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Permutations and Combinations

🎓 Class 11📖 Mathematics📖 8 notes🧠 15 Q&A⏱️ ~12 min

Permutations and CombinationsStudy Notes

NCERT-aligned · 8 notes · 3 shown free

6.1 Introduction

Explanation

6.1 Introduction

This introductory section presents a practical problem involving a suitcase with a number lock having 4 wheels, each labelled with digits from 0 to 9. The lock opens only if 4 specific digits are arranged in a particular sequence without repetition. The problem posed is: if only the first digit (7) is remembered, how many sequences of the remaining 3 digits must be checked to open the lock? Listing all possible arrangements manually is tedious due to the large number of possibilities. This motivates the need for counting techniques that allow us to determine the number of arrangements or selections without explicit enumeration. The chapter aims to introduce these fundamental counting methods, starting with the Fundamental Principle of Counting, which forms the basis for permutations and combinations. The historical note mentions Jacob Bernoulli, a pioneer in the formal study of permutations and combinations.

  • Problem of counting sequences in a number lock with 4 wheels and digits 0-9.
  • Need for systematic counting techniques to avoid tedious listing.
  • Counting arrangements without repetition is the core problem.
  • Introduction to Fundamental Principle of Counting as a foundation.
  • Historical reference to Jacob Bernoulli's contribution.
  • Sets the stage for permutations and combinations study.
  • 📌 Permutation: arrangement of objects in a definite order.
  • 📌 Combination: selection of objects where order does not matter.

6.2 Fundamental Principle of Counting

Concept

6.2 Fundamental Principle of Counting

This section introduces the Fundamental Principle of Counting, also known as the multiplication principle. It states that if one event can occur in m ways and a subsequent event can occur in n ways, then the total number of ways both events can occur in sequence is m × n. This principle generalizes to any finite number of successive events, where the total number of outcomes is the product of the number of ways each event can occur. Two illustrative problems are presented: Mohan choosing pants and shirts, and Sabnam choosing school bags, tiffin boxes, and water bottles. The section explains how to count the total number of possible pairs or triplets by multiplying the number of choices at each step. Several examples demonstrate the application of this principle to counting arrangements of letters, flags, and digits with or without repetition. The principle is fundamental to understanding permutations and combinations, as it provides a systematic way to count complex arrangements by breaking them down into simpler successive choices.

  • Fundamental Principle of Counting states total ways = product of ways of successive events.
  • If event 1 can occur in m ways and event 2 in n ways, total ways = m × n.
  • Generalizes to multiple events: total ways = m × n × p × ...
  • Illustrated by examples of choosing pants and shirts, and school bags, tiffin boxes, water bottles.
  • Applied to counting arrangements of letters and digits with/without repetition.
  • Foundation for permutations and combinations.
  • 📌 Multiplication Principle: counting principle for successive events.
  • 📌 Event: an occurrence or choice in a counting problem.

6.3 Permutations

Definition

6.3 Permutations

This section formally introduces permutations as arrangements of objects in a definite order, taken some or all at a time. It builds on earlier examples such as forming words from letters of ROSE or NUMBER. The key idea is that order matters in permu

Practice QuestionsPermutations and Combinations

Includes NCERT exercise questions with answers

Q1.1. How many 3-digit numbers can be formed from the digits 1, 2, 3, 4 and 5 assuming that (i) repetition of the digits is allowed? (ii) repetition of the digits is not allowed?

Answer:

(i) When repetition of digits is allowed, each of the 3 places can be filled with any of the 5 digits. So, number of 3-digit numbers = 5 × 5 × 5 = 125. (ii) When repetition is not allowed, the first digit can be chosen in 5 ways, the second in 4 ways (excluding the first digit), and the third in 3 ways. So, number of 3-digit numbers = 5 × 4 × 3 = 60.

Explanation:

For (i), since repetition is allowed, each digit place is independent and can be any of the 5 digits. For (ii), since repetition is not allowed, the number of choices reduces by one for each subsequent digit.

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Q2.2. How many 3-digit even numbers can be formed from the digits 1, 2, 3, 4, 5, 6 if the digits can be repeated?

Answer:

To form a 3-digit even number, the last digit must be even. Digits available: 1,2,3,4,5,6 Even digits: 2,4,6 (3 choices for last digit) Since repetition is allowed: First digit: 6 choices (1 to 6) Second digit: 6 choices (1 to 6) Third digit (last digit): 3 choices (2,4,6) Total numbers = 6 × 6 × 3 = 108.

Explanation:

The last digit must be even to make the number even. Repetition allowed means each digit place can be chosen independently. Multiply the number of choices for each digit place.

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Q3.3. How many 4-letter code can be formed using the first 10 letters of the English alphabet, if no letter can be repeated?

Answer:

First 10 letters: A, B, C, D, E, F, G, H, I, J No repetition allowed. Number of 4-letter codes = Number of permutations of 10 letters taken 4 at a time = {}^{10}P_4 = 10 × 9 × 8 × 7 = 5040.

Explanation:

Since no letter can be repeated, the number of choices reduces by one for each subsequent letter. Use permutation formula for n distinct objects taken r at a time.

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Q4.4. How many 5-digit telephone numbers can be constructed using the digits 0 to 9 if each number starts with 67 and no digit appears more than once?

Answer:

The telephone number is 5-digit starting with 6 and 7. Digits used: 0 to 9 First two digits fixed: 6 and 7 Remaining 3 digits to be chosen from digits excluding 6 and 7 (digits left = 8) No repetition allowed. Number of ways = permutations of 8 digits taken 3 at a time = {}^{8}P_3 = 8 × 7 × 6 = 336.

Explanation:

Since first two digits are fixed, only last 3 digits vary. No repetition means digits already used (6 and 7) cannot be used again. Use permutation formula for remaining digits.

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Q5.5. A coin is tossed 3 times and the outcomes are recorded. How many possible outcomes are there?

Answer:

Each toss has 2 possible outcomes: Head (H) or Tail (T). Number of tosses = 3 Total possible outcomes = 2 × 2 × 2 = 2^3 = 8.

Explanation:

Each toss is independent with 2 outcomes. Multiply the number of outcomes for each toss to get total outcomes.

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Q6.6. Given 5 flags of different colours, how many different signals can be generated if each signal requires the use of 2 flags, one below the other?

Answer:

Number of flags = 5 Each signal uses 2 flags arranged one below the other (order matters). Number of signals = permutations of 5 flags taken 2 at a time = {}^{5}P_2 = 5 × 4 = 20.

Explanation:

Since order matters (one below the other), use permutations. Choose first flag in 5 ways, second in 4 ways.

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Q7.1. Evaluate (i) 8! (ii) 4! - 3!

Answer:

Solution: (i) 8! = 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 = 40320 (ii) 4! - 3! = (4 × 3 × 2 × 1) - (3 × 2 × 1) = 24 - 6 = 18

Explanation:

Calculate factorial values stepwise and subtract as required.

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Q8.2. Is 3! + 4! = 7! ?

Answer:

Solution: Calculate each factorial: 3! = 6 4! = 24 Sum = 6 + 24 = 30 7! = 5040 Since 30 ≠ 5040, the statement is false.

Explanation:

Compute factorial values and compare the sum with 7! to verify equality.

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