*According to the CBSE Syllabus 2023-24, this chapter has been renumbered as Chapter 6.
NCERT Solutions for Class 11 Maths Chapter 7 Ex 7.1 are helpful to improve the analysing abilities among students in solving problems related to permutations and combinations. All the questions of this exercise have been solved by subject experts according to the latest CBSE Syllabus 2023-24 and its guidelines. Exercise 7.1 of NCERT Solutions for Class 11 Maths Chapter 7 – Permutations and Combinations is based on the following topics:
- Introduction to Permutations and Combinations
- Fundamental Principle of Counting
These solutions are prepared by subject matter experts at BYJU’S, describing the complete method of solving problems. By understanding the concepts used in NCERT Solutions for Class 11 Maths, students will be able to clear all their doubts related to the chapters of Class 11 and can ace their board exams.
NCERT Solutions for Class 11 Maths Chapter 7 – Permutations and Combinations Exercise 7.1
Solutions for Class 11 Maths Chapter 7 – Exercise 7.1
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)Â 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)Â 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-digit 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 be even, the digits possible at C are 2 or 4 or 6. That is, the number of possible digits at C is 3.
Now, as 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 of possible 3-digit numbers = 6 × 6 × 3 = 108
3. How many 4-letter codes 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.
In the first place, the number of letters possible is 10.
Let’s suppose any 1 of the ten occupies place 1.
Now, as 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-digit 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 the first 2 digits of each number are 67. Therefore, the number is 67CDE.
As 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
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:
Given, 5 flags of different colours.
We know the signal requires 2 flags.
The number of flags possible for the upper flag is 5.
Now, as one of the flags is taken, the number of flags remaining for the lower flag in the signal is 4.
The number of ways in which signal can be given = 5 × 4 = 20
Access Other Exercise Solutions of Class 11 Maths Chapter 7 – Permutations and Combinations
Exercise 7.2 Solutions 5 Questions
Exercise 7.3 Solutions 11 Questions
Exercise 7.4 Solutions 9 Questions
Miscellaneous Exercise on Chapter 7 Solutions 11 Questions
Also explore – NCERT Class 11 Solutions
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