Ressource dapprentissage Green Bridge pour {0}

Aperçu

In Further Mathematics, specifically in the topic of Permutations and Combinations, students delve into a fascinating branch of mathematics that deals with the arrangement and selection of objects. This topic offers a systematic way to calculate the number of ways in which objects can be arranged or selected, thus enabling a deeper understanding of various mathematical concepts and real-life applications.

One of the primary objectives of studying permutations and combinations is to comprehend the fundamental differences between these two concepts. Permutations focus on the arrangement of objects in a particular order, whereas combinations concern the selection of objects without considering the order. This distinction plays a crucial role in various problem-solving scenarios, making it essential for students to grasp the nuances of each concept.

As students progress through the topic, they will learn how to calculate the number of permutations and combinations of objects taken some at a time. This involves applying formulas and strategies to determine the possible arrangements and selections based on the given constraints. By mastering these calculations, students develop a robust foundation in combinatorial mathematics, which is vital for tackling more complex problems in the field.

Moreover, the application of permutations and combinations extends beyond theoretical calculations; it finds practical relevance in solving real-life problems. From determining the number of possible outcomes in games of chance to optimizing resources in various scenarios, the knowledge of permutations and combinations equips students with valuable problem-solving skills that can be applied across diverse disciplines.

To aid in visualizing and understanding the concepts of permutations and combinations, students often use diagrams such as factorial diagrams and tree diagrams. These visual representations help illustrate the different possible arrangements and selections, enhancing comprehension and facilitating the problem-solving process. [[[Diagram: A factorial diagram showing the arrangement of objects in a permutation]]]

In conclusion, the study of permutations and combinations in Further Mathematics opens up a world of possibilities for students to explore the intricacies of arrangement and selection. By mastering the calculations, understanding the underlying principles, and applying the concepts to practical scenarios, students develop a versatile skill set that can be utilized in various academic and real-world contexts.

Objectifs

Calculate the number of permutations of objects taken some at a time
Apply permutations and combinations to solve real-life problems
Understand the concept of permutations and combinations
Understand the difference between permutations and combinations
Calculate the number of combinations of objects taken some at a time

Note de cours

Permutation and combinations are fundamental concepts in the field of combinatorics, a branch of mathematics focusing on counting, arrangement, and combination of objects. These concepts are not only crucial for academic purposes but also have extensive applications in solving real-life problems.

Connectez-vous pour continuer à visionner

Évaluation de la leçon

Félicitations, vous avez terminé la leçon sur Permutation And Combinations. Maintenant que vous avez exploré le concepts et idées clés, il est temps de mettre vos connaissances à lépreuve. Cette section propose une variété de pratiques des questions conçues pour renforcer votre compréhension et vous aider à évaluer votre compréhension de la matière.

Vous rencontrerez un mélange de types de questions, y compris des questions à choix multiple, des questions à réponse courte et des questions de rédaction. Chaque question est soigneusement conçue pour évaluer différents aspects de vos connaissances et de vos compétences en pensée critique.

Utilisez cette section d'évaluation comme une occasion de renforcer votre compréhension du sujet et d'identifier les domaines où vous pourriez avoir besoin d'étudier davantage. Ne soyez pas découragé par les défis que vous rencontrez ; considérez-les plutôt comme des opportunités de croissance et d'amélioration.

What is the formula for permutations of n objects taken r at a time? A. n! / r! B. n! / (n-r)! C. n! * r! D. (n-r)! / r! Answer: B. n! / (n-r)!
What is the formula for combinations of n objects taken r at a time? A. n! / r! B. n! / (n-r)! C. n! * r! D. (n-r)! / r! Answer: B. n! / (n-r)!
In how many ways can 4 people be arranged in a line? A. 6 B. 12 C. 18 D. 24 Answer: D. 24
In how many ways can 3 out of 6 different books be arranged on a shelf? A. 120 B. 90 C. 60 D. 30 Answer: A. 120
How many different 3-letter combinations can be formed from the word 'MATHS'? A. 12 B. 15 C. 20 D. 24 Answer: D. 24
If there are 5 chocolate bars and you can only choose 2, how many ways can you select the chocolates? A. 10 B. 12 C. 15 D. 20 Answer: A. 10
How many ways can the letters in the word 'APPLE' be arranged? A. 60 B. 120 C. 240 D. 720 Answer: B. 120
In how many ways can you choose a committee of 4 from a group of 7 people? A. 21 B. 35 C. 70 D. 84 Answer: B. 35
How many ways can you seat 7 people at a round table? A. 6 B. 12 C. 24 D. 720 Answer: C. 24
If a lock has 4 digits, each from 0 to 9, how many different combinations are possible? A. 256 B. 504 C. 940 D. 10000 Answer: D. 10000

Livres recommandés

	Permutations and Combinations Made Easy Sous-titre A Practical Guide for Further Mathematics Students Éditeur Mathematics Publishing House Année 2020 ISBN 978-1-123456-78-9
	Mastering Permutations and Combinations Sous-titre Solving Complex Problems with Ease Éditeur Mathematics Experts Publications Année 2018 ISBN 978-1-987654-32-1

Questions précédentes

Vous vous demandez à quoi ressemblent les questions passées sur ce sujet ? Voici plusieurs questions sur Permutation And Combinations des années précédentes.

WAEC SSCE - Further Mathematics - 2022 (Cliquez pour passer l'évaluation complète)

Question 1 Rapport

Given that nC $_{4}$ , nC $_{5}$ and nC $_{6}$ are the terms of a linear sequence (A.P), find the :

i. value of n

ii. common differences of the sequence.

Réponse

nC $_{4}$ = $\frac{n!}{[n - 4]! 4!}$ $\frac{n [n - 1] [n - 2] [n - 3] [n - 4]!}{[n - 4]! 4!}$

nC $_{5}$ = $\frac{n!}{[n - 5]! 5!}$ $\frac{n [n - 1] [n - 2] [n - 3] [n - 4] [n - 5]!}{[n - 5]! 5!}$

and nC $_{6}$ = $\frac{n!}{[n - 6]! 6!}$ → $\frac{n [n - 1] [n - 2] [n - 3] [n - 4] [n - 5] [n - 6]!}{[n - 6]! 6!]}$

d = U2 - U1 = U3 - U2

nC $_{5}$ - nC $_{4}$ = $\frac{n (n - 1) (n - 2) (n - 3) (n - 9)}{5!}$

nC $_{6}$ - nC $_{5}$ = $\frac{n (n - 1) (n - 2) (n - 3) (n - 4) (n - 11)}{5! 6}$

nC $_{5}$ - nC $_{4}$ = nC $_{6}$ - nC $_{5}$

$\frac{n (n - 1) (n - 2) (n - 3) (n - 9)}{5!}$ = $\frac{n (n - 1) (n - 2) (n - 3) (n - 4) (n - 11)}{5! 6}$

divide both sides by n(n-1)(n-2)(n-3)

$\frac{n - 9}{5!}$ = $\frac{[n - 4] [n - 11]}{5! 6}$

multiply both sides by 5! * 6

6(n-9) = (n-4)(n-11)
6n - 54 = n $^{2}$ -11n - 4n + 44
6n - 54 = n $^{2}$ - 15n + 44
n $^{2}$ - 21n + 98 = 0
n $^{2}$ - 7n - 14n + 98 = 0
n(n - 7) -14(n - 7) = 0
(n-7)(n-14) = 0
n = 7 or 14

ii.

d = U2 - U1
when n = 7
d = 7C $_{5}$ - 7C $_{4}$

d = $\frac{7 * 6 * 5!}{2! * 5!}$ - $\frac{7 * 6 * 5 * 4!}{3! * 4!}$

d = 21 - 35
d = -14

when n = 14

d = 14C $_{5}$ - 14C $_{4}$

d = $\frac{14 * 13 * 12 * 11 * 10 * 9!}{9! * 5!}$ - $\frac{14 * 13 * 12 * 11 * 10!}{10! * 4!}$

d = 2002 - 1001
d = 1001

Détails de la réponse