Exercise Solutions Available
Step-by-step answers for all textbook questions in this chapter
Chapter Overview
Permutations and Combinations is Chapter 6 of CBSE Class 11 Mathematics and belongs to the Algebra unit (Unit II). It teaches you how to count — systematically and without missing or double-counting arrangements and selections. This chapter is crucial for probability, combinatorics, and competitive exam problem-solving.
This guide covers every NCERT topic with formulae, comparison tables, worked examples, common mistakes, and exam tips.
Key Definitions
| Term | Definition |
|---|---|
| Factorial | n! = n × (n−1) × (n−2) × … × 2 × 1. By definition, 0! = 1. |
| Permutation | An arrangement of objects in a specific order. Order matters. |
| Combination | A selection of objects where order does not matter. |
| Fundamental Principle of Counting (Multiplication) | If one task can be done in m ways and another in n ways, both together can be done in m × n ways. |
| Addition Principle | If one task can be done in m ways OR another in n ways (mutually exclusive), the total is m + n. |
Fundamental Principle of Counting
Multiplication Principle (AND)
If an event A can occur in m ways and, after A has occurred, event B can occur in n ways, then A and B together can occur in m × n ways.
Example: A restaurant has 5 starters and 3 main courses. Number of meal combinations = 5 × 3 = 15.
Addition Principle (OR)
If event A can occur in m ways and event B can occur in n ways (and they cannot both occur together), then A or B can occur in m + n ways.
Example: 4 buses or 3 trains to reach a city. Total ways = 4 + 3 = 7.
When to Use Which
| Situation | Principle | Keyword |
|---|---|---|
| Do this AND then that | Multiply (m × n) | AND |
| Do this OR that | Add (m + n) | OR |
| Arrange in order | Permutation (nPr) | Order matters |
| Select without order | Combination (nCr) | Order does not matter |
Permutations
Formula: nPr
The number of permutations of r objects chosen from n distinct objects:
nPr = n!/(n − r)!
where 0 ≤ r ≤ n.
- nP0 = 1 (choosing nothing — one way)
- nP1 = n (choosing one object — n ways)
- nPn = n! (arranging all n objects)
Permutations of All n Objects
Arranging all n distinct objects: n! ways.
Example: Arrange 5 books on a shelf: 5! = 120 ways.
Permutations with Repetition
The number of permutations of n objects taken r at a time with repetition allowed:
nʳ
Each of the r positions can be filled in n ways independently.
Example: Number of 4-digit codes using digits 0–9 (with repetition): 10⁴ = 10,000.
Permutations of Objects Not All Distinct
If n objects have p alike of one kind, q alike of another kind, r alike of another kind, then:
Number of distinct permutations = n!/(p! × q! × r! × …)
Example: Permutations of the letters in MISSISSIPPI. Total letters = 11; I = 4, S = 4, P = 2, M = 1. Number of arrangements = 11!/(4! × 4! × 2! × 1!) = 11!/(4! × 4! × 2!)
Combinations
Formula: nCr
The number of combinations of r objects chosen from n distinct objects (order does not matter):
nCr = n!/[r!(n − r)!]
where 0 ≤ r ≤ n.
- nC0 = 1 (choosing nothing — one way)
- nC1 = n (choosing one object — n ways)
- nCn = 1 (choosing all — one way)
Key Properties of Combinations
| Property | Formula |
|---|---|
| Symmetry | nCr = nC(n−r) |
| Pascal's identity | nCr + nC(r−1) = (n+1)Cr |
| Relation to permutation | nPr = r! × nCr |
| Sum of all combinations | nC0 + nC1 + nC2 + … + nCn = 2ⁿ |
| nCr maximum | nC(n/2) when n is even; nC((n−1)/2) = nC((n+1)/2) when n is odd |
Permutation vs Combination — Comparison
| Feature | Permutation | Combination |
|---|---|---|
| Order | Matters | Does not matter |
| Formula | nPr = n!/(n−r)! | nCr = n!/[r!(n−r)!] |
| Relation | nPr = r! × nCr | nCr = nPr/r! |
| Keyword | Arrange, seat, rank | Choose, select, team |
| Example | 5 books on a shelf | 5 books from 10 |
| n=5, r=3 | 5P3 = 60 | 5C3 = 10 |
When to Use Permutation vs Combination
| Problem Type | Use | Reason |
|---|---|---|
| Arrange books on a shelf | Permutation | Order matters |
| Select a committee | Combination | Order does not matter |
| Form a 4-digit number | Permutation | 1234 ≠ 4321 |
| Choose 3 subjects from 8 | Combination | Order does not matter |
| Seat people around a table | Permutation | Different seating = different arrangement |
| Pick a team of 5 from 11 | Combination | Order of selection does not matter |
Important Numericals
1. nPr Calculation
Find 6P4.
6P4 = 6!/(6−4)! = 6!/2! = (6 × 5 × 4 × 3 × 2 × 1)/(2 × 1) = 720/2 = 360
2. nCr Calculation
Find 10C3.
10C3 = 10!/(3! × 7!) = (10 × 9 × 8)/(3 × 2 × 1) = 720/6 = 120
3. Word Arrangement
How many words can be formed using the letters of the word DELHI?
All 5 letters are distinct. Number of arrangements = 5! = 120
How many words can be formed using the letters of BANANA?
Total = 6 letters; A = 3, N = 2, B = 1. Number of arrangements = 6!/(3! × 2!) = 720/(6 × 2) = 60
4. Team Selection
A committee of 5 is to be formed from 6 men and 4 women. In how many ways can it be done if the committee must include at least 1 woman?
Total ways (no restriction) = 10C5 = 252 Ways with NO women = 6C5 = 6 Ways with at least 1 woman = 252 − 6 = 246
5. Distribution Problem
In how many ways can 5 different books be distributed among 3 students if each student gets at least one book?
This is a case of surjective functions from 5 books to 3 students: Using inclusion-exclusion: 3⁵ − 3C1 × 2⁵ + 3C2 × 1⁵ = 243 − 96 + 3 = 150
6. Digits Problem
How many 3-digit numbers can be formed using digits 1, 2, 3, 4, 5 without repetition?
Hundreds place: 4 choices (not 0, but all digits available since none is 0) Tens place: 4 choices (remaining digits) Units place: 3 choices Total = 5P3 = 60
7. Combination with Constraint
From 7 men and 3 women, choose a team of 5 with exactly 2 women.
Choose 2 women from 3: 3C2 = 3 Choose 3 men from 7: 7C3 = 35 Total = 3 × 35 = 105
Circular Permutations
Arranging n distinct objects around a circle:
Number of circular permutations = (n − 1)!
This is because fixing one object eliminates rotations that look the same. In a line, ABCD and DABC are different. In a circle, they are the same arrangement (rotated).
Example: Arrange 5 people around a circular table: (5 − 1)! = 4! = 24 ways.
With clockwise and anticlockwise different: (n − 1)! for a necklace (flip matters) or (n − 1)!/2 when clockwise = anticlockwise.
Practice Problems
Problem 1
How many 4-digit even numbers can be formed using digits 1, 2, 3, 4, 5, 6 without repetition?
For even numbers, the last digit must be even (2, 4, or 6): 3 choices. Remaining 3 positions from 5 remaining digits: 5P3 = 5 × 4 × 3 = 60. Total = 3 × 60 = 180
Problem 2
In how many ways can a cricket team of 11 players be selected from 15 players if a particular player must be included?
If one player is fixed, we need to choose 10 more from the remaining 14. Ways = 14C10 = 14C4 = (14 × 13 × 12 × 11)/(4 × 3 × 2 × 1) = 1001
Problem 3
Find the number of arrangements of the letters of the word MATHEMATICS.
Total letters = 11; M = 2, A = 2, T = 2, H = 1, E = 1, I = 1, C = 1, S = 1. Arrangements = 11!/(2! × 2! × 2!) = 11!/(2 × 2 × 2) = 11!/8
Problem 4
If nP3 : nP2 = 5 : 1, find n.
nP3/nP2 = 5/1 → [n!/(n−3)!] / [n!/(n−2)!] = 5 → (n − 2)!/(n − 3)! = 5 → (n − 2) = 5 → n = 7
NCERT Exercise Mapping
| Exercise | Focus | Key Question Types |
|---|---|---|
| Exercise 7.1 | Fundamental principle of counting | Count ways using multiplication/addition |
| Exercise 7.2 | Permutations (nPr) | Arrangements, word arrangements, digits |
| Exercise 7.3 | Combinations (nCr) | Selections, team formation, committees |
| Exercise 7.4 | Mixed problems | Distribution, constraints, properties |
| Miscellaneous | Comprehensive problems | Complex counting, proofs of identities |
Common Mistakes
1. Using permutation when order does not matter: Selecting a committee is a combination, not a permutation. The number of ways to choose 3 from 10 is 10C3 = 120, not 10P3 = 720. 2. Dividing by r! when it should not be divided: When arranging objects in a specific order (permutation), do NOT divide by r!. Dividing by r! converts a permutation to a combination. 3. Forgetting that 0! = 1: 0! is defined as 1, not 0. This matters in formulae like nPn = n!/(n−n)! = n!/0! = n!/1 = n!. 4. Overcounting with identical objects: MISSISSIPPI has repeated letters. Using 11! overcounts by a factor of 4! × 4! × 2!. Always divide by factorials of repeated counts. 5. Confusing "at least" with "exactly": "At least 2 women" means 2 or 3 or 4 … women. "Exactly 2 women" means only 2 women. Read the question carefully. 6. Wrong use of nCr when positions are distinct: If 3 prizes are given to 3 different students (1st, 2nd, 3rd), order matters → use permutation. If 3 identical prizes are given, order does not matter → use combination. 7. Not using the complement method: When "at least one" is asked, it is often easier to compute total minus "none" rather than adding all cases: nC1 + nC2 + … = 2ⁿ − nC0 = 2ⁿ − 1.
Exam Tips
- 1-mark questions: Evaluate nPr or nCr, or state the fundamental principle of counting.
- 2-mark questions: Find the number of arrangements of letters in a word, or compute nCr using properties.
- 4-mark questions: Team selection with constraints (at least, exactly, at most), or word arrangement with repeated letters.
- 6-mark questions: Distribution problems, digit problems with constraints, or proving nCr + nC(r−1) = (n+1)Cr algebraically.
- Always identify whether the problem asks for arrangement (permutation) or selection (combination) before choosing the formula.
- For word problems, underline keywords: "arrange" → permutation, "select/choose" → combination.
- Use the complement method for "at least" problems to save time.
- Write the formula before substituting values — this earns method marks even if the calculation has an error.
Important Results and Shortcuts
| Result | Formula |
|---|---|
| nPr = r! × nCr | Convert between permutation and combination |
| nCr + nC(r−1) = (n+1)Cr | Pascal's identity |
| nCr = nC(n−r) | Symmetry — use smaller r for easier computation |
| nC0 + nC1 + … + nCn = 2ⁿ | Sum of all combinations |
| nPn = n! | Arrange all objects |
| Circular arrangement | (n − 1)! for n distinct objects |
| Arrangement with restrictions | Place restricted objects first, then fill remaining |
Study Plan for This Chapter
| Day | Task | Duration |
|---|---|---|
| Day 1 | Read NCERT, learn multiplication and addition principles | 45 min |
| Day 2 | Practice nPr calculations, word arrangements | 45 min |
| Day 3 | Learn nCr, practice team selection problems | 45 min |
| Day 4 | Permutations with identical objects, circular permutations | 50 min |
| Day 5 | NCERT exercise questions — focus on mixed problems | 50 min |
| Day 6 | Revise with this guide, attempt timed problem sets | 45 min |
Quick Revision Notes
- n! = n × (n−1) × … × 1. 0! = 1.
- Multiplication principle: AND → multiply. Addition principle: OR → add.
- nPr = n!/(n−r)!. Permutation: order matters.
- nCr = n!/[r!(n−r)!]. Combination: order does not matter.
- nPr = r! × nCr.
- Permutation with repetition: nʳ.
- Identical objects: n!/(p! q! …) for p alike, q alike, etc.
- nCr = nC(n−r). Symmetry.
- nCr + nC(r−1) = (n+1)Cr. Pascal's identity.
- Sum: nC0 + nC1 + … + nCn = 2ⁿ.
- "At least one" = Total − "none" = 2ⁿ − 1.
- Keywords: arrange/rank/order → permutation; choose/select/team → combination.