NCERT Solutions for Class 11 Maths Chapter 6 Exercise 6.1 Permutations and Combinations help students understand the basic principles of counting techniques used in mathematics. This exercise introduces students to the concept of arranging objects and calculating the number of possible outcomes.
Prepared according to the latest CBSE Class 11 Maths syllabus, Exercise 6.1 focuses on understanding the fundamental principle of counting and applying it to solve simple problems. It also introduces the concept of factorial notation (n!), which is widely used in permutations and combinations.
NCERT Solutions for Class 11 Maths Chapter 6 Permutations and Combinations Exercise 6.1
Q.
How many words, with or without meaning, can be formed using all the letters of the word EQUATION, using each letter exactly once?
Q.
How many words, with or without meaning can be made from the letters of the word MONDAY, assuming that no letter is repeated, if.
(i) 4 letters are used at a time,
(ii) all letters are used at a time,
(iii) all letters are used but first letter is a vowel?
Q.
Determine the number of 5 card combinations out of a deck of 52 cards if there is exactly one ace in each combination.
Q.
How many words, with or without meaning, each of 2 vowels and 3 consonants can be formed from the letters of the word DAUGHTER?
Q.
How many words, with or without meaning, can be formed using all the letters of the word EQUATION at a time so that the vowels and consonants occur together?
Q.
A committee of 7 has to be formed from 9 boys and 4 girls. In how many ways can this be done when the committee consists of:
(i) exactly 3 girls ?
(ii) atleast 3 girls ?
(iii) atmost 3 girls ?
Q.
If the different permutations of all the letter of the word EXAMINATION are listed as in a dictionary, how many words are there in this list before the first word starting with E?
Q.
How many 6-digit numbers can be formed from the digits 0, 1, 3, 5, 7 and 9 which are divisible by 10 and no digit is repeated?
Q.
It is required to seat 5 men and 4 women in a row so that the women occupy the even places. How many such arrangements are possible?
NCERT Solutions for Class 11 Maths Chapter 6 Permutations and Combinations Exercise 6.1
The solutions are explained in a clear and step-by-step format so students can easily understand the logic and apply the correct methods in exams.
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?
Ans:
There would be many ways as there are methods of filling 3 empty spots in progression by the given five digits. In this case, repetition of digits is allowed.
Therefore, the units spot can be filled in by any of the given five digits. Similarly, tens and hundred digits can be filled in by any of the given five digits.
Hence, by the multiplication principle, the number of ways in which three-digit numbers can be formed from the given digits is:
5 × 5 × 5 = 125
(ii) repetition of the digits is not allowed?
Ans:
In this case, repetition of digits is not allowed. Here, if the unit place is filled first, it can be filled by any of the given five digits.
Therefore, the number of ways of filling the units spot of the three-digit number is 5.
Then, the tens spot can be filled with any of the remaining four digits and the hundreds spot can be filled with any of the remaining three digits.
Thus, by the multiplication principle, the number of ways in which three-digit numbers can be formed without repeating the given digits is:
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?
Ans:
There will be as many ways as there are ways of filling 3 vacant places in succession by the given six digits.
For this situation, the units spot can be filled by 2 or 4 or 6 only, i.e., the units spot can be filled in 3 ways.
The tens spot can be filled by any of the 6 digits in 6 ways and the hundreds spot can also be filled by any of the 6 digits in 6 ways, as the digits can be repeated.
Thus, by the multiplication principle, the required number of three-digit even numbers is:
3 × 6 × 6 = 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?
Ans:
There are as many codes as there are ways of filling 4 vacant places in succession by the first 10 letters of the English alphabet, keeping in mind that repetition of letters is not allowed.
The first spot can be filled in 10 ways by any of the first 10 letters of the English alphabet.
The second spot can then be filled in 9 ways.
The third spot can be filled in 8 ways.
The fourth spot can be filled in 7 ways.
Therefore, by the multiplication principle, the required number of ways in which 4 empty places can be filled is:
10 × 9 × 8 × 7 = 5040
Hence, 5040 four-letter codes can be formed using the first 10 letters of the English alphabet, if no letter is repeated.
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?
Ans:
Since it is given that the 5-digit telephone numbers always start with 67.
Therefore, there will be as many phone numbers as there are methods of filling 3 empty spots in:
6, 7, __, __, __
by the digits 0–9, keeping in mind that the digits cannot be repeated.
Now, after using 6 and 7, there are 8 digits left for the third place.
Then 7 digits remain for the fourth place.
Then 6 digits remain for the fifth place.
Therefore, by the multiplication principle, the required number of ways in which 5-digit telephone numbers can be constructed is:
8 × 7 × 6 = 336
5. A coin is tossed 3 times and the outcomes are recorded. How many possible outcomes are there?
Ans:
When a coin is tossed once, the number of outcomes is 2 (Head and Tail), i.e., in each throw, the number of ways of showing a different face is 2.
Thus, by the multiplication principle, the required number of possible outcomes is:
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?
Ans:
Since each signal requires the use of 2 flags.
There will be as many signals as there are methods of filling 2 empty spots in succession by the given 5 flags of different colours.
The upper empty spot can be filled in 5 ways by any of the 5 flags.
The lower empty spot can then be filled in 4 ways by any of the remaining 4 flags.
Thus, by the multiplication principle, the number of different signals that can be generated is:
5 × 4 = 20
FAQs – Class 11 Maths Chapter 6 Exercise 6.1 Permutations and Combinations
Q1. What is the focus of Exercise 6.1?
Exercise 6.1 focuses on the fundamental principle of counting and basic counting techniques.
Q2. What is the fundamental principle of counting?
If one task can be done in m ways and another task in n ways, then both tasks together can be completed in m × n ways.
Q3. What is factorial notation?
Factorial of a number n is written as n! and defined as:
n! = n × (n−1) × (n−2) × ... × 1
Q4. Why is Exercise 6.1 important for exams?
This exercise forms the base of permutations and combinations, which is an important topic for higher mathematics and competitive exams.
Q5. How can students prepare effectively for Exercise 6.1?
Students should understand the counting principle clearly, practice factorial-based problems, and solve different types of questions to strengthen their concepts.