Class 11 NCERT Solutions Chapter 7 Permutations And Combinations Exercise 7.1 (original) (raw)

Last Updated : 23 Jul, 2025

Chapter 7 of the Class 11 NCERT Mathematics textbook, titled "Permutations and Combinations," introduces students to fundamental concepts in counting and arranging objects. This chapter focuses on the principles of permutations and combinations, which are essential for solving various problems in probability and combinatorial mathematics. Exercise 7.1 deals with basic problems on permutations and combinations, helping students understand how to count arrangements and selections.

NCERT Solutions for Class 11 - Chapter 7 Permutations and Combinations - Exercise 7.1

This section provides detailed solutions for Exercise 7.1 from Chapter 7 of the Class 11 NCERT Mathematics textbook. The exercise includes problems that involve calculating permutations and combinations in different scenarios. The solutions are explained step-by-step to help students grasp the methods for solving these problems and applying the principles of permutations and combinations effectively.

**Problem 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?**

**Solution:

**Answer: 125.

**Method:

Here, Total number of digits = 5

Let 3-digit number be XYZ.

Now the number of digits available for X = 5,

As repetition is allowed,

So the number of digits available for Y and Z will also be 5 (each).

Thus, The total number of 3-digit numbers that can be formed = 5×5×5 = 125.

****(ii) repetition of the digits is not allowed?**

**Solution:

**Answer: 60.

**Method:

Here, Total number of digits = 5

Let 3-digit number be XYZ.

Now the number of digits available for X = 5,

As repetition is not allowed,

So the number of digits available for Y = 4 (As one digit has already been chosen at X),

Similarly, the number of digits available for Z = 3.

Thus, The total number of 3-digit numbers that can be formed = 5×4×3 = 60.

**Problem 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?

**Solution:

**Answer: 108.

**Method:

Here, Total number of digits = 6

Let 3-digit number be XYZ.

Now, as the number should be even so the digits at unit place must be even, so number of digits available for Z = 3 (As 2,4,6 are even digits here),

As the repetition is allowed,

So the number of digits available for X = 6,

Similarly, the number of digits available for Y = 6.

Thus, The total number of 3-digit even numbers that can be formed = 6×6×3 = 108.

**Problem 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:

**Answer: 5040

**Method:

Here, Total number of letters = 10

Let the 4-letter code be 1234.

Now, the number of letters available for 1st place = 10,

As repetition is not allowed,

So the number of letters possible at 2nd place = 9 (As one letter has already been chosen at 1st place),

Similarly, the number of letters available for 3rd place = 8,

and the number of letters available for 4th place = 7.

Thus, The total number of 4-letter code that can be formed = 10×9×8×7 = 5040.

**Problem 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:

**Answer: 336

**Method:

Here, Total number of digits = 10 (from 0 to 9)

Let 5-digit number be ABCDE.

Now, As the number should start from 67 so the number of possible digits at A and B = 1 (each),

As repetition is not allowed,

So the number of digits available for C = 8 ( As 2 digits have already been chosen at A and B),

Similarly, the number of digits available for D = 7,

and the number of digits available for E = 6.

Thus, The total number of 5-digit telephone numbers that can be formed = 1×1×8×7×6 = 336.

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

**Solution:

**Answer: 8

**Method:

We know that, the possible outcome after tossing a coin is either head or tail (2 outcomes),

Here, a coin is tossed 3 times and outcomes are recorded after each toss,

Thus, the total number of outcomes = 2×2×2 = 8.

**Problem 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?

**Solution:

**Answer: 20.

**Method:

Here, Total number of flags = 5

As each signal requires 2 flag and signals should be different so repetition will not be allowed,

So, the number of flags possible for the upper place = 5,

and the number of flags possible for the lower place = 4.

Thus, the total number of different signals that can be generated = 5×4 = 20.

Summary

Chapter 7 of the Class 11 NCERT Mathematics textbook, "Permutations and Combinations," covers the fundamental principles of counting arrangements and selections. Exercise 7.1 focuses on basic problems involving permutations and combinations, providing students with practice in calculating the number of ways to arrange or select objects. The exercise aims to reinforce students' understanding of these concepts through a series of step-by-step problems.