Chapter 7

Chapter 7: Permutations and Combinations

Chapter Overview

In this chapter, we will delve into the world of permutations and combinations, which are fundamental concepts in mathematics. Permutations and combinations are used to calculate the number of ways in which a set of objects can be arranged or selected. These concepts have numerous applications in various fields, including probability, statistics, and computer science. To understand permutations and combinations, it's essential to grasp the concept of factorials, as they are used extensively in the formulas.

💡 Pro Tip: To understand permutations and combinations, it's essential to grasp the concept of factorials, as they are used extensively in the formulas.

Learning Objectives

• Understand the concept of permutations and combinations
• Learn to calculate the number of permutations and combinations
• Apply the concepts of permutations and combinations to solve problems 🧠 Trick to Remember: Permutations are like "people" in a line, where order matters, while combinations are like "groups" of people, where order doesn't matter.

Important Concepts

Permutations

Permutations refer to the arrangement of objects in a specific order. The number of permutations of n distinct objects taken r at a time is denoted by P(n, r) and is calculated using the formula:

P(n, r) = n! / (n-r)!

where n! represents the factorial of n.

Example: Suppose we have 5 distinct objects, and we want to arrange them in a line. We can use the permutation formula to calculate the number of possible arrangements.

P(5, 5) = 5! / (5-5)! = 5! / 0! = 5! (since 0! = 1) = 5 × 4 × 3 × 2 × 1 = 120

So, there are 120 possible arrangements of the 5 distinct objects.

💡 Pro Tip: When calculating permutations, make sure to consider the order of the objects, as it affects the result.

Combinations

Combinations refer to the selection of objects without considering the order. The number of combinations of n distinct objects taken r at a time is denoted by C(n, r) and is calculated using the formula:

C(n, r) = n! / (r!(n-r)!)

Example: Suppose we have 5 distinct objects, and we want to select 3 of them without considering the order. We can use the combination formula to calculate the number of possible selections.

C(5, 3) = 5! / (3!(5-3)!) = 5! / (3!2!) = (5 × 4 × 3 × 2 × 1) / (3 × 2 × 1 × 2 × 1) = (5 × 4) / 2 = 10

So, there are 10 possible selections of 3 objects from the 5 distinct objects.

🧠 Trick to Remember: Combinations can be thought of as "choosing" a certain number of objects from a larger set, without regard to order.

Properties of Permutations and Combinations

• The number of permutations of n distinct objects taken r at a time is equal to the number of combinations of n distinct objects taken r at a time multiplied by r!.
• Permutations and combinations are used in various fields, such as probability, statistics, and computer science.

Theorem: The number of permutations of n distinct objects taken r at a time is equal to the number of combinations of n distinct objects taken r at a time multiplied by r!.

Proof:

P(n, r) = n! / (n-r)! = n! / (r!(n-r)!) × r! = C(n, r) × r!

Therefore, the number of permutations of n distinct objects taken r at a time is equal to the number of combinations of n distinct objects taken r at a time multiplied by r!.

💡 Pro Tip: Understanding the properties of permutations and combinations can help you solve problems more efficiently.

Advanced Section: Deep-Dive Case Studies and Real-Life Applications

Permutations and combinations have numerous applications in real-life scenarios, such as:

• Counting the number of ways to arrange a deck of cards: Suppose we have a standard deck of 52 cards, and we want to calculate the number of possible arrangements of the cards. We can use the permutation formula to calculate the number of possible arrangements.

P(52, 52) = 52! / (52-52)! = 52! / 0! = 52! (since 0! = 1) = 52 × 51 × 50 × ... × 1 = 8.0658 × 10^67

So, there are approximately 8.0658 × 10^67 possible arrangements of the 52 cards.

• Calculating the number of possible outcomes in a game or experiment: Suppose we have a game where we roll 3 dice, and we want to calculate the number of possible outcomes. We can use the combination formula to calculate the number of possible outcomes.

C(6, 3) = 6! / (3!(6-3)!) = 6! / (3!3!) = (6 × 5 × 4 × 3 × 2 × 1) / (3 × 2 × 1 × 3 × 2 × 1) = (6 × 5 × 4) / 3 = 20

So, there are 20 possible outcomes in the game.

• Determining the number of ways to select a team or committee: Suppose we have a group of 10 people, and we want to select a team of 5 people. We can use the combination formula to calculate the number of possible selections.

C(10, 5) = 10! / (5!(10-5)!) = 10! / (5!5!) = (10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1) / (5 × 4 × 3 × 2 × 1 × 5 × 4 × 3 × 2 × 1) = (10 × 9 × 8 × 7 × 6) / 5 = 252

So, there are 252 possible selections of a team of 5 people from the group of 10 people.

💡 Pro Tip: Recognizing the applications of permutations and combinations can help you appreciate their importance in real-life situations.

Step-by-Step Problem Solving Strategies & Detailed Proofs

Problem: Calculate the number of permutations of 5 distinct objects taken 3 at a time.

Solution:

1. Write down the formula for permutations:

P(n, r) = n! / (n-r)!

2. Plug in the values:

P(5, 3) = 5! / (5-3)! = 5! / 2! = (5 × 4 × 3 × 2 × 1) / (2 × 1) = (5 × 4 × 3) / 1 = 60

So, there are 60 possible permutations of 5 distinct objects taken 3 at a time.

Proof:

P(n, r) = n! / (n-r)! = n! / (r!(n-r)!) × r! = C(n, r) × r!

Therefore, the number of permutations of n distinct objects taken r at a time is equal to the number of combinations of n distinct objects taken r at a time multiplied by r!.

💡 Pro Tip: Using step-by-step problem solving strategies and detailed proofs can help you solve problems more efficiently.

Higher-Order Thinking Skills (HOTS) Questions

Question 1: A committee of 5 people is to be selected from a group of 10 people. If the order of selection does not matter, how many possible committees can be formed?

Solution:

We can use the combination formula to calculate the number of possible committees.

C(10, 5) = 10! / (5!(10-5)!) = 10! / (5!5!) = (10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1) / (5 × 4 × 3 × 2 × 1 × 5 × 4 × 3 × 2 × 1) = (10 × 9 × 8 × 7 × 6) / 5 = 252

So, there are 252 possible committees that can be formed.

Question 2: A deck of 52 cards is shuffled, and 3 cards are drawn. If the order of the cards does not matter, how many possible combinations of 3 cards can be drawn?

Solution:

We can use the combination formula to calculate the number of possible combinations of 3 cards.

C(52, 3) = 52! / (3!(52-3)!) = 52! / (3!49!) = (52 × 51 × 50 × 49 × 48 × 47 × ... × 1) / (3 × 2 × 1 × 49 × 48 × 47 × ... × 1) = (52 × 51 × 50) / 3 = 22100

So, there are 22100 possible combinations of 3 cards that can be drawn.

💡 Pro Tip: Using HOTS questions can help you develop critical thinking skills and problem-solving strategies.

Previous Year Questions (PYQs) with solutions

PYQ 1: Calculate the number of permutations of 4 distinct objects taken 2 at a time.

Solution:

We can use the permutation formula to calculate the number of permutations.

P(4, 2) = 4! / (4-2)! = 4! / 2! = (4 × 3 × 2 × 1) / (2 × 1) = (4 × 3) / 1 = 12

So, there are 12 possible permutations of 4 distinct objects taken 2 at a time.

PYQ 2: A committee of 4 people is to be selected from a group of 8 people. If the order of selection does not matter, how many possible committees can be formed?

Solution:

We can use the combination formula to calculate the number of possible committees.

C(8, 4) = 8! / (4!(8-4)!) = 8! / (4!4!) = (8 × 7 × 6 × 5 × 4 × 3 × 2 × 1) / (4 × 3 × 2 × 1 × 4 × 3 × 2 × 1) = (8 × 7 × 6 × 5) / 4 = 70

So, there are 70 possible committees that can be formed.

💡 Pro Tip: Using PYQs can help you develop problem-solving strategies and critical thinking skills.

NCERT Textbook Questions & Detailed Answers

Question 1: Calculate the number of permutations of 5 distinct objects taken 3 at a time.

Solution:

We can use the permutation formula to calculate the number of permutations.

P(5, 3) = 5! / (5-3)! = 5! / 2! = (5 × 4 × 3 × 2 × 1) / (2 × 1) = (5 × 4 × 3) / 1 = 60

So, there are 60 possible permutations of 5 distinct objects taken 3 at a time.

Question 2: A committee of 5