Class 12 Mathematics PERMUTATION AND COMBINATION Notes

Class 12 Mathematics PERMUTATION AND COMBINATION Notes
Class 12 Mathematics Permutation and Combination Notes | Nepal eNotes

Class 12 Mathematics

Unit 1

Algebra

Chapter 1

Permutation and Combination

Class 12 Mathematics – Permutation and Combination Notes PDF

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Contents / Quick Navigation

1. Introduction 2. Basic Principle of Counting 3. Factorial Notation 4. Permutations 5. Non-distinct Objects 6. Circular Permutations 7. Repeated Use 8. Combinations 9. Properties of Combinations 10. Permutation vs Combination 11. Worked Examples 12. Common Mistakes 13. Important Exam Questions 14. One-Minute Revision 15. Diagram Practice

NEB / CDC Focus

This chapter develops systematic methods for counting arrangements and selections. The core ideas are the basic counting principle, permutations, combinations, circular arrangements, arrangements with repeated objects, repeated use, and standard properties of combinations.

1. Introduction

Definition

Permutation means an arrangement in which order matters.

Combination means a selection in which order does not matter.

For example, choosing a captain and vice-captain from a group is an arrangement: choosing A as captain and B as vice-captain differs from choosing B as captain and A as vice-captain. But choosing two committee members A and B gives the same pair as B and A.

Permutation Combination A B AB and BA are different A B {A,B} is one selection
Figure 1: The key decision — does changing order create a new outcome?

2. Basic Principle of Counting

2.1 Multiplication Principle

Multiplication Principle

If one operation can be performed in m ways and, after that, a second operation can be performed in n ways, then the two operations can be performed successively in m × n ways.

More generally, if successive stages have n1, n2, …, nk choices, the total number of outcomes is n1n2…nk.

2 shirts × 3 trousers = 6 outfits Start S₁ S₂ T₁T₂T₃ T₁T₂T₃ Each branch represents one independent choice at the next stage.
Figure 2: Tree representation of the multiplication principle.

2.2 Addition Principle

If one task can be performed in m mutually exclusive ways and another alternative task can be performed in n mutually exclusive ways, then one of the two tasks can be performed in m+n ways.

Worked Example

A student may choose one book from 5 Mathematics books or 4 Physics books.

Total choices = 5 + 4 = 9.

3. Factorial Notation

Formula
n! = n(n − 1)(n − 2)…3·2·1, for n ≥ 1 0! = 1

Examples: 5! = 120, 7!/5! = 7×6 = 42.

Remember

Do not treat 0! as zero. By definition, 0! = 1.

4. Permutation of Distinct Objects

Definition

A permutation is an ordered arrangement. The number of permutations of n distinct objects taken r at a time is denoted by nPr.

4.1 Derivation of nPr

For the first position there are n choices. After one object is used, there are n−1 choices for the second, then n−2 for the third, and so on.

Formula
nPr = n(n−1)(n−2)…(n−r+1) = n!(n−r)! nPn = n! nP0 = 1
Arrange r objects from n distinct objects n n−1 n−2 … 1st position 2nd position 3rd position rth position Multiply the available choices in all slots.
Figure 3: Slot method leading to nPr.
Worked Example: Three positions from seven students

How many ways can president, secretary and treasurer be chosen from 7 students?

  1. The posts are different, so order matters.
  2. We arrange 3 students from 7.
  3. 7P3 = 7×6×5 = 210.

Answer: 210 ways.

5. Permutation When Objects Are Not All Different

When some objects are identical, interchanging identical objects does not create a new arrangement. Therefore, ordinary n! overcounts.

Formula

If among n objects, p are alike of one kind, q are alike of another kind, r are alike of a third kind, and so on, then:

Number of distinct arrangements = n!p!q!r!…

5.1 Why divide by factorials of repeated objects?

If two A’s are identical, every visible arrangement is counted 2! times when the two A’s are temporarily treated as different. With p identical objects, each arrangement is counted p! times.

Example: LEVEL L E V E L 5 letters; L repeats twice and E repeats twice. Distinct arrangements = 5! / (2!·2!) = 30
Figure 4: Repeated letters reduce the number of distinct permutations.
Worked Example: MISSISSIPPI

The word MISSISSIPPI has 11 letters: M = 1, I = 4, S = 4, P = 2.

Number of distinct arrangements = 11!/(4!4!2!) = 34,650.

6. Circular Permutation

In a circle there is no fixed first position. Rotating the whole arrangement does not create a new seating order. To remove rotational duplication, fix one object and arrange the remaining n−1.

Formula
Circular arrangements of n distinct objects = (n−1)!
Fix one person; arrange the remaining n−1 A fixed BCDE For 5 people: (5−1)! = 4! = 24 arrangements
Figure 5: Fixing one reference position converts a circular arrangement into a linear arrangement of the remaining objects.
Important Note

If clockwise and anticlockwise arrangements are also considered identical (as in some necklace/garland problems), divide by 2: (n−1)!/2. Use this only when the problem explicitly treats mirror-image directions as the same.

Worked Example: Round-table seating

In how many ways can 8 people sit around a round table?

(8−1)! = 7! = 5040.

Answer: 5040 ways.

7. Permutation with Repeated Use Allowed

If each of r positions can be filled by any of n available objects and repetition is allowed, then every position has n choices.

Formula
Number of ordered selections of length r with repetition = nr
Worked Example: 4-digit PIN using 0–9

Each of 4 positions has 10 choices, and repetition is allowed.

Total = 10×10×10×10 = 104 = 10,000.

If the question says “4-digit number” rather than PIN, the first digit cannot be zero; then the count changes. Always read restrictions carefully.

8. Combination of Distinct Objects

Definition

A combination is a selection where order is ignored. The number of ways to choose r objects from n distinct objects is denoted by nCr.

8.1 Derivation of the combination formula

First choose r objects: nCr ways. Each chosen group can then be arranged in r! ways. Therefore:

nPr = nCr · r!

Formula
nCr = nPrr! = n!r!(n−r)! nC0 = nCn = 1
Choose 3 from {A, B, C, D, E} ABCDE Selected group: {A, C, E} ACE, AEC, CAE… are not new combinations.
Figure 6: A combination records membership of a selected group, not its internal order.
Worked Example: Committee selection

How many 4-member committees can be formed from 10 students?

The positions are identical, so order does not matter.

10C4 = 10!/(4!6!) = (10×9×8×7)/(4×3×2×1) = 210.

Answer: 210 committees.

9. Properties of Combination

Property Meaning / Use
nCr = nCn−r Choosing r objects is equivalent to deciding which n−r objects are left out.
nCr + nCr−1 = n+1Cr Pascal identity.
r·nCr = n·n−1Cr−1 Useful for simplifying expressions involving rC terms.
(n−r)·nCr = n·n−1Cr Companion identity obtained algebraically.
nCr / nCr−1 = (n−r+1)/r Useful for successive combination coefficients.
nC0 + nC1 + … + nCn = 2n Total number of subsets of an n-element set.

9.1 Proof of symmetry

nCr = n!/[r!(n−r)!] = n!/[(n−r)!r!] = nCn−r.

9.2 Pascal identity — counting argument

Suppose a group has n+1 people and one person is distinguished, say A. To form a committee of r:

  • If A is excluded: choose all r members from the other n → nCr.
  • If A is included: choose the remaining r−1 members from the other n → nCr−1.

Hence nCr + nCr−1 = n+1Cr.

Pascal Triangle and nCr 1 11 121 1331 14641 15101051 Each interior entry is the sum of the two entries immediately above it.
Figure 7: Pascal identity visually represents a standard property of combinations.

10. Permutation vs Combination

FeaturePermutationCombination
Main ideaArrangementSelection
Does order matter?YesNo
NotationnPrnCr
Formula n!/(n−r)! n!/[r!(n−r)!]
Typical wordsarrange, rank, seat, code, postschoose, select, committee, team, group
RelationshipnPr = nCr·r!
Exam Important: Quick Decision Rule

Ask: “If I swap the selected objects, does the outcome change?” If yes, use permutation. If no, use combination.

11. Worked Numerical Examples

Example 1: Forming numbers without repetition

How many 3-digit numbers can be formed from 1, 2, 3, 4, 5 without repetition?

  1. All five digits are non-zero, so any digit can occupy the hundreds place.
  2. We arrange 3 distinct digits from 5.
  3. 5P3 = 5×4×3 = 60.

Answer: 60.

Example 2: Forming numbers when zero is available

How many 4-digit numbers can be formed from 0, 1, 2, 3, 4 without repetition?

  1. First digit cannot be zero: 4 choices (1,2,3,4).
  2. Second digit: 4 remaining choices, including zero.
  3. Third digit: 3 choices.
  4. Fourth digit: 2 choices.

Total = 4×4×3×2 = 96.

Answer: 96.

Example 3: Arrangements with a restriction

In how many ways can 5 boys and 4 girls sit in a row if all girls sit together?

  1. Treat the 4 girls as one block. Then there are 6 units: 5 boys + 1 block.
  2. Arrange these 6 units in 6! ways.
  3. Arrange the 4 girls within their block in 4! ways.
  4. Total = 6!×4! = 720×24 = 17,280.

Answer: 17,280 ways.

Block Method: all girls together B₁ B₂ B₃ B₄ B₅ Girls block G₁ G₂ G₃ G₄ 6 units → 6! arrangements; inside block → 4! arrangements Total = 6! × 4!
Figure 8: Treat objects that must stay together as one temporary block.
Example 4: Committee with a compulsory member

A committee of 5 is chosen from 9 people. In how many ways can it be formed if one particular person must be included?

  1. Include the compulsory person.
  2. Choose the remaining 4 members from the other 8.
  3. 8C4 = 70.

Answer: 70 committees.

Example 5: Committee with at least one girl

From 6 boys and 4 girls, how many 3-member committees contain at least one girl?

  1. Total committees: 10C3 = 120.
  2. Committees with no girl: choose all 3 from 6 boys → 6C3 = 20.
  3. Required = 120 − 20 = 100.

Answer: 100 committees.

Example 6: Circular seating with two people together

In how many ways can 6 people sit around a round table if A and B must sit together?

  1. Treat A and B as one block. There are now 5 circular units.
  2. Arrange them around the circle: (5−1)! = 4!.
  3. A and B can switch positions inside their block in 2! ways.
  4. Total = 4!×2 = 48.

Answer: 48 ways.

Example 7: Find r from an equality

If 10Cr = 10C3, find possible values of r.

Using nCr = nCn−r, either r=3 or r=10−3=7.

Answer: r = 3 or 7.

12. Step-by-Step Strategy for Solving Problems

  1. Identify the objects: distinct or repeated?
  2. Decide whether order matters: permutation or combination?
  3. Check repetition: allowed or not allowed?
  4. Check shape: linear arrangement or circular arrangement?
  5. Read restrictions: together, apart, compulsory, excluded, first digit non-zero, etc.
  6. Choose the formula only after modelling the situation.
  7. Simplify factorials before multiplying large numbers.
  8. Check reasonableness: a restricted count should not exceed the unrestricted count.

13. Common Mistakes and Warnings

Mistake 1: Using nPr for a committee

A committee has no ranking unless posts are specified. Normally use nCr.

Mistake 2: Forgetting zero restrictions

A numeral cannot begin with zero, though a code or PIN may.

Mistake 3: Using n! for repeated letters

Divide by factorials corresponding to identical repetitions.

Mistake 4: Using n! for a round table

For n distinct people around a circle, use (n−1)!.

Mistake 5: “At least one” by long casework

Often it is shorter to use: total − none.

Mistake 6: Cancelling factorials incorrectly

(n−r)! is not n!−r!.

14. Exam-Important Formula Sheet

Core Formulas
n! = n(n−1)(n−2)…2·1,   0! = 1 nPr = n!/(n−r)! Permutation of n objects with repetitions p, q, r,… = n!/(p!q!r!…) Circular permutation of n distinct objects = (n−1)! Ordered selections with repetition allowed = nr nCr = n!/[r!(n−r)!] nPr = nCrr! nCr = nCn−r nCr + nCr−1 = n+1Cr

15. Important Exam Questions

Short-Answer Questions

  1. State the multiplication principle of counting with one example.
  2. Define permutation and combination. Give one situation for each.
  3. Show that nPr = nCrr!.
  4. Evaluate 8P3 and 8C3.
  5. Why is the number of circular arrangements of n distinct objects equal to (n−1)!?
  6. Find the number of distinct arrangements of the letters of NEPAL.
  7. Find the number of distinct arrangements of the letters of SUCCESS.
  8. Prove nCr = nCn−r.
  9. State Pascal’s identity for combinations.

Long-Answer / Derivation Questions

  1. Derive the formula nPr = n!/(n−r)! using the multiplication principle.
  2. Derive the formula for arrangements of n objects when p, q, r, … objects are identical.
  3. Derive nCr = n!/[r!(n−r)!] from permutations.
  4. Prove Pascal’s identity nCr + nCr−1 = n+1Cr.
  5. Explain circular permutation and derive the number of arrangements of n distinct people around a round table.

Numerical Questions

  1. How many 5-digit numbers can be formed from given digits without repetition when zero is included?
  2. How many arrangements of a word containing repeated letters are possible?
  3. From a mixed group of boys and girls, form a committee satisfying “at least”, “at most”, or compulsory-member conditions.
  4. Find the number of arrangements when specified people must sit together or must not sit together.
  5. Find circular seating arrangements with one or more adjacency restrictions.
  6. Solve equations involving nPr or nCr using factorial identities and symmetry.

Diagram / Explanation Questions

  1. Use a tree diagram to explain the multiplication principle.
  2. Use the slot method to explain the derivation of nPr.
  3. Draw a circular seating diagram to explain why rotation does not make a new arrangement.
  4. Use Pascal’s triangle to illustrate a property of combinations.

16. One-Minute Revision

Quick Revision
  • Use multiplication when choices occur in successive stages.
  • Use addition for mutually exclusive alternatives.
  • 0! = 1.
  • Permutation means arrangement; order matters.
  • nPr = n!/(n−r)!.
  • For repeated identical objects, divide by the factorial of each repetition count.
  • For a round table, n distinct objects have (n−1)! arrangements.
  • With repetition allowed in r ordered positions, the number is nr.
  • Combination means selection; order does not matter.
  • nCr = n!/[r!(n−r)!].
  • nPr = nCrr!.
  • nCr = nCn−r.
  • For “at least one”, complement counting (total − none) is often fastest.
  • For objects that must remain together, treat them temporarily as one block.
  • Always check whether zero, repetition, circular order, or restrictions change the basic formula.

17. Diagram Practice

Students should be able to redraw and explain these simple teaching diagrams:

  • Tree diagram for the multiplication principle.
  • Slot diagram for deriving nPr.
  • Repeated-letter arrangement model.
  • Circular seating diagram with one position fixed.
  • Selection diagram showing why order is ignored in combinations.
  • Pascal triangle illustrating combination identities.
  • Block method diagram for objects that must remain together.

Also Visit

Class 12 Mathematics Notes Class 12 Notes Practical File Question Papers Important Questions
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