 # NCERT Solutions for Class 11 Maths Chapter 7- Permutations and Combinations Exercise 7.3

The solutions for the questions present in the third exercise of the seventh chapter of Class 11 Maths are given here. The solution PDFs can be downloaded from the links given below. Exercise 7.3 of NCERT Solutions for Class 11 Maths Chapter 7- Permutations and Combinations is based on the following topics:

1. Permutations
1. Derivation of the formula for nPr
2. Permutations when all the objects are not distinct objects

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 topics present in the textbook.

### Download PDF of NCERT Solutions for Class 11 Maths Chapter 7- Permutations and Combinations Exercise 7.3      ### Solutions for Class 11 Maths Chapter 7 – Exercise 7.3

1. How many 3-digit numbers can be formed by using the digits 1 to 9 if no digit is repeated?

Solution: 2. How many 4-digit numbers are there with no digit repeated?

Solution:

To find four digit number (digits does not repeat)

Now we will have 4 places where 4 digits are to be put.

So, at thousand’s place = There are 9 ways as 0 cannot be at thousand’s place = 9 ways

At hundredth’s place = There are 9 digits to be filled as 1 digit is already taken = 9 ways

At ten’s place = There are now 8 digits to be filled as 2 digits are already taken = 8 ways

At unit’s place = There are 7 digits that can be filled = 7 ways

Total Number of ways to fill the four places = 9 × 9 × 8 × 7 = 4536 ways.

So a total of 4536 four digit numbers can be there with no digits repeated.

3. How many 3-digit even numbers can be made using the digits 1, 2, 3, 4, 6, 7, if no digit is repeated?

Solution:

Even number means that last digit should be even,

Number of possible digits at one’s place = 3 (2, 4 and 6)

⇒ Number of permutations= One of digit is taken at one’s place, Number of possible digits available = 5

⇒ Number of permutations= Therefore, total number of permutations =3 × 20=60.

4. Find the number of 4-digit numbers that can be formed using the digits 1, 2, 3, 4, 5 if no digit is repeated. How many of these will be even?

Solution:

Total number of digits possible for choosing = 5

Number of places for which a digit has to be taken = 4

As there is no repetition allowed,

⇒ Number of permutations = The number will be even when 2 and 4 are at one’s place.

The possibility of (2, 4) at one’s place = 2/5 = 0.4

Total number of even number = 120 × 0.4 = 48.

5. From a committee of 8 persons, in how many ways can we choose a chairman and a vice chairman assuming one person cannot hold more than one position?

Solution:

Total number of people in committee = 8

Number of positions to be filled = 2

⇒ Number of permutations = 6. Find n if n-1P3nP3 = 1: 9.

Solution:  7. Find r if

(i)5Pr = 26Pr-1

(ii) 5Pr = 6Pr-1

Solution:  8. How many words, with or without meaning, can be formed using all the letters of the word EQUATION, using each letter exactly once?

Solution:

Total number of different letters in EQUATION = 8

Number of letters to be used to form a word = 8

⇒ Number of permutations = 9. 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?

Solution:

(i) Number of letters to be used =4

⇒ Number of permutations = (ii) Number of letters to be used = 6

⇒ Number of permutations = (iii) Number of vowels in MONDAY = 2 (O and A)

⇒ Number of permutations in vowel = Now, remaining places = 5

Remaining letters to be used =5

⇒ Number of permutations = Therefore, total number of permutations = 2 × 120 =240.

10. In how many of the distinct permutations of the letters in MISSISSIPPI do the four I’s not come together?

Solution:

Total number of letters in MISSISSIPPI =11

Letter Number of occurrence

 M 1 I 4 S 4 P 2

⇒ Number of permutations = We take that 4 I’s come together, and they are treated as 1 letter,

∴ Total number of letters=11 – 4 + 1 = 8

⇒ Number of permutations = Therefore, total number of permutations where four I’s don’t come together = 34650-840=33810

11. In how many ways can the letters of the word PERMUTATIONS be arranged if the

(ii) Vowels are all together,
(iii) There are always 4 letters between P and S?

Solution:

(i) Total number of letters in PERMUTATIONS =12

Only repeated letter is T; 2times

First and last letter of the word are fixed as P and S respectively.

Number of letters remaining =12 – 2 = 10

⇒ Number of permutations = (ii) Number of vowels in PERMUTATIONS = 5 (E, U, A, I, O)

Now, we consider all the vowels together as one.

Number of permutations of vowels = 120

Now total number of letters = 12 – 5 + 1= 8

⇒ Number of permutations = Therefore, total number of permutations = 120 × 20160 = 2419200

(iii) Number of places are as 1 2 3 4 5 6 7 8 9 10 11 12

There should always be 4 letters between P and S.

Possible places of P and S are 1 and 6, 2and 7, 3 and 8, 4 and 9, 5 and 10, 6 and 11, 7 and 12

Possible ways =7,

Also, P and S can be interchanged,

No. of permutations =2 × 7 =14

Remaining 10 places can be filled with 10 remaining letters,

∴ No. of permutations = Therefore, total number of permutations = 14 × 1814400 =25401600.

### Access other exercise solutions of Class 11 Maths Chapter 7- Permutations and Combinations

Exercise 7.1 Solutions 6 Questions

Exercise 7.2 Solutions 5 Questions

Exercise 7.4 Solutions 9 Questions

Miscellaneous Exercise On Chapter 7 Solutions 11 Questions