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.

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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 :

$\frac{8!}{6!\times 2!}$

Ans.

$\frac{8!}{6!\times 2!}$

$=\frac{8\times 7\times 6!}{6!\times 2\times 1}$

$=\frac{56}{2}=28$

$Q4.If\: \frac{1}{6!}+\frac{1}{7!}=\frac{x}{8!},find\: x$

Ans.

$\frac{1}{6!}+\frac{1}{7!}=\frac{x}{8!}$

$\frac{1}{6!}+\frac{1}{7\times 6!}=\frac{x}{8\times 7\times 6!}$

$\frac{1}{6!}\left ( 1+\frac{1}{7} \right )=\frac{x}{56\times 6!}$

$\frac{8}{7}=\frac{x}{56}$

x = 64

$Q5.Evaluate\: \frac{n!}{\left ( n-r \right )!},when$

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

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

$\frac{n!}{\left ( n-r \right )!}=\frac{6!}{\left ( 6-2 \right )!}=\frac{6!}{4!}=\frac{6\times 5\times 4!}{4!}=30$

(ii) n = 9, r = 5

$\frac{n!}{\left ( n-r \right )!}=\frac{9!}{\left ( 9-5 \right )!}=\frac{9!}{4!}=\frac{9\times 8\times 7\times 6\times 5\times 4!}{4!}=15120$

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

$=\frac{9!}{\left ( 9-3 \right )!}=\frac{9!}{6!}=\frac{9\times 8\times 7\times 6!}{6!}=9\times 8\times 7=504$

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

$\frac{10!}{\left ( 10-4 \right )!}=\frac{10!}{6!}=\frac{10\times 9\times 8\times 7\times \times 6!}{6!}=5040$

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

$\frac{9!}{\left ( 9-3 \right )!}=\frac{9!}{6!}=\frac{9\times 8\times 7\times 6!}{6!}=9\times 8\times 7=504$

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

$=\frac{3!}{\left ( 3-1 \right )!}=\frac{3!}{2!}=\frac{3\times 2\times 1}{2\times 1}=\frac{6}{2}=3$

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

$=\frac{5!}{\left ( 5-2 \right )!}=\frac{5!}{3!}=\frac{5\times 4\times 3!}{3!}=20$

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

$=\frac{5!}{\left ( 5-4 \right )!}=\frac{5!}{1!}=\frac{5\times 4\times 3\times 2\times 1!}{1!}=120$

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

$=\frac{2!}{\left ( 2-1 \right )!}=\frac{2!}{1!}=\frac{2\times 1}{1}=2$

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

$=\frac{4!}{\left ( 4-3 \right )!}=\frac{4!}{1!}=\frac{4\times 3\times 2}{1}=24$

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.

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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

$=\frac{8!}{\left ( 8-2 \right )!}=\frac{8!}{6!}=\frac{8\times 7\times 6!}{6!}=8\times 7=56$

Q6. Find n if ^n-1P₃: ⁿP₄= 1 : 9

Ans. ^n-1P₃: ⁿP₄= 1 : 9

$\frac{\left ( n-1 \right )!}{\left ( n-4 \right )!}:\frac{n!}{\left ( n-4 \right )!}=\frac{1}{9}$

$\frac{\left ( n-1 \right )!}{n!}=\frac{1}{9}$

$\frac{\left ( n-1 \right )!}{n\left ( n-1 \right )!}=\frac{1}{9}$

$\frac{1}{n}=\frac{1}{9}$

n = 9

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

Ans.

(i) ⁵P_r= 2.⁶P_r-1

$\frac{5!}{\left ( 5-r \right )!}=2.\frac{6!}{\left ( 6-r+1 \right )!}$

$\frac{5!}{\left ( 5-r \right )!}=2.\frac{6\times 5!}{\left ( 7-r \right )!}$

$\frac{5!}{\left ( 5-r \right )!}=2.\frac{6\times 5!}{\left ( 7-r \right )\left ( 6-r \right )\left ( 5-r \right )!}$

$\frac{12}{\left ( 7-r \right )\left ( 6-r \right )}=1$

(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_r,0≤r ≤n, therefore the value of r is 3

(ii) ⁵P_r= ⁶P_r-1

$\frac{5!}{\left ( 5-r \right )!}=\frac{6!}{\left ( 7-r \right )!}$

$\frac{5!}{\left ( 5-r \right )!}=\frac{6\times 5!}{\left ( 7-r \right )\left ( 6-r \right )\left ( 5-r \right )!}$

$\frac{6}{\left ( 7-r \right )\left ( 6-r \right )}=1$

(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_r,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

$\frac{8!}{\left ( 8-8 \right )!}=\frac{8!}{0!}=8!=8\times 7\times 6\times 5\times 4\times 3\times 2=40320$

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

$\frac{6!}{\left ( 6-6 \right )!}=\frac{6!}{0!}=\frac{6\times 5\times 4\times 3\times \times 2\times 1}{1}=6\times 5\times 4\times 3\times 2=720$

(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

= ²P₁= 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

$\frac{11!}{4!4!2!}=\frac{11\times 10\times 9\times 8\times 7\times 6\times 5\times 4\times 3\times 2}{4\times 3\times 2\times 4\times 3\times 2\times 2}=34650$

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

$=\frac{8!}{4!2!}=\frac{8\times 7\times 6\times 5\times 4\times 3\times 2 }{4\times 3\times 2\times 2}=840$

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

(i) words start with P and end with S

(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

$=\frac{10!}{2!}=\frac{10\times 9\times 7\times 6\times 5\times 4\times 3\times 2 }{2}=1814400$

(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

$=\frac{8!}{2!}=\frac{8\times 7\times 6\times 5\times 4\times 3\times 2 }{2}=20160$

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.

$P- - --S------$

$- P----S-----$

$- -P----S----$

$- --P----S---$

$- ---P----S--$

$- ----P----S-$

$- -----P----S$

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

$^{n}P_{r}=\frac{n!}{\left ( n-r \right )!}$

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

$=\frac{n!}{t!m!}$

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 9 maths

Chapter 1- Number System	Chapter 9-Areas of parallelogram and triangles
Chapter 2-Polynomial	Chapter 10-Circles
Chapter 3- Coordinate Geometry	Chapter 11-Construction
Chapter 4- Linear equations in two variables	Chapter 12-Heron’s Formula
Chapter 5- Introduction to Euclid’s Geometry	Chapter 13-Surface Areas and Volumes
Chapter 6-Lines and Angles	Chapter 14-Statistics
Chapter 7-Triangles	Chapter 15-Probability
Chapter 8- Quadrilateral

NCERT Solutions for class 9 science

Chapter 1-Matter in our surroundings	Chapter 9- Force and laws of motion
Chapter 2-Is matter around us pure?	Chapter 10- Gravitation
Chapter3- Atoms and Molecules	Chapter 11- Work and Energy
Chapter 4-Structure of the Atom	Chapter 12- Sound
Chapter 5-Fundamental unit of life	Chapter 13-Why do we fall ill ?
Chapter 6- Tissues	Chapter 14- Natural Resources
Chapter 7- Diversity in living organism	Chapter 15-Improvement in food resources
Chapter 8- Motion	Last years question papers & sample papers

NCERT Solutions for class 10 maths

Chapter 1-Real number	Chapter 9-Some application of Trigonometry
Chapter 2-Polynomial	Chapter 10-Circles
Chapter 3-Linear equations	Chapter 11- Construction
Chapter 4- Quadratic equations	Chapter 12-Area related to circle
Chapter 5-Arithmetic Progression	Chapter 13-Surface areas and Volume
Chapter 6-Triangle	Chapter 14-Statistics
Chapter 7- Co-ordinate geometry	Chapter 15-Probability
Chapter 8-Trigonometry

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 10 Science

Chapter 1- Chemical reactions and equations	Chapter 9- Heredity and Evolution
Chapter 2- Acid, Base and Salt	Chapter 10- Light reflection and refraction
Chapter 3- Metals and Non-Metals	Chapter 11- Human eye and colorful world
Chapter 4- Carbon and its Compounds	Chapter 12- Electricity
Chapter 5-Periodic classification of elements	Chapter 13-Magnetic effect of electric current
Chapter 6- Life Process	Chapter 14-Sources of Energy
Chapter 7-Control and Coordination	Chapter 15-Environment
Chapter 8- How do organisms reproduce?	Chapter 16-Management of Natural Resources

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