# Permutations and Combinations Chapter 7 Class 11 Maths NCERT Textbook With Solutions PDF

NCERT Solutions for Class 11 Maths Chapter 7′ PDF Quick download link is given at the bottom of this article. You can see the PDF demo, size of the PDF, page numbers, and direct download Free PDF of ‘Ncert Class 11 Maths Chapter 7 Exercise Solution’ using the download button.

### Chapter 7: Permutations and Combinations

#### 7.1 Introduction

Suppose you have a suitcase with a number lock. The number lock has 4 wheels each labeled with 10 digits from 0 to 9.

The lock can be opened if 4 specific digits are arranged in a particular sequence with no repetition. Somehow, you have forgotten this specific sequence of digits.

You remember only the first digit which is 7. In order to open the lock, how many sequences of 3-digits you may have to check with?

To answer this question, you may, immediately, start listing all possible arrangements of 9 remaining digits taken 3 at a time.

But, this method will be tedious, because the number of possible sequences may be large.

Here, in this Chapter, we shall learn some basic counting techniques which will enable us to answer this question without actually listing 3-digit arrangements.

In fact, these techniques will be useful in determining the number of different ways of arranging and selecting objects without actually listing them.

As a first step, we shall examine a principle that is most fundamental to the learning of these techniques.

#### 7.2 Fundamental Principle of Counting

Let us consider the following problem. Mohan has 3 pairs of pants and 2 shirts. How many different pairs of a pant and a shirt, can he dress up with?

There are 3 ways in which a pant can be chosen because there are 3 pairs of pants available. Similarly, a shirt can be chosen in 2 ways.

For every choice of a pant, there are 2 choices of a shirt. Therefore, there are 3 × 2 = 6 pairs of a pant and a shirt

#### 7.3 Permutations

In Example 1 of the previous section, we are actually counting the different possible arrangements of the letters such as ROSE, REOS, …, etc.

Here, in this list, each arrangement is different from the others. In other words, the order of writing the letters is important.

Each arrangement is called a permutation of 4 different letters taken all at a time. Now, if we have to determine the number of 3-letter words, with or without meaning,

which can be formed out of the letters of the word NUMBER, where the repetition of the letters is not allowed, we need to count the arrangements NUM, NMU, MUN, NUB, …, etc.

Here, we are counting the permutations of 6 different letters taken 3 at a time. The required number of words = 6 × 5 × 4 = 120 (by using the multiplication principle).

### NCERT Solutions Class 11 Maths Chapter 7 Permutations and Combinations

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?

Solution:

(i) Let the 3-digit number be ABC, where C is at the units place, B at the tens place and A at the hundreds place.

Now when repetition is allowed,

The number of digits possible at C is 5. As repetition is allowed, the number of digits possible at B and A is also 5 at each.

Hence, the total number possible 3-digit numbers =5 × 5 × 5 =125

(ii) Let the 3-digit number be ABC, where C is at the units place, B at the tens place and A at the hundreds place.

Now when repetition is not allowed,

The number of digits possible at C is 5. Let’s suppose one of 5 digits occupies place C, now as the repletion is not allowed, the possible digits for place B are 4 and similarly there are only 3 possible digits for place A.

Therefore, The total number of possible 3-digit numbers=5 × 4 × 3=60

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

Solution:

Let the 3-digit number be ABC, where C is at the unit’s place, B at the tens place and A at the hundreds place.

As the number has to even, the digits possible at C are 2 or 4 or 6. That is number of possible digits at C is 3.

Now, as the repetition is allowed, the digits possible at B is 6. Similarly, at A, also, the number of digits possible is 6.

Therefore, The total number possible 3 digit numbers = 6 × 6 × 3 = 108.

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

Solution:

Let the 4 digit code be 1234.

At the first place, the number of letters possible is 10.

Let’s suppose any 1 of the ten occupies place 1.

Now, as the repetition is not allowed, the number of letters possible at place 2 is 9. Now at 1 and 2, any 2 of the 10 alphabets have been taken. The number of alphabets left for place 3 is 8 and similarly the number of alphabets possible at 4 is 7.

Therefore the total number of 4 letter codes=10 × 9 × 8 × 7=5040.

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?

Solution:

Let the five-digit number be ABCDE. Given that first 2 digits of each number is 67. Therefore, the number is 67CDE.

As the repetition is not allowed and 6 and 7 are already taken, the digits available for place C are 0,1,2,3,4,5,8,9. The number of possible digits at place C is 8. Suppose one of them is taken at C, now the digits possible at place D is 7. And similarly, at E the possible digits are 6.

∴The total five-digit numbers with given conditions = 8 × 7 × 6 = 336.

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

Solution:

Given A coin is tossed 3 times and the outcomes are recorded

The possible outcomes after a coin toss are head and tail.

The number of possible outcomes at each coin toss is 2.

∴The total number of possible outcomes after 3 times = 2 × 2 × 2 = 8.