Green Bridge Leermiddel Voor Permutation And Combinations

Overzicht

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.

Doelstellingen

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

Lesnotitie

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.

Lesevaluatie

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Je zult een mix van vraagtypen tegenkomen, waaronder meerkeuzevragen, korte antwoordvragen en essayvragen. Elke vraag is zorgvuldig samengesteld om verschillende aspecten van je kennis en kritisch denkvermogen te beoordelen.

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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

Aanbevolen Boeken

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

Eerdere Vragen

Benieuwd hoe eerdere vragen over dit onderwerp eruitzien? Hier zijn een aantal vragen over Permutation And Combinations van voorgaande jaren.

WAEC SSCE - Further Mathematics - 2022 (Klik om de volledige beoordeling te maken)

Vraag 1 Verslag

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.

Antwoord

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

Antwoorddetails