Loading PDF...
Chapter Analysis
Intermediate24 pages • EnglishQuick Summary
The chapter on permutations and combinations introduces fundamental counting principles and explores permutations as arrangements of objects, with detailed exploration of permutations with distinct and non-distinct objects. Combinations, as selections of items without regard to order, are analyzed using binomial coefficients. The chapter includes formulas and example problems to solidify understanding of calculating permutations and combinations efficiently.
Key Topics
- •Fundamental Principle of Counting
- •Permutations of distinct objects
- •Permutations of non-distinct objects
- •Combinations and Binomial Coefficients
- •Factorial notation
- •Applications of permutations and combinations
Learning Objectives
- ✓Understand the fundamental principle of counting
- ✓Apply permutations for ordering distinct and non-distinct items
- ✓Calculate combinations using binomial coefficients
- ✓Employ factorial notation to simplify permutation and combination expressions
- ✓Solve real-world problems involving permutations and combinations
Questions in Chapter
Evaluate (i) 8 ! (ii) 4 ! – 3 !
Page 107
Is 3 ! + 4 ! = 7 ! ?
Page 107
Compute 8! / (6! * 2!)
Page 107
If 1/6! + 1/7! = 1/8!, find x
Page 107
Evaluate n! / (n - r)!, when n = 9, r = 5
Page 107
Additional Practice Questions
How many different ways can you arrange 4 books on a shelf?
easyAnswer: The number of permutations of 4 distinct books is 4! = 4 × 3 × 2 × 1 = 24.
In how many ways can a committee of 4 be chosen from a group of 10 people?
mediumAnswer: The number of combinations is C(10, 4) = 10! / (4! × (10 - 4)!) = 210.
How many different 3-digit numbers can be formed from the digits 1, 2, 3, 4, 5 if each digit is used only once?
mediumAnswer: The number of permutations is 5P3 = 5 × 4 × 3 = 60.
Find the number of permutations of the letters of the word MATHEMATICS where vowels are always together.
hardAnswer: Consider vowels AEAI together as one unit; permutations are then based on MATHTICS and AEAI. Number = 8! / (2! × 2!) × 4! / (2!) = 10080.
How many ways can 5 people be seated in a row if two specific people must sit next to each other?
mediumAnswer: Calculate permutations as treating the two people as one unit: result is 4! × 2! = 48.