Lesson Notes By Weeks and Term v5 - Grade 12

Lesson Video

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Performance objectives

• Apply the fundamental counting principle, permutations, and combinations.
• Distinguish between permutations and combinations.
• Calculate probability using counting techniques.
• Solve problems with restrictions and repetitions.
• Apply knowledge to real-world scenarios.

Lesson summary

Counting and probability are fundamental concepts in mathematics with far-reaching applications. They allow us to quantify uncertainty and make informed decisions in various situations. From understanding the odds of winning the Lotto to assessing the risk associated with business ventures, these skills are essential for navigating the complexities of modern life. In South Africa, a country with diverse challenges and opportunities, a solid understanding of counting and probability empowers individuals to make better financial decisions, evaluate statistical information critically, and participate more effectively in democratic processes.

Lesson notes

2.1 Fundamental Counting Principle (FCP) The fundamental counting principle states that if there are m ways to do one thing and n ways to do another, then there are m x n ways to do both. This principle extends to any number of independent events. It forms the basis for more complex counting techniques.

Example: A restaurant in Durban offers 3 choices for starters, 5 choices for main courses, and 2 choices for desserts. How many different three-course meals can a customer choose?

Solution: Using the FCP, the total number of meals is 3 x 5 x 2 = 30. 2.2 Permutations A permutation is an arrangement of objects in a specific order. The order of arrangement matters. The number of permutations of n distinct objects taken r at a time is denoted by nPr or P(n, r) and is calculated as: nPr = n! / (n-r)! where n! (n factorial) is the product of all positive integers from 1 to n (e.g., 5! = 5 x 4 x 3 x 2 x 1 = 120).

Example: In a class of 10 students, how many ways can a teacher choose a class captain, vice-captain, and treasurer?

Solution: Since the order of selection is important (captain, vice-captain, treasurer are distinct roles), this is a permutation problem. We have n = 10 and r = 3. 10P3 = 10! / (10-3)! = 10! / 7! = (10 x 9 x 8 x 7!) / 7! = 10 x 9 x 8 =

7

2

0. Therefore, there are 720 ways to choose a class captain, vice-captain, and treasurer. 2.3 Combinations A combination is a selection of objects where the order of selection does not matter. The number of combinations of n distinct objects taken r at a time is denoted by nCr or C(n, r) or (n choose r) and is calculated as: nCr = n! / (r! * (n-r)!)

Example: A committee of 3 needs to be formed from a group of 8 people. How many different committees can be formed?

Solution: Since the order of selection does not matter (a committee of Alice, Bob, and Carol is the same as a committee of Carol, Bob, and Alice), this is a combination problem. We have n = 8 and r = 3. 8C3 = 8! / (3! (8-3)!) = 8! / (3! 5!) = (8 x 7 x 6 x 5!) / (3 x 2 x 1 x 5!) = (8 x 7 x 6) / (3 x 2 x 1) =

5

6. Therefore, there are 56 different committees that can be formed. 2.4 Probability and Counting Techniques Probability is the measure of the likelihood of an event occurring.

It is calculated as: Probability (Event) = Number of favorable outcomes / Total number of possible outcomes Counting techniques are often used to determine the number of favorable outcomes and the total number of possible outcomes.

Example: A bag contains 5 red balls and 3 blue balls. If two balls are drawn at random without replacement, what is the probability that both balls are red?

Solution: Total number of ways to choose 2 balls from 8: 8C2 = 8! / (2! 6!) = 28 Number of ways to choose 2 red balls from 5: 5C2 = 5! / (2! 3!) = 10 Probability (both red) = 10 / 28 = 5 / 14 2.5 Arrangements with Restrictions and Repetitions Some problems involve restrictions on the arrangement of objects (e.g., certain objects must be together or apart) or repetitions of objects (e.g., arranging the letters in the word "MISSISSIPPI").

Arrangements with Restrictions: Consider these restrictions by arranging the restricted items first, then the rest.

Arrangements with Repetitions: For arranging n objects where some are identical, the formula is: n! / (n1! n2! ... * nk!) where n1, n2, ..., nk are the number of repetitions of each distinct object.

Example (Repetitions): How many different arrangements can be made using all the letters of the word "MATHEMATICS"?

Solution: There are 11 letters in total. M appears twice, A appears twice, T appears twice. The number of arrangements is 11! / (2! 2! 2!) = 4,989,600 Guided Practice (With Solutions)

Question 1: A password for a website must be 8 characters long. It must consist of 3 letters (A-Z, case insensitive), 2 numbers (0-9), and 3 special characters (!@#$%^&*). How many different passwords can be created if repetition is allowed?

Solution: Number of choices for each letter: 26 (A-Z)

Number of choices for each number: 10 (0-9) Number of choices for each special character: 8 (!@#$%^&)

Using the FCP: 26 x 26 x 26 x 10 x 10 x 8 x 8 x 8 = 26^3 10^2 8^3 = 91,313,920

Commentary: We use the fundamental counting principle because each character choice is independent of the others. Since repetition is allowed, we have the same number of choices for each position.

Question 2: A committee of 5 people is to be chosen from 6 men and 4 women. What is the probability that the committee consists of exactly 3 men and 2 women?

Solution: Total number of ways to choose a committee of 5 from 10 people: 10C5 = 252 Number of ways to choose 3 men from 6: 6C3 = 20 Number of ways to choose 2 women from 4: 4C2 = 6 Number of ways to choose 3 men and 2 women: 6C3 4C2 = 20 * 6 = 120 Probability (3 men and 2 women) = 120 / 252 = 10 / 21

Commentary: This problem combines combinations (choosing the committee members) with probability.