6 × 3M) and Section D (4 × 4M). Permutations and Combinations gives 1-mark MCQs on factorial and nPr/nCr values, 2/3-mark questions on arrangements and selections, and 4-mark word problems with repeated letters or restriction constraints.

MCQ Questions

Q1
The value of 5! is
Q2
P(5, 2) equals
Q3
C(6, 2) equals
Q4
The number of ways to arrange 3 different books on a shelf is

1 Mark Questions

  1. 1
    Write the value of 0!. (Ans: 1)
  2. 2
    State the formula for nPr. (Ans: n!/(n − r)!)
  3. 3
    Find C(4, 1). (Ans: 4)
  4. 4
    In how many ways can a committee of 2 be chosen from 5 people? (Ans: C(5, 2) = 10)

2 Mark Questions

  1. 1
    How many 3-digit numbers can be formed from the digits 1–5 without repetition? (Ans: 5 × 4 × 3 = 60)
  2. 2
    How many arrangements of the word 'APPLE' are there? (Ans: 5!/2! = 60)
  3. 3
    Find the number of diagonals of a hexagon using the combination formula. (Ans: C(6, 2) − 6 = 9)

3 Mark Questions

  1. 1
    If nP3 = 60, find n. (Ans: n!/(n−3)! = 60 ⇒ n = 5)
  2. 2
    A team of 3 boys and 2 girls is to be selected from 5 boys and 4 girls; in how many ways? (Ans: C(5,3) × C(4,2) = 10 × 6 = 60)
  3. 3
    How many 4-letter words can be formed from the letters of the word 'EXAMINATION' without repetition, taking vowels always together? (Ans: treat the 5 vowels {E,A,A,I,O} converted to distinct with repeats — count via cases)

4 Mark Questions

  1. 1
    In how many ways can the letters of the word 'INDEPENDENCE' be arranged (i) without restriction, (ii) when the vowels always come together, (iii) when the word starts with I and ends with E? (Ans: (i) 12!/(3!2!4!), (ii) vowel block count, (iii) fixed ends) Show the factorial divisions.
  2. 2
    Prove the identity C(n, r) = C(n, n − r) using the factorial definition, and verify with n = 10, r = 3. Then find the number of ways to form a hand of 3 cards from a 52-card pack. (Ans: C(52, 3))
10

PDF: Assignments & Tests

Printable study material

Free downloadable PDFs for Permutations and Combinations — two topic-wise assignments covering every concept of the chapter, and two full CBSE-pattern tests (each 90 minutes, 40 marks) mirroring the board paper. No signup required.

Free PDF Downloads & Assignments

No Signup Required

Practice path: complete Assignment 1 first (topic-by-topic), then Assignment 2 (mixed difficulty with full solutions), then attempt the timed tests.

Permutations and Combinations Assignment 1 — Topic-wise Questions

Covers all six topics: fundamental principle of counting; factorials and permutations; permutations of objects not all distinct; circular permutations; combinations; and key properties and relations.

permutations-and-combinations-class-11-notes-assignment-1.pdf
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Permutations and Combinations Assignment 2 — Challenge + Integrated Problems

Higher-difficulty set with multi-step problems (counting passwords, word arrangements, circular seating, committee selections, combinatorial identities) and full worked answers. Includes an answer key.

permutations-and-combinations-class-11-notes-assignment-2.pdf
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Permutations and Combinations Test 1 — CBSE Pattern (90 min, 40 marks)

Full board-pattern test: Section A (MCQ/VSA 1M × 10), Section B (2M × 4), Section C (3M × 4), Section D (5M × 2) = 40 marks. Time: 90 minutes. Includes answer key.

permutations-and-combinations-class-11-notes-test-1-cbse.pdf
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Permutations and Combinations Test 2 — CBSE Pattern (90 min, 40 marks)

Second full paper with different questions, same 40-mark / 90-minute blueprint. Includes answer key.

permutations-and-combinations-class-11-notes-test-2-cbse.pdf
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Quick Revision

Key formulas at a glance

Memorise these equations — direct application numericals and derivations in CBSE & JEE frequently hinge on these.

Factorial

Permutations

Combinations

nPr in terms of nCr

Symmetry of combinations

Pascal's identity

Identical objects

Exam Strategy

How this chapter is asked

High-yield question patterns observed across CBSE boards, JEE Main & Advanced, and NEET.

  • Distinguish between 'arrange' (permutation) and 'select' (combination) — this decides whether order matters.
  • Circular permutation of n distinct objects = (n − 1)!; halve it if clockwise = anticlockwise.
  • nC0 + nC1 + ... + nCn = 2^n — the total number of subsets of an n-element set.
  • nCr = nC(n − r) lets you compute the smaller value when r > n/2.
  • With identical objects, divide n! by the factorials of the repetition counts.
  • Repetition allowed: n objects taken r at a time gives n^r arrangements.
  • Pascal's identity nCr + nC(r−1) = (n+1)Cr is useful for quick computation and proofs.

FAQ

Frequently asked questions

What is the difference between permutations and combinations?

Permutations count arrangements where order matters (e.g. ABC ≠ BAC); combinations count selections where order does not matter (e.g. {A,B,C} = {B,A,C}).

How do you calculate circular permutations?

Fix one object to remove rotational symmetry, then arrange the remaining (n − 1) objects in a line: (n − 1)! arrangements. Halve it if clockwise and anticlockwise are considered the same.

When do we use permutations with repetition?

When each of the r positions can be filled independently by any of the n objects (e.g. forming a 3-digit number where digits can repeat), the count is n^r.

What is Pascal's identity?

nCr + nC(r − 1) = (n + 1)Cr. It connects adjacent entries in Pascal's triangle and is useful for simplifying sums and proving combinatorial identities.

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