Lesson Notes By Weeks and Term v3 - Senior Secondary 2
Permutation and combination cont
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Permutation and combination cont
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Subject: Further Mathematics
Class: Senior Secondary 2
Term: 1st Term
Week: 5
Theme: Statistics
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This lesson builds upon foundational knowledge of basic permutations and combinations, introducing more complex scenarios and applications within the realm of discrete mathematics, which is crucial for advanced studies in statistics and probability. Understanding these concepts enables learners to systematically approach problems involving arrangements and selections, a skill vital in fields like data analysis, logistics, and resource allocation. For Nigerian learners, these skills are applicable in everyday decision-making, from planning community events and resource distribution to understanding lottery probabilities or even arranging goods in a market stall for optimal display.
included, how many different delegations are possible?
Solution:
1. Total traders, $n = 10$. Delegation size, $r = 4$.
2. Number of unreliable traders to exclude = 3.
3. Number of reliable traders available for selection = $10 - 3 = 7$.
4. Number of ways to choose 4 reliable traders from 7 = $_7C_4 = \frac{7!}{4!(7-4)!} = \frac{7!}{4!3!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 7 \times 5 = 35$.
Answer: 35 different delegations are possible.
Case C: At least / At most restrictions.
Strategy: This often involves summing up combinations for different possibilities. E.g., "at least 2 boys" means (2 boys and remaining girls) OR (3 boys and remaining girls) OR ...
Worked Example 8 (Nigerian Context): A class has 8 boys and 7 girls. A team of 5 students is to be selected. How many ways can the team be formed if there must be at least 3 girls?
Solution:
1. Total boys = 8, Total girls =
7. Team size = 5. 2. "At least 3 girls" means we can have: 3 girls and 2 boys: $_7C_3 \times _8C_2 = (\frac{7!}{3!4!}) \times (\frac{8!}{2!6!}) = (35) \times (28) = 980$. 4 girls and 1 boy: $_7C_4 \times _8C_1 = (\frac{7!}{4!3!}) \times (\frac{8!}{1!7!}) = (35) \times (8) = 280$. 5 girls and 0 boys: $_7C_5 \times _8C_0 = (\frac{7!}{5!2!}) \times (1) = (21) \times (1) = 21$.
3. Total ways = Sum of possibilities = $980 + 280 + 21 = 1281$.
Answer: There are 1281 ways to form the team with at least 3 girls. 2.
5. Combined Permutation and Combination Problems Some problems require both selection (combination) and arrangement (permutation).
Strategy: First, use combinations to select the items. Then, use permutations to arrange the selected items.
Worked Example 9 (Nigerian Context): From a list of 10 available dishes for an event, a caterer must choose 4 dishes and arrange them on a display table. How many ways can this be done?
Solution:
1. Select 4 dishes from 10: This is a combination problem as the order of choosing doesn't matter initially. $_ {10}C_4 = \frac{10!}{4!6!} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 10 \times 3 \times 7 = 210$.
2. Arrange the 4 selected dishes on the display table: This is a permutation problem as the order of arrangement matters. The 4 chosen dishes can be arranged in $4! = 24$ ways.
3. Total ways = (Number of ways to choose) $\times$ (Number of ways to arrange) = $210 \times 24 = 5040$.
Answer: There are 5040 ways to choose and arrange the dishes. --- in a circle where clockwise and anti-clockwise arrangements are considered identical (e.g., beads on a necklace, keys on a ring).
Formula: $\frac{(n-1)!}{2}$ Explanation: This accounts for symmetry where flipping the arrangement doesn't create a new distinct one. 2.
4. Restricted Permutations and Combinations These involve conditions or constraints on the arrangement or selection of objects. 2.4.
1. Permutations with Restrictions Case A: Objects must be together.
Strategy: Treat the group of objects that must stay together as a single 'block' or unit. Calculate the permutations of these blocks (including the block itself). Then, multiply by the permutations possible within the block.
Worked Example 4 (Nigerian Context): In how many ways can 5 students (Ahmed, Bola, Chika, Dayo, Emeka) be arranged in a row for a group photograph if Ahmed and Bola must always sit together?
Solution:
1. Treat Ahmed and Bola as a single unit (AB).
Now we have 4 'items' to arrange: (AB), Chika, Dayo, Emeka.
2. Permutations of these 4 items = $4! = 24$.
3. Within the (AB) unit, Ahmed and Bola can be arranged in $2! = 2$ ways (AB or BA).
4. Total arrangements = (Permutations of units) $\times$ (Permutations within the unit) = $24 \times 2 = 48$.
Answer: There are 48 ways to arrange the students if Ahmed and Bola must sit together.
Case B: Objects must NOT be together.
Strategy: Calculate the total number of unrestricted arrangements. Then, subtract the number of arrangements where the objects are together (using Case A strategy).
Worked Example 5 (Nigerian Context): In how many ways can 5 students (Ahmed, Bola, Chika, Dayo, Emeka) be arranged in a row if Ahmed and Bola must NOT sit together?
Solution:
1. Total unrestricted arrangements of 5 students = $5! = 120$.
2. Arrangements where Ahmed and Bola are together (from Worked Example 4) = 48.
3. Arrangements where Ahmed and Bola are NOT together = Total arrangements - Arrangements where they are together = $120 - 48 = 72$.
Answer: There are 72 ways to arrange the students if Ahmed and Bola must not sit together. 2.4.
2. Combinations with Restrictions Case A: Specific items must be included.
Strategy: Subtract the number of 'must-be-included' items from both the total number of items 'n' and the number of items to be chosen 'r'. Then, calculate the combination.
Worked Example 6 (Nigerian Context): A community wants to form a 5-member committee from a group of 12 people. If the Igwe (traditional ruler) and the Youth Leader must be on the committee, how many different committees can be formed?
Solution:
1. Total people, $n = 12$. Committee size, $r = 5$.
2. The Igwe and Youth Leader (2 people) must be included.
3. Remaining people to choose from = $12 - 2 = 10$.
4. Remaining committee members to choose = $5 - 2 = 3$.
5. Number of ways to choose the remaining 3 members from 10 people = $_{10}C_3 = \frac{10!}{3!(10-3)!} = \frac{10!}{3!7!} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 10 \times 3 \times 4 = 120$.
Answer: 120 different committees can be formed.
Case B: Specific items must be excluded.
Strategy: Subtract the number of 'must-be-excluded' items from the total number of items 'n'. Then, calculate the combination with the original 'r'.
Worked Example 7 (Nigerian Context): From a group of 10 market traders, 3 are known to be unreliable. If a delegation of 4 traders is to be formed, and none of the unreliable traders should be included, how many different delegations are possible?
Solution:
1. Total traders, $n = 10$. Delegation size, $r = 4$.
2. Number of unreliable traders to exclude = 3.
3. Number of reliable traders available for selection = $10 - 3 = 7$.
4. Number of ways to choose 4 reliable traders from 7 = $_7C_4 = \frac{7!}{4!(7-4)!} = \frac{7!}{4!3!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 7 \times 5 = 35$.
Answer: 35 different delegations are possible. * Case C: At least / At most restrictions.** This section extends the basic understanding of permutations (arrangement) and combinations (selection) to cover more intricate scenarios. 2.
1. Recap of Basic Concepts (For Teacher Reference) Factorial (!): The product of an integer and all the integers below it down to
1. E.g., $5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$. $0! = 1$. Permutation ($P(n, r)$ or $_nP_r$): The number of ways to arrange 'r' distinct objects selected from 'n' distinct objects, where order matters.
Formula: $_nP_r = \frac{n!}{(n-r)!}$. Combination ($C(n, r)$ or $_nC_r$): The number of ways to select 'r' distinct objects from 'n' distinct objects, where order does not matter.
Formula: $_nC_r = \frac{n!}{r!(n-r)!}$. 2.
2. Permutation with Repetition (Identical Objects) When finding the number of distinct permutations of 'n' objects where there are $n_1$ identical objects of one type, $n_2$ identical objects of a second type, ..., $n_k$ identical objects of a k-th type, the formula is: $\frac{n!}{n_1! n_2! \dots n_k!}$ where $n = n_1 + n_2 + \dots + n_k$.
Explanation: The standard permutation formula assumes all objects are distinct. When objects are identical, swapping two identical objects does not create a new distinct arrangement.
Therefore, we divide by the factorial of the number of identical objects for each type to correct for overcounting.
Worked Example 1 (Nigerian Context): How many distinct arrangements can be made from the letters of the word "ABUJA"?
Solution:
1. Count the total number of letters, $n = 5$.
2. Identify repeated letters: 'A' appears 2 times ($n_1 = 2$). Other letters ('B', 'U', 'J') appear once.
3. Apply the formula: $\frac{n!}{n_1!} = \frac{5!}{2!} = \frac{5 \times 4 \times 3 \times 2 \times 1}{2 \times 1} = \frac{120}{2} = 60$.
Answer: There are 60 distinct arrangements of the letters in "ABUJA".
Worked Example 2: Find the number of distinct ways to arrange the letters of the word "LAGOSSTATISTICS".
Solution:
1. Total letters, $n = 15$.
2. Identify repeated letters: 'L': 1 'A': 1 'G': 1 'O': 1 'S': 4 ($n_1 = 4$) 'T': 3 ($n_2 = 3$) 'I': 2 ($n_3 = 2$) 'C': 1
3. Apply the formula: $\frac{15!}{4! \times 3! \times 2!} = \frac{1,307,674,368,000}{(24 \times 6 \times 2)} = \frac{1,307,674,368,000}{288} = 4,547,487,312.5$ (This is not an integer, indicating a calculation error somewhere.) Let's re-count "LAGOSSTATISTICS": L(1), A(1), G(1), O(1), S(4), T(3), I(2), C(1). Sum = 1+1+1+1+4+3+2+1 =
1
4. Ah, the word is "LAGOSSTATISTICS", not "LAGOS STATISTICS".
Let's count: L: 1 A: 1 G: 1 O: 1 S: 4 (S, S, S, S) T: 3 (T, T, T) I: 2 (I, I) C: 1 Total letters, $n=1+1+1+1+4+3+2+1 = 14$.
4. Apply the formula: $\frac{14!}{4! \times 3! \times 2!} = \frac{87,178,291,200}{(24 \times 6 \times 2)} = \frac{87,178,291,200}{288} = 302,702,400$.
Answer: There are 302,702,400 distinct arrangements. 2.
3. Circular Permutations The number of ways to arrange 'n' distinct objects in a circle.
Formula: $(n-1)!$ Explanation: In a linear arrangement, each position is distinct. In a circular arrangement, rotations of the same arrangement are considered identical. We fix one object's position, and then arrange the remaining $(n-1)$ objects linearly.
Worked Example 3 (Nigerian Context): A family of 6 wants to sit around a circular dining table for Sunday dinner. How many distinct seating arrangements are possible?
Solution:
1. Total people, $n = 6$.
2. Apply the circular permutation formula: $(n-1)! = (6-1)! = 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$.
Answer: There are 120 distinct seating arrangements.
Special Case: Objects arranged in a circle where clockwise and anti-clockwise arrangements are considered identical (e.g., beads on a necklace, keys on a ring).
Formula: $\frac{(n-1)!}{2}$ Explanation: This accounts for symmetry where flipping the arrangement doesn't create a new distinct one. 2.
4. Restricted Permutations and Combinations These involve conditions or constraints on the arrangement or selection of objects. 2.4.
1. Permutations with Restrictions Case A: Objects must be together.
Strategy: Treat the group of objects that must stay together as a single 'block' or unit. Calculate the permutations of these blocks (including the 3.
1. Teacher Activities: Introduction & Review (10 mins): Teacher begins by asking students to recall definitions and formulas for basic permutation and combination (e.g., "What is the difference between permutation and combination?", "When do we use factorial?"). Teacher quickly solves a simple permutation and combination problem on the board to refresh memory. Teacher introduces the topic "Permutation and Combination Cont." and briefly explains that it involves more complex scenarios with restrictions and identical items. Concept Explanation & Guided Examples (30 mins): Teacher systematically explains Permutations with Repetition (identical objects) using the word "ABUJA" or "LAGOSSTATISTICS", writing out the formula and step-by-step solution. Teacher explains Circular Permutations, demonstrating with a simple drawing of people around a table. Teacher explains Restricted Permutations (together/apart) with a clear, relatable example (e.g., seating students). Teacher explains Restricted Combinations (include/exclude, at least/at most) using examples of forming committees or selecting teams. Teacher explains Combined P&C problems using the dishes example. For each concept, the teacher works through one example on the board, encouraging student participation and questions.
Collaborative Problem Solving (20 mins): Teacher divides the class into small groups (3-4 students). Teacher provides a set of problems (mix of different types) for groups to solve. Teacher circulates, monitors group progress, provides hints, and clarifies misconceptions. Group Presentations & Discussion (15 mins): Teacher selects a few groups to present their solutions to specific problems on the board. Teacher facilitates a class discussion, comparing different approaches and ensuring correct understanding.
Conclusion & Assignment (5 mins): Teacher summarizes the key concepts covered (identical objects, circular, restricted, combined P&C). Teacher assigns independent practice questions as homework. 3.
2. Student Activities: Active Participation: Students answer questions during the review and concept explanation phases.
Note-Taking: Students copy definitions, formulas, and worked examples into their notebooks.
Group Work: Students collaborate in groups to solve the assigned problems, discussing strategies and solutions.
Presentation: Selected students present their group's solutions to the class, explaining their reasoning.
Questioning: Students ask clarifying questions when they encounter difficulties or do not understand a concept.
Independent Practice: Students complete assigned homework problems to reinforce learning. ---
Logistics and Scheduling (Nigerian Road Transportation): Permutations and combinations are crucial for planning efficient routes for commercial vehicles (e.g., Okada riders, Keke Napep operators, inter-state luxury buses) or scheduling deliveries.
Application: A logistics company transporting goods across states might need to determine the optimal sequence of stops in different cities (permutation) or select which trucks to use from a fleet based on cargo type and capacity (combination). If certain routes must be taken together or avoided due to road conditions (restrictions), P&C principles help in calculating the viable options. Committee Formation and Resource Allocation (Community Development): Local government areas and community development associations frequently need to form committees or distribute resources.
Application: When forming a 7-member community health committee from 15 available volunteers, with restrictions like "at least 3 nurses" or "the community leader must be included," combinations are used to determine the number of possible committee structures. This ensures fair representation and adherence to specific mandates, crucial for effective governance and project implementation in a diverse Nigerian community. Security Codes and Password Generation (Digital Nigeria): With increasing digitalization in Nigeria (e.g., online banking, NIN registration, JAMB portal), secure codes and passwords are vital.
Application: Understanding how many unique permutations of characters exist for a 6-digit PIN or an alphanumeric password (considering repetitions, length, and character sets) helps in appreciating password strength and the importance of complex passwords. This directly relates to cybersecurity awareness for students as digital citizens. ---