Future Study Point

# NCERT solutions for class 11 maths exercise 7.2 of chapter 7-Permutations and Combinations

### Download PDF of NCERT solutions for class 11 maths exercise 7.2 of chapter 7-Permutations and Combinations

Download PDF-NCERT solutions for class 11 maths exercise 7.2 of chapter 7-Permutations and Combinations

NCERT solutions for class 11 maths exercise 7.2 of chapter 7-Permutations and Combinations are prepared for helping the students in their maths study so that they could clear their maths concepts.NCERT solutions for class 11 maths exercise 7.2 of chapter 7-Permutations and Combinations are based on permutations.A permutation is the number of arrangements formed by choosing a certain number of objects at a time from a set of objects.

Click for online shopping

Future Study Point.Deal: Cloths, Laptops, Computers, Mobiles, Shoes etc

## NCERT Solutions for Class 11 Maths  Chapter 7 Permutation and Combination

Exercise 7.1- Permutations and Combinations

Exercise 7.2 -Permutation and Combination

Exercise 7.3-Permutation and Combination

Exercise 7.4-Permutation and Combination

Miscellaneous Exercise-Permutation and Combination

NCERT solutions of class 11 maths

 Chapter 1-Sets Chapter 9-Sequences and Series Chapter 2- Relations and functions Chapter 10- Straight Lines Chapter 3- Trigonometry Chapter 11-Conic Sections Chapter 4-Principle of mathematical induction Chapter 12-Introduction to three Dimensional Geometry Chapter 5-Complex numbers Chapter 13- Limits and Derivatives Chapter 6- Linear Inequalities Chapter 14-Mathematical Reasoning Chapter 7- Permutations and Combinations Chapter 15- Statistics Chapter 8- Binomial Theorem Chapter 16- Probability

### CBSE Class 11-Question paper of maths 2015

CBSE Class 11 – Second unit test of maths 2021 with solutions

Study notes of Maths and Science NCERT and CBSE from class 9 to 12

Q1. Evaluate

(i) 8 !           (ii) 4 ! – 3 !

Ans.(i) 8 ! = 8 ×7 ×6 ×5 ×4 ×3 ×2 ×1 = 40320

(i) 4 !- 3!  = 4 ×3 ×2 ×1-3×2×1= 24-6 = 18

Q2. Is 3 ! + 4 ! = 7! ?

Ans. Taking LHS

3 ! + 4 ! =3×2×1 +4× 3×2×1  = 6 +24 =30

RHS= 7! = 7×6×5 ×4× 3×2×1=5040

∴ 3 ! + 4 ! ≠ 7!

Q3. Compute :

Ans.

Ans.

x = 64

(i) n = 6,r= 2        (ii) n = 9, r = 5

Ans.(i) n = 6,r= 2

(ii) n = 9, r = 5

Exercise 7.3

Q1. How many 3-digit numbers can be formed by using the digits 1 to 9 if no digit is repeated?

Ans. The given digits are 9, i.e  1,2,3….9

The number of 3 digits numbers using 9 digits = Permutations of  9 different digits choosing 3 digits at a time

Hence required number of 3 digit numbers

9P3

Q2. How many 4-digit numbers are there with no digit repeated?

Ans. The  digits utilized to form 4 digits numbers are 0,1,2,3….9

Number of 4-digit numbers with no digit repeated = Number of permutations of 9 different  digits choosing 4 digits at a time

Hence required number of 4 digit numbers

= 10P4

Since the number of 5040 also includes the numbers which starts from 0 i.e 0123,0324 etc,but actually these are three digits numbers

Therefore the number of permutations of 9 different  digits (o is excluded) choosing 3 digits at a time

= 9P4

Therefore the required 4 digit numbers are = 5040 – 504 = 3536

Q3. How many 3-digit even numbers can be made using the digits 1, 2, 3, 4, 6,7 if no digit is repeated.

Ans. Let’s find out 3-digit even numbers using the digits 1, 2, 3, 4,6,7

For a three-digit even number , any of the 3 digits 2,4, or 6  is required in its unit place.

The number of ways of filling  unit place is = The number of permutations of 3 different  digits  choosing 1 digits (2,4,or 6) at a time

= 3P1

Since digits are not allowed to be repeated then the number of ways of filling tens and hundred places are = Number of permutations of  remaining 5 digits choosing 2 digits at a time

= 5P2

According to the multiplication principle the number of ways creating three-digit even numbers are = 3 × 20 = 60

Therefore required 3 digit even numbers are = 60

Q4. 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?

Ans.The number of 4-digit numbers that can be formed using the digits 1, 2, 3, 4, 5 if no digit is repeated = Number of permutations of  5 digits choosing 4 digits at a time

= 5P4

Let’s find out 4-digit even numbers using the digits 1, 2, 3, 4,5

For a four-digit even number, any of the 2 digits 2,4  is required in its unit place.

The number of ways of filling  unit place is = The number of permutations of 2 different  digits  choosing 1 digits (2,4,) at a time

= 2P1

Since digits are not allowed to be repeated then the number of ways of filling tens and hundred and thousand  places are = Number of permutations of  remaining 4 digits choosing 3 digits at a time

= 4P3

According to the multiplication principle the number of ways of creating four-digit even numbers are = 2 × 24 =48

Therefore required 4 digit even numbers are  48 and number of 4-digit numbers that can be formed are 120.

Q5. From a committee of 8 persons, in how many ways can we choose a chairman and a vice-chairman assuming one person can not hold more than one position?

Ans. From a committee of 8 persons, let’s find out the number of ways of choosing a chairman and a vice-chairman assuming one person can not hold more than one position

The number of ways of choosing a chairman and a vice-chairman out of a committee of 8 persons = The number of permutations of 8 different  persons choosing 2 persons at a time

= 8P2

Q6. Find n if    n-1P: nP4 = 1 : 9

Ans. n-1P: nP4 = 1 : 9

n = 9

Q7. Find r if (i) 5P= 2.6Pr-1  (ii) 5P= 6Pr-1

Ans.

(i) 5P= 2.6Pr-1

(7 -r)(6 -r) = 12

42 – 7r – 6r + r² = 12

r²- 13r + 30 = 0

r² – 10 r – 3r + 30 = 0

r( r- 10) – 3(r – 10) =0

( r- 10)(r – 3) = 0

r = 10,3

In  nP,0≤r ≤n, therefore the value of r is 3

(ii) 5P= 6Pr-1

(7 – r)(6 – r) = 6

42 – 7r – 6r + r² = 6

r² – 13r + 36 = 0

r² – 9r – 4r – 36 = 0

r(r – 9)  – 4(r – 9) = 0

(r – 9) (r – 4) = 0

r =9,4

In  nP,0≤r ≤n, therefore the value of r is 4

Q8. How many words can be, with or without meaning, can be formed using all the letters of the word EQUATION, using each letter once?

Ans. The number of letters in the word EQUATION are = 8

The number of  ways,the words with or without meaning, can be formed using all 8 letters = Number of permutations of   8 letters choosing 8 letters at a time

= 8P8

Q9. How many words can be, 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 vowel?

Ans.(i) The number of letters in the word MONDAY are = 6

The number of  ways,the words with or without meaning, can be formed using  4 letters at a time out of 6 letters= Number of permutations of   6 letters choosing 4 letters at a time

= 6P4

(ii) The number of  ways,the words with or without meaning, can be formed using  all 6 letters at a time out of 6 letters= Number of permutations of   6 letters choosing 6 letters at a time

= 6P6

(iii)  The number of vowels in the word MONDAY  is = 2 (i.e O, A)

The number of  ways the words  formed using 1 vowel at a first place out of 2 vowels= Number of permutations of   2 vowels choosing 1 vowel at a time

= 2P= 2

Since one place is already occupied by one vowel therefore the number of  ways,of selecting the remaining 5 letters at a time = Number of permutations of   5 letters choosing 5 letters at a time

= 5P5 = 5! = 5×4×3×2= 120

According to the multiplication principle, the number of ways of creating 6 letters words  are = 2 × 120 =240

Q10. In how many of the distinct permutations of the letters in MISSISSIPPI do the four I’s not come together?

Ans. The total number of words in MISSISSIPPI are = 11

The frequency of I = 4

The frequency of S = 4

The frequency of P = 2

We have to determine

Distinct permutations of the letters in word when 4I’s not come together= Distinct permutations of the letters in the word -Distinct permutations of the letters in word when 4I’s  come together

Number of words formed by using all the letters of the word =The distinct permutations of the letters in the word

When 4I’s  come together should be treated as a single object, therefore in this case total letters would be supposed as 8, with the frequency of S =4 and P=2

The distinct permutations of the letters in the word when 4 I’s are together

Distinct permutations of the letters in word when 4I’s not come together =34650-840= 33810

Q11. 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?

Ans. (i) The number of letters in word PERMUTATIONS are = 12

Let’s find out the words start with P and end with S, means first and last letter are fixed, permutations of letters of the remaining 10 letters in which there are 2 T ‘s

Therefore in this case the number of ways of arrangement

(ii) In this case when vowels are all together,so all vowels together are treated as single letter.There are 5 vowels A,E,I,O,U, reamaining 7 letters and 5 vowels treated as 8 letters with 2 T’s

Therefore in this case the number of ways of arrangement

The number of ways of arrangement of 5 vowels = 5! = 5×4×3×2=120

Hence according to multiplication principle total number of arrangements in this case =20160 × 120 = 2419200

(iii) There are always 4 letters between P and S, it can be shown as below.

It can be observed that the number of ways of placing 4 letters between P and S are = 7 ways

The way of arrangement of P and S is =2! = 2

The remaining 10 letters can be arranged  with 2 I’s by  (10!)/2!

According to the multiplication principle, the number of ways of placing 4 letters between P and S   are = 7×2 × (10!)/2! =25401600

See the video of the solutions of question number 11

## Permutations

A permutation is the number of arrangements formed by choosing a certain number of objects at a time from a set of objects.Let there are n objects and we have to creat the arrangement of the objects with r objects taken at a time then permutations of  the n objects r taken at a time

If certain objects repeated by m and t times then permutations of all objects taken at a time

Permutations are the way of arrangement of the things when order also matters, which means in the arrangement if we have taken 23 then 32 is also counted.

## NCERT Solutions of Science and Maths for Class 9,10,11 and 12

### NCERT Solutions for class 10 maths

CBSE Class 10-Question paper of maths 2021 with solutions

CBSE Class 10-Half yearly question paper of maths 2020 with solutions

CBSE Class 10 -Question paper of maths 2020 with solutions

CBSE Class 10-Question paper of maths 2019 with solutions

### NCERT Solutions for class 11 maths

 Chapter 1-Sets Chapter 9-Sequences and Series Chapter 2- Relations and functions Chapter 10- Straight Lines Chapter 3- Trigonometry Chapter 11-Conic Sections Chapter 4-Principle of mathematical induction Chapter 12-Introduction to three Dimensional Geometry Chapter 5-Complex numbers Chapter 13- Limits and Derivatives Chapter 6- Linear Inequalities Chapter 14-Mathematical Reasoning Chapter 7- Permutations and Combinations Chapter 15- Statistics Chapter 8- Binomial Theorem Chapter 16- Probability

CBSE Class 11-Question paper of maths 2015

CBSE Class 11 – Second unit test of maths 2021 with solutions

### NCERT solutions for class 12 maths

 Chapter 1-Relations and Functions Chapter 9-Differential Equations Chapter 2-Inverse Trigonometric Functions Chapter 10-Vector Algebra Chapter 3-Matrices Chapter 11 – Three Dimensional Geometry Chapter 4-Determinants Chapter 12-Linear Programming Chapter 5- Continuity and Differentiability Chapter 13-Probability Chapter 6- Application of Derivation CBSE Class 12- Question paper of maths 2021 with solutions Chapter 7- Integrals Chapter 8-Application of Integrals

Class 12 Solutions of Maths Latest Sample Paper Published by CBSE for 2021-22 Term 2

Class 12 Maths Important Questions-Application of Integrals

Class 12 Maths Important questions on Chapter 7 Integral with Solutions for term 2 CBSE Board 2021-22

Solutions of Class 12 Maths Question Paper of Preboard -2 Exam Term-2 CBSE Board 2021-22

Solutions of class 12  maths question paper 2021 preboard exam CBSE Solution

Scroll to Top
Optimized by Seraphinite Accelerator
Turns on site high speed to be attractive for people and search engines.