##### VMC VIQ Scholarship Test

ApplyRegister for Vidyamandir Intellect Quest. Get Scholarship and Cash Rewards.

Edited By Ravindra Pindel | Updated on Sep 12, 2022 05:54 PM IST

**NCERT Exemplar Class 11 Maths Chapter 7** solutions covers Permutations and Combinations. Permutations are explained as arrangements of objects, all taken at the same time or at different times, but in a definite order. Combination refers to the pairing of two or more objects with each other to know its suitability and results. NCERT Exemplar Class 11 Maths Solutions Chapter 7 cover arrangement and pairing over objects for the derivation of answers. Along with these two concepts, they have certain denotations that help in making up of formulas and understanding the way of formula derivation.

**JEE Main Scholarship Test Kit (Class 11): Narayana | Physics Wallah | **

**Suggested: ****JEE Main: high scoring chapters | ****Past 10 year's papers**

**Scholarship Test:** **Vidyamandir Intellect Quest (VIQ)**

This Story also Contains

- NCERT Exemplar Class 11 Maths Solutions Chapter 7: Exercise - 1.3
- Important Notes From NCERT Exemplar Class 11 Maths Solutions Chapter 7 Permutations and Combinations
- Main Subtopics in NCERT Exemplar Class 11 Maths Chapter 7 Solutions
- What will the students learn from NCERT Exemplar Class 11 Maths Chapter 7 Solutions?
- NCERT Solutions for Class 11 Mathematics Chapters
- Important Topics in NCERT Exemplar Class 11 Maths Chapter 7 Solutions Permutations and Combinations

Question:1

Answer:

Let the women choose the chairs from amongst the chairs numbered 1 to 4.

Two women can be arranged in 4 chairs in ways

In remaining 6 chairs, 3 men can be arranged in ways

Total possible arrangements=

Question:2

Answer:

Alphabetical order of letters of word RACHIT is A, C, H, I, R,T .

Number of words starting with A=

Number of words starting with C=

Number of words starting with H=

Number of words starting with I=

For R, the first word is RACHIT

So, rank of words=

Question:3

Answer:

Given :Candidates acn not attempt more than 5 questions from either group and can only attempt

a minimum of 2 questions from either group.

Possible number of questions attempted from each group are :

a) 5 from Group I and 2 from Group II

b) 4 from Group I and 3 from Group II

c) 3 from Group I and 4 from Group II

d) 2 from Group I and 5 from Group II

Total ways=

Question:4

Answer:

Number of straight lines formed by joining 18 points taking 2 at a time=

Number of straight lines formed by joining 5 collinear points taking 2 at a time=

5 collinear points when joined pairwise give only one line.

Required number of straight lines=

Question:5

Answer:

Number of persons to be chosen out of 8=6

CASE I:When A is chosen, B must be chosen

Number of ways of selecting 4 more persons from remaining 6 persons=

CASE II:When A is not chosen

Number of ways of selecting 6 persons from remaining 7 persons=

Total ways=

Question:6

Answer:

Out of 12 persons, a chairperson can be selected in =12 ways.

The remaining 4 persons can be selected out of 11 persons in =330 ways

Total number of committe=

Question:7

Answer:

Each plate contains 2 different letters followed by 3 different digits

Arrangement of 26 letters taken 2 at a time=

Arrangementof 10 digits taken 3 at a time=

Total number of license plates=

Question:8

Answer:

2 black balls can be selected from 5 black balls in ways

3 red balls can be selected from 6 red balls in ways

Total number of ways=

Question:9

Answer:

Combination of n things taken r at a time in which 3 things always occur together=

3 things taken together is considered 1 group Number of arrangement of 3 things=

Now, number of arrangement of r-2 objects=

Total number of arrangements=

Question:10

Answer:

Consonants in TRIANGLE are T, R, N, G, L . Vowels are I, A, E

Five consonants can be arranged in 5! ways XCXCXCXCXCX

Arrangements of consonants as C above creates 6 gaps marked as X

Now, 3 vowels can be arranged in any three of these 6 gaps in ways

So, total number of arrangements=

Question:11

Answer:

A number is divisible by 5, if at the unit place of the number there is 0 or 5. So, unit digit can be filled in 2 ways. Since, we have to form 4-digit numbers greater than 6000 and less than 7000. ‘6’ can fill the thousandth place only.

The hundredth and tenth place can be filled together in 8*7 =56 ways. So, total number of ways = 56*2 =112

Question:12

Answer:

Given that

persons out of which 5 are to be arranged but must occur

whereas P_{4}and P_{5} never occur. So, now we only have to select 4 out of 7 persons

Number of selections=

Number of the arrangement of 5 persons=

Question:13

Answer:

The hall can be illuminated if atleast one lamp is switched out of 10 lamps Total number of ways

Question:14

Answer:

Out of 2 white, 3 black and 4 red balls we have to draw 3 balls such that atleast one black is included.

The possibilities are:

a.1 black ball and 2 others b.2 black balls and 1 other (c)3 black balls

Number of selections =

=64

Question:15

If , and , then find .

[Hint: From equation using to find the value of r.]

Answer:

Equation (i) + 2 Equation(ii)

10r - 3n +4n - 10r = 3 + 6

n = 9

r =3

Question:16

Answer:

Since, repetition is not allowed a maximum of 5- digit numbers can be formed with digits 3, 5,7, 8 and 9.

Since, all the 5-digit numbers are greater than 7000 the number of 5-digitintegers=

A 4-digit number is greater than 7000 if thousandth place has any of the 7,8 and 9. The remaining 3 places can be filled from remaining 4 digits in *ways*

So, total number of 4-digit integers =

Total number of integers=

Question:17

Answer:

If no two lines are parallel, all lines are intersecting, and no three lines are concurrent.

Two straight lines create one point of intersection

Number of points of intersection = Number of combinations of 20 straight lines taken two at a time

Question:18

Answer:

If first two digit is 41, the remaining 4 digits can be arranged in ways=

In addition, if first two digits are 42, 46, 62 or 64 the remaining 4 can be arranged in ways= ways.

So, total number of telephone numbers having all 6 digits different =

Question:19

Answer:

Since, 2 questions are compulsory, the other 2 can be selected from the remaining 3 questions in ways = 3 ways

Question:20

Answer:

Let the convex polygon have n sides.

Number of diagonals = number of ways of selecting two vertices- number of sides

Question:21

Answer:

Since, 18 mice were placed equally in three groups. Each group had 6 mice

Number of ways of forming 3 groups each with 6 mice=

But this includes the order in which the three groups are formed i.e. 3! which should not be

counted.

So, actual number of ways=

Question:22

Answer:

Total number of marbles=

a.If they can be of any color we have to select 4 out of 11 marbles=

b.2 white marbles can be selected in ways and 2 red marbles in .

Total number of ways=

c. If they all are of same color 4 white out of 6 marbles can be selected in ways and

4 red out of 5 can be selected in ways

Required number of ways=

Question:23

Answer:

Number of ways to select a team of 11 players out of 16=

a. If two particular players are include then 9 more can be selected from remaining 14 players=

b. If two particular players are excluded then all 11 players can be selected from remaining 14

players in

Question:24

Answer:

We have to select atleast 5 students from each class out of 20 students in each class . So, we

can either select 5 students from class XI and 6 from class XII or vice versa

Total number of ways=

Question:25

Answer:

i.Team having no girls=

ii.Team having atleast one boy and one girl=

iii.Team having atleast 3 girls =

Question:27

The number of possible outcomes when a coin is tossed 6 times is

A. 36

B. 64

C. 12

D. 32

Answer:

The answer is the option (b) number of outcomes when a coin is tossed= 2 (head or tail)

Total possible outcomes when a coin tossed 6 times =

Question:28

The number of different four digit numbers that can be formed with the digits 2, 3, 4, 7 and using each digit only once is

A. 120

B. 96

C. 24

D. 100

Answer:

The answer is the option (c)

We have to form 4-digit numbers using 2,3,4 and 7.

Required number of ways =

Question:29

The sum of the digits in unit place of all the numbers formed with the help of 3, 4, 5 and 6 taken all at a time is

A. 432

B. 108

C. 36

D. 18

Answer:

The answer is the option (b)

If we fix 3 at the unit place, the remaining places can be filled in 3! Ways.

Thus ‘3’ appears in unit place in 3! Times.

Similarly for each of the digits 4,5 and 6. So, sum of digits in unit place =

Question:30

Total number of words formed by 2 vowels and 3 consonants taken from 4 vowels and 5 consonants is equal to

A. 60

B. 120

C. 7200

D. 720

Answer:

The answer is the option (c)

With 4 vowels and 5 consonants we have to form words with 2 vowels and 3 consonants.

Total number of ways =

Question:31

A five digit number divisible by 3 is to be formed using the numbers 0, 1, 2, 3, 4 and 5 without repetitions. The total number of ways this can be done is

A. 216

B. 600

C. 240

D. 3125

[Hint: 5 digit numbers can be formed using digits 0, 1, 2, 4, 5 or by using digits 1, 2, 3, 4, 5 since sum of digits in these cases is divisible by 3.]

Answer:

The answer is the option (a).

A number is divisible by 3 if the sum of its digit is divisible by 3. So, to form a number of 5-digit which is divisible by 3 we can remove either 0 or 3. If digits 1,2, 3, 4, 5 are used then the number of required numbers = 5!

If digits 0,1,2,4,5 are used then first place from left can be filled in 4 ways and remaining 4 places can be filled in 4! Ways. So, 4*4! Ways.

Total numbers = 120+96 =216

Question:32

Everybody in a room shakes hands with everybody else. The total number of handshakes is 66. The total number of persons in the room is

A. 11

B. 12

C. 13

D. 14

Answer:

The answer is the option *b.*

Question:33

The number of triangles that are formed by choosing the vertices from a set of 12 points, seven of which lie on the same line is

A. 105

B. 15

C. 175

D. 185

Answer:

The answer is the option (d).

Number of ways of selecting 3 points from given 12 points=

But 3 points from given 7 collinear points does not form triangle.

Number of ways of selecting 3 points from seven collinear points=

Total number of triangles=

Question:34

The number of parallelograms that can be formed from a set of four parallel lines intersecting another set of three parallel lines is

A. 6

B. 18

C. 12

D. 9

Answer:

The answer is the option (b).

We need to select a pair of line from a set of 4 lines and another pair from another set of 3 lines to form a parallelogram

Required number =

Question:35

Answer:

The answer is the option (c).

We have to select 11 players out of 22.

After excluding particular 4 of them, 18 are available. Amongst these 18, 2 particular players are always included. Thus, we have to select 9 more players from the remaining 16 players in ways.

Question:36

The number of 5-digit telephone numbers having at least one of their digits repeated is

A. 90,000

B. 10,000

C. 30,240

D. 69,760

Answer:

The answer is the option (d).

Total number of telephone numbers when there is no restriction =

Number of telephone numbers with all digits different =

Required number of ways=

Question:37

The number of ways in which we can choose a committee from four men and six women so that the committee includes at least two men and exactly twice as many women as men is

A. 94

B. 126

C. 128

D. None

Answer:

The answer is the option (a).

To satisfy the condition mentioned in the question we can either select 2 men and 4 women or 3 men and 6 women =

Question:38

The total number of 9-digit numbers which have all different digits is

A. 10!

B. 9!

C. 9 × 9!

D. 10×10!

Answer:

The answer is the option (c).

We have to form 9-digit number, which has all different digits.

First digit from the left can be filled in 9 ways excluding 0. After this, nine digits are left including ‘0’. Remaining 8 places can be filled with these nine digits in ways. So, total numbers=

Question:39

The number of words which can be formed out of the letters of the word ARTICLE, so that vowels occupy the even place is

A. 1440

B. 144

C. 7!

D.

Answer:

The answer is the option (b)

Vowels in ARTICLE are A,I,E and consonants are R,T,C,L . Vowels occupy 3 even places in 3! Ways and in remaining 4 places 4 consonants can be arranged in 4! Ways

Total number of words =

Question:40

Given 5 different green dyes, four different blue dyes and three different red dyes, the number of combinations of dyes which can be chosen taking at least one green and one blue dye is

A. 3600

B. 3720

C. 3800

D. 3600

[Hint: Possible numbers of choosing or not choosing 5 green dyes, 4 blue dyes and 3 red dyes are , respectively.]

Answer:

The answer is the option (b).

At least one green dye can be chosen in ways

At least one blue dye can be chosen in ways

Any number of red dyes can be chosen in ways

So, total number of selection=

Question:44

Answer:

. Vowels in INTERMEDIATE are I, E, E, I, A, E and consonants N, T, R, M, D, T

First six consonants are arranged in ways.

In the 7 gaps, six vowels are arranged in ways

Total number of words

Question:45

Answer:

Possible selection can be either '2red and 1 other' or ' 3 red'.

Required number of ways=

Question:46

Fill in the Blanks

The number of six-digit numbers, all digits of which are odd is ______.

Answer:

Odd digits are 1, 3, 5, 7, 9

The required number of numbers

Question:47

Answer:

. Let the number of teams participating in the championship be n.

It is given that every 2 teams played one match with another.

Total match=

Question:48

Answer:

. Number of ways of arrangement of six+signs is 1.

In the 7 gaps created, 4 are to be chosen for '-' signs in ways.

These signs can only be arranged in 1 way as they are identical.

Total number of ways ways

Question:49

Answer:

*Number of ways of forming a committee of 6 persons containing atleast 3 men and 2 women*

* Total number of committee when two particular women are never together*

*=Total number of ways-Number of ways when 2 particular women are together*

*= *

*= *

Question:50

Required number of ways such that 3 balls are drawn and atleast one black ball is included

in the draw=

=

Question:51

Answer:

False. Required number of lines=

Question:52

Answer:

False.

Total number of ways of posting 3 letters =

Question:53

Answer:

False. In arrangement of n things taken r at a time in which m occur together, first we

select objects from objects in ways.

If we consider these m things as 1 group number of objects excluding these m objects=

Now, we have to arrange objects

Number of arrangements=

Also, m objects which are considered as 1 group can be arranged in m!ways

Required number of arrangements=

Question:54

Answer:

True. In each stall any oneof the three animals can be shipped

So, total number of ways of loading

Question:55

Answer:

True.

If some or all objects taken at a time, then the number of combinations would be

Question:56

Answer:

False.

Number of ways of selecting any number of objects from n given identical objects is 1. Now, selecting 0 or more red ball from 4 identical red balls=

Selecting at least 1 black ball from 5 identical black balls =

So, total ways=

Question:57

Answer:

True

Let the two sides of table be A and B with 9 seats each.

Let on side A 4 particular guests and on side B 3 particular guests be seated.

Now, for side A, 5 more guests can be selected can be selected from remaining 11 guests in

ways.

Also, on each side nine guests can be arranged in 9!ways.

So, total number of ways of arrangement=

Question:58

Answer:

False.

A candidate can attempt questions in either of the following ways :

a.2 from group A and 5 from group B

b.3 from group A and 4 from group B

c.4 from group A and 3 from group B

d.5 from group A and 2 from group B

Number of ways of attempting 7 questions

Question:59

Answer:

True. We can select 3 scheduled caste candidate out of 5 in ways.

And we can select 9 other candidates out of 22 in ways.

Total number of selections=

Question:60

Answer:

Number of book of mathematics=3

Number of book of physics=4

Number of book of English=5

a.One book of each subject=

b.At least one book of each subject=

c. Adding one book of English=

So, a-(ii), b-(iii), c-(i)

Question:61

Answer:

Given that number of boys=5 and number of girls=5

a.Boys and girls alternate=

b.No two girls sit together=

c.All the girls sit together=

d .All the girls are never together :

Total number of boys and girls=

Number of ways=

Question:62

a)In how many ways committee:can be formed | i) |

b)In how many ways a particular: professor is included | ii) |

c)In how many ways a particular: lecturer is included | iii) |

d)n how many ways a particular: lecturer is excluded | iv) |

Answer:

Given that number of professors=10and number of lecturers=20

a.In how many ways committee can be formed=

b.In how many ways a particular professor is included=

c.In how many ways a aprticular lecturer is included=

d.In how many ways a particular lecturer is excluded=

So a-iv, b-iii, c-ii, d-i

Question:63

Using the digits 1, 2, 3, 4, 5, 6, 7, a number of 4 different digits is formed. Find

Answer:

Given :Total number of digit to be formed=4

a. How many numbers are formed=

b. How many numbers are exactly divisible by 2?

Number ending with 2=

Number ending with 4 =

Numbers ending with 6 =

So, total numbers=

c. Numbers exactly divisible by 25=40

d. The numbers which have last 2 digit divisible by 4 are 12, 16 , 24 , 32, 36 , 44 ,52, 56, 64, 72, 76 So =

Question:64

Answer:

Given that Total number of letter of MONDAY=6

Total vowel=2

a. When 4 letters are used at a time=

b. All the letters are used at a time =

c. All letters are used but the first is a vowel=

The letter ‘n’ is important when it comes to formulas, where any number to be found said to be it. It is the first natural number and the product procured by ‘n’ is denoted as ‘n!’, known as factorial notations.’ Through NCERT Exemplar Class 11 Maths Chapter 7 solutions, students will learn about the formula derivation, along with the letter used and its role.

Combinations could be made in any manner without any restrictions, all answers will be correct. On the other hand, in permutations case, they need to be arranged properly and in a correct order to get definite results. All of this is explained in the NCERT Exemplar solutions for Class 11 Maths chapter 7. By using NCERT Exemplar Class 11 Maths Chapter 7 solutions PDF Download, students can make learning even more convenient. Here, this guide would be easy for students to solve their issues and easily understand this topic.

- Introduction
- Fundamental Principle of Counting
- Permutations
- Permutations when all the objects are distinct
- Factorial notation
- Derivation of the formula for nPr
- Permutations when all the objects are not distinct
- Combinations

NCERT Exemplar Class 11 Chapter 7 solutions deals with a subcategory of probability. Students here will know about the difference between permutations and combinations, which has a fine line of separation. Not to be confused, this solution will help students identify the arrangements and pairing portions.

Whenever probability will come in picture, whether basic or advanced both of these matters come in their role-play and play their part. Resolving issues using these formulae will be helpful.

As per NCERT Exemplar Class 11 Maths Chapter 7 solutions, Questions on compound events could be easily solved using these techniques. In combination and permutation, the issue is resolved by knowing that one factor, about arranging the objects correctly in the right order is main, and the pairing could be done in any manner.

In NCERT Exemplar Class 11 Maths Solutions Chapter 7 the students will learn about probability solving questions, along with the number of combinations that could be made from it, that could be very useful for future uses. Easy examples help kids do it better and cleverly.

· Class 11 Maths NCERT Exemplar solutions Chapter 7 has detailed that Principles of counting, permutations, its formula and other subtopics and combinations are important topics which students should pay extra attention to.

· As per NCERT Exemplar Class 11 Maths solutions Chapter 7, Permutations and Combinations are said to be the ‘fundamentals principle of counting’- leading to arranging and combining objects. It is about how the students will understand the different situations occurring at different periods, and what could be the possible answers derived to those questions.

**Check Chapter-Wise NCERT Solutions of Book**

Chapter-1 | |

Chapter-2 | |

Chapter-3 | |

Chapter-4 | |

Chapter-5 | |

Chapter-6 | |

Chapter-7 | |

Chapter-8 | |

Chapter-9 | |

Chapter-10 | |

Chapter-11 | |

Chapter-12 | |

Chapter-13 | |

Chapter-14 | |

Chapter-15 | |

Chapter-16 |

**Read more NCERT Solution subject wise -**

- NCERT Solutions for Class 11 Maths
- NCERT Solutions for Class 11 Physics
- NCERT Solutions for Class 11 Chemistry
- NCERT Solutions for Class 11 Biology

**Also, read NCERT Notes subject wise -**

- NCERT Notes for Class 11 Maths
- NCERT Notes for Class 11 Physics
- NCERT Notes for Class 11 Chemistry
- NCERT Notes for Class 11 Biology

JEE Main Highest Scoring Chapters & Topics

Just Study 40% Syllabus and Score upto 100%

Download EBook**Also Check NCERT Books and NCERT Syllabus here:**

1. Is this chapter crucial for higher education?

Yes, permutations and combinations are one of the most important chapters of maths that will create a base for board exams and entrance exams.

2. How many questions are solved in these solutions?

In total, 42 questions are solved from four exercises and miscellaneous exercises from the NCERT chapter in NCERT Exemplar Class 11 Maths Chapter 7 Solutions.

3. Are these solutions as per the marking scheme?

Yes, these solutions are designed stepwise as per the CBSE pattern so that one can see how the marks are divided among each step.

4. Are these solutions accessible offline?

Yes, these NCERT Exemplar Class 11 Maths Solutions Chapter 7 are downloadable in PDF format so that one can refer to solutions while solving without being online.

Sep 17, 2024

Application Date:05 September,2024 - 20 September,2024

Admit Card Date:13 September,2024 - 07 October,2024

Admit Card Date:13 September,2024 - 07 October,2024

Application Date:17 September,2024 - 30 September,2024

Get answers from students and experts

Register for Vidyamandir Intellect Quest. Get Scholarship and Cash Rewards.

Register for Tallentex '25 - One of The Biggest Talent Encouragement Exam

As per latest 2024 syllabus. Physics formulas, equations, & laws of class 11 & 12th chapters

As per latest 2024 syllabus. Chemistry formulas, equations, & laws of class 11 & 12th chapters

Accepted by more than 11,000 universities in over 150 countries worldwide

Register now for PTE & Unlock 10% OFF : Use promo code: 'C360SPL10'. Limited Period Offer! Trusted by 3,500+ universities globally

News and Notifications

Back to top