Lesson Note on Permutations and Combinations: Repetition, Combination and Probability Uses for SSS 2

Class: Senior Secondary School 2 (SS2 / SSS2)
Term: 3rd Term
Week: 5
Age: 16 years
Duration: 45 minutes
Subject: Further Mathematics
Curriculum Theme: Algebra and Combinatorics
Previous Lesson: Permutations and Combinations: Meaning and Types of Permutations.
Topic: PERMUTATIONS AND COMBINATIONS
Subject Matter: Arrangements in which repetitions are allowed, introduction to combination on selection, conditional arrangements and selection, probability arrangement problems involving arrangement and selection

Specific Objectives

By the end of the lesson, pupils should be able to:

Cognitive Domain:

• Define permutations and combinations with repetition.
• State the formulas for permutations with repetition and combinations.
• Solve problems involving conditional arrangements.
• Solve problems involving conditional selections.
• Apply permutation and combination concepts to solve probability problems.

Affective Domain:

• Appreciate the difference between permutation and combination in problem-solving.
• Develop confidence in tackling complex arrangement and selection problems.

Psychomotor Domain:

• Calculate the number of possible arrangements when repetitions are allowed.
• Determine the number of selections possible under various conditions.
• Accurately compute probabilities in scenarios requiring permutations or combinations.

Social Domain:

• Collaborate with peers to solve challenging problems in permutations and combinations.

Reference Materials

The following resources were used in planning this lesson:

• Senior Secondary Schools Education Curriculum
• State Unified Scheme of Work
• Further Mathematics for Senior Secondary Schools by P.N. Okeke and others.
• https://www.mathsisfun.com/combinatorics/combinations-permutations.html
• https://byjus.com/maths/permutation-and-combination/

Instructional Materials

The teacher will teach this lesson with the aid of:

• Charts showing functional notation (factorials).
• Whiteboard and markers.
• Further Mathematics textbook.
• Calculators.

Rationale for the Lesson

This lesson helps pupils understand how to count arrangements and selections, which is important for solving problems in probability and real-life scenarios like scheduling, coding, and committee formation. It enables pupils to differentiate between situations where order matters and where it does not.

Prerequisite/Previous Knowledge

Pupils should have prior knowledge of basic permutations, combinations, factorial notation, and fundamental concepts of probability.

Lesson Content/Board Summary

PERMUTATIONS AND COMBINATIONS

1. Permutations with Repetition

Permutations with repetition refer to arrangements where items can be used more than once. If there are ‘n’ types of items to choose from, and we choose ‘r’ of them, repetitions are allowed, and order matters.

Formula: The number of permutations of ‘r’ items chosen from ‘n’ types with repetition allowed is given by:
P(n, r) = n^r

Example 1: How many 3-digit numbers can be formed using the digits 1, 2, 3, 4, 5 if repetition is allowed?

Solution:

• Step 1: Identify n and r. Here, n = 5 (digits available: 1, 2, 3, 4, 5) and r = 3 (number of digits to form).
• Step 2: Apply the formula for permutations with repetition.
• P(5, 3) = 5³ = 5 × 5 × 5 = 125

Therefore, 125 three-digit numbers can be formed.

2. Introduction to Combination

A combination is a selection of items from a larger set where the order of selection does not matter. It answers “how many ways can we choose ‘r’ items from a set of ‘n’ items without regard to their arrangement.”

Formula: The number of combinations of ‘r’ items chosen from ‘n’ distinct items is given by:
C(n, r) = nCr = n! / (r! * (n-r)!)

Where:
n = total number of items available.
r = number of items to choose.
! = factorial (e.g., 5! = 5 × 4 × 3 × 2 × 1).

Example 2: In how many ways can a committee of 3 members be selected from a group of 7 people?

Solution:

• Step 1: Identify n and r. Here, n = 7 (total people) and r = 3 (members to select).
• Step 2: Apply the combination formula.
• C(7, 3) = 7! / (3! * (7-3)!)
• = 7! / (3! * 4!)
• = (7 × 6 × 5 × 4 × 3 × 2 × 1) / ((3 × 2 × 1) × (4 × 3 × 2 × 1))
• = (7 × 6 × 5) / (3 × 2 × 1)
• = 210 / 6 = 35

Therefore, a committee of 3 members can be selected in 35 ways.

3. Conditional Arrangements

Conditional arrangements involve finding the number of permutations under specific conditions, such as certain items always being together, or apart, or in fixed positions.

Example 3: In how many ways can the letters of the word ‘DAUGHTER’ be arranged so that the vowels always come together?

Solution:

• Step 1: Identify the vowels and consonants. Vowels: A, U, E (3). Consonants: D, G, H, T, R (5).
• Step 2: Treat the group of vowels as a single unit. So, we have (AUE), D, G, H, T, R. This gives 6 units to arrange.
• Step 3: Arrange these 6 units. The number of ways is 6! = 720.
• Step 4: Arrange the vowels within their group. The number of ways to arrange A, U, E is 3! = 6.
• Step 5: Multiply the arrangements from Step 3 and Step 4.
• Total arrangements = 6! × 3! = 720 × 6 = 4320.

Therefore, the letters of ‘DAUGHTER’ can be arranged in 4320 ways with vowels together.

4. Conditional Selections

Conditional selections involve finding the number of combinations under specific conditions, such as including or excluding certain items from the selection.

Example 4: A committee of 4 is to be chosen from 6 men and 5 women. In how many ways can this be done if it must contain exactly 2 men?

Solution:

• Step 1: Determine the number of men to be chosen. Exactly 2 men from 6 men. This is C(6, 2).
• C(6, 2) = 6! / (2! * 4!) = (6 × 5) / (2 × 1) = 15 ways.
• Step 2: Determine the number of women to be chosen. Since the committee has 4 members and 2 are men, 2 must be women. This is C(5, 2) from 5 women.
• C(5, 2) = 5! / (2! * 3!) = (5 × 4) / (2 × 1) = 10 ways.
• Step 3: Multiply the number of ways to choose men and women.
• Total selections = C(6, 2) × C(5, 2) = 15 × 10 = 150.

Therefore, the committee can be chosen in 150 ways with exactly 2 men.

5. Probability Problems Involving Permutation and Combination

Probability is the likelihood of an event occurring, calculated as the ratio of favorable outcomes to the total possible outcomes. Permutations and combinations are used to determine the number of favorable outcomes and the total possible outcomes.

Formula: P(Event) = (Number of favorable outcomes) / (Total number of possible outcomes)
P(E) = n(E) / n(S)

Example 5: A bag contains 4 red balls and 6 blue balls. If 3 balls are drawn at random, what is the probability that all three are red?

Solution:

• Step 1: Calculate the total number of possible outcomes (n(S)).
  Total balls = 4 red + 6 blue = 10 balls.
  We are drawing 3 balls from 10. Since the order doesn’t matter, use combination.
  n(S) = C(10, 3) = 10! / (3! * 7!) = (10 × 9 × 8) / (3 × 2 × 1) = 120.
• Step 2: Calculate the number of favorable outcomes (n(E)).
  We want all 3 balls to be red. There are 4 red balls, and we choose 3.
  n(E) = C(4, 3) = 4! / (3! * 1!) = 4 / 1 = 4.
• Step 3: Calculate the probability.
  P(all three are red) = n(E) / n(S) = 4 / 120 = 1/30.

Therefore, the probability that all three balls drawn are red is 1/30.

Example 6: The letters of the word ‘MATHS’ are arranged randomly. What is the probability that the arrangement starts with ‘M’?

Solution:

• Step 1: Calculate the total number of possible arrangements (n(S)).
  The word ‘MATHS’ has 5 distinct letters.
  n(S) = 5! = 5 × 4 × 3 × 2 × 1 = 120.
• Step 2: Calculate the number of favorable arrangements (n(E)).
  If the arrangement starts with ‘M’, then the remaining 4 letters (A, T, H, S) can be arranged in 4! ways.
  n(E) = 4! = 4 × 3 × 2 × 1 = 24.
• Step 3: Calculate the probability.
  P(starts with ‘M’) = n(E) / n(S) = 24 / 120 = 1/5.

Therefore, the probability that the arrangement starts with ‘M’ is 1/5.

Teaching Methods/Instructional Techniques

Discussion, Lecture, Demonstration, Question and Answer, Visual Aids

Instructional Procedures

Step 1: Introduction

Time: 5 minutes
Teaching Skill: Set Induction
Teacher’s Activity: The teacher reviews the previous lesson on basic permutations and combinations and asks pupils to define factorial notation.
Pupils’ Activity: Pupils recall and define factorial notation and answer questions on basic permutations.
Learning Point: Pupils activate prior knowledge relevant to the new topic.

Step 2: Permutations with Repetition

Time: 7 minutes
Teaching Skill: Explanation/Demonstration
Teacher’s Activity: The teacher explains permutations where repetitions are allowed, providing the formula P(n, r) = n^r, and solves Example 1 from the board summary.
Pupils’ Activity: Pupils listen, take notes, and ask questions for clarification.
Learning Point: Pupils understand and can apply the formula for permutations with repetition.

Step 3: Introduction to Combination

Time: 7 minutes
Teaching Skill: Explanation/Illustration
Teacher’s Activity: The teacher introduces combinations as selections where order does not matter, presents the formula C(n, r) = n! / (r! * (n-r)!), and solves Example 2.
Pupils’ Activity: Pupils differentiate between permutation and combination and practice applying the combination formula.
Learning Point: Pupils grasp the concept and formula for combinations.

Step 4: Conditional Arrangements and Selections

Time: 8 minutes
Teaching Skill: Problem-Solving/Guided Practice
Teacher’s Activity: The teacher explains conditional arrangements and selections, demonstrating how to solve problems with specific constraints using Examples 3 and 4.
Pupils’ Activity: Pupils follow the steps, ask questions, and attempt similar problems.
Learning Point: Pupils can solve problems involving specific conditions for arrangements and selections.

Step 5: Probability Problems Involving Permutation and Combination

Time: 8 minutes
Teaching Skill: Application/Integration
Teacher’s Activity: The teacher explains how permutations and combinations are used to find favorable and total outcomes in probability problems and solves Examples 5 and 6.
Pupils’ Activity: Pupils observe how to integrate P&C into probability calculations.
Learning Point: Pupils understand the application of permutations and combinations in solving probability problems.

Step 6: Class Practice / Consolidation

Time: 5 minutes
Teaching Skill: Collaborative Learning
Teacher’s Activity: The teacher provides additional practice problems from the textbook or scheme of work for pupils to solve in groups or individually.
Pupils’ Activity: Pupils work on the practice problems, discussing solutions with peers.
Learning Point: Pupils reinforce their understanding and problem-solving skills through practice.

Step 7: Evaluation/Review

Time: 5 minutes
Teaching Skill: Questioning/Assessment
Teacher’s Activity: The teacher evaluates the learning by asking the following questions:

1. Define a permutation with repetition.
2. State the formula for combinations of ‘r’ items from ‘n’ distinct items.
3. In how many ways can the letters of the word ‘PENCIL’ be arranged if the vowels must be together?
4. A committee of 3 is to be chosen from 5 men and 4 women. How many committees can be formed if it must contain exactly 1 woman?

Pupils’ Activity: Pupils answer orally and in writing.

Learning Point: Pupils demonstrate understanding of the lesson.

Step 8: Conclusion

Time: Remaining minutes
Teaching Skill: Summarization
Teacher’s Activity: The teacher summarizes the key concepts of permutations with repetition, combinations, conditional arrangements, and probability applications. The teacher gives homework.
Pupils’ Activity: Pupils listen to the summary and copy down the homework assignment.
Learning Point: Pupils consolidate their learning and prepare for independent practice.

Lesson Keywords

• Permutation – An arrangement of items where order matters.
• Combination – A selection of items where order does not matter.
• Repetition – The act of using an item multiple times in an arrangement or selection.
• Factorial – The product of an integer and all the integers below it (e.g., 5! = 5x4x3x2x1).
• Conditional – Subject to one or more requirements or restrictions.

Differentiation

For struggling learners, the teacher will provide simpler problems focusing on basic formula application and offer individual support. Advanced learners will be given more complex problems involving multiple conditions or real-world scenarios to challenge their understanding and problem-solving skills.

Note for teachers using this lesson plan

Ensure that pupils clearly understand the distinction between permutations (order matters) and combinations (order does not matter), especially when dealing with conditional problems and probability applications. Emphasize step-by-step problem-solving and encourage pupils to justify their choice of formula (permutation or combination) for each problem.