NCERT Solutions for Miscellaneous Exercise Chapter 7 Class 11 - Permutations and Combinations

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

Edited By Vishal kumar | Updated on Nov 16, 2023 02:13 PM IST

## NCERT Solutions for Class 11 Maths Chapter 7 Permutations and Combinations Miscellaneous Exercise- Download Free PDF

In this chapter, you have already learned about the fundamental principles of counting, factorial notation, permutaions, combinations, etc. In this article, you will get NCERT solutions for Class 11 Maths chapter 7 miscellaneous exercise. As the name suggests miscellaneous exercises contain a mixture of questions from all the topics of this chapter. If you have done all the previous exercises of this chapter, you can solve miscellaneous exercise chapter 7 Class 11. The miscellaneous exercise is considered to be a bit difficult as compared to other exercises of this chapter.

You can go through the Class 11 Maths chapter 7 miscellaneous exercise solutions, so it will be easy for you to solve the NCERT exercise problems. The miscellaneous exercises questions are more related to the real-life problems of permutations and combinations so you will get to know about the practical applications of permutations and combinations. These class 11 chapter 7 maths miscellaneous solutions are meticulously crafted by subject experts at Careers360. They are presented in an easily understandable manner, accompanied by detailed explanations. The availability of PDF versions further enhances accessibility, allowing students the flexibility to complete tasks at their convenience. These class 11 maths ch 7 miscellaneous exercise solutions resources are designed to assist students in comprehensively understanding and mastering the concepts, contributing to effective exam preparation. If you are looking for NCERT Solutions, you can click on the given link.

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**In the CBSE Syllabus for 2023-24, this miscellaneous exercise class 11 chapter 7 has been renumbered as Chapter 6.

## Access Permutation And Combinations Class 11 Chapter 7- Miscellaneous Exercise

In the word DAUGHTER, we have

vowels = 3(A,E,U)

consonants = 5(D,G,H,T,R)

Number of ways of selecting 2 vowels $=^3C_2$

Number of ways of selecting 3 consonants $=^5C_3$

Therefore, the number of ways of selecting 2 vowels and 3 consonants $=^3C_2 .^5C_3$

$=3\times 10=30$

Each of these 30 combinations of 2 vowels and 3 consonants can be arranged in $5!$ ways.

Thus, the required number of different words $= 5!\times 30=120\times 30=3600$

In the word EQUATION, we have

vowels = 5(A,E,I,O,U)

consonants = 3(Q,T,N)

Since all the vowels and consonants occur together so (AEIOU) and (QTN) can be assumed as single objects.

Then, permutations of these two objects taken at a time $=^2P_2=2!=2$

Corresponding to each of these permutations, there are $5!$ permutations for vowels and $3!$ permutations for consonants.

Thus, by multiplication principle, required the number of different words $= 2\times 5!\times 3!=2\times 120\times 6=1440$

exactly 3 girls?

There are 9 boys and 4 girls. A committee of 7 has to be formed.

Given : Girls =3, so boys in committee= 7-3=4

Thus, the required number of ways $=^4C_3.^9C_4$

$=\frac{4!}{3!1!}\times \frac{9!}{4!5!}$

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

$= 9\times 8\times 7=504$

at least 3 girls?

There are 9 boys and 4 girls. A committee of 7 has to be formed.

(ii) at least 3 girls, there can be two cases :

(a) Girls =3, so boys in committee= 7-3=4

Thus, the required number of ways $=^4C_3.^9C_4$

$=\frac{4!}{3!1!}\times \frac{9!}{4!5!}$

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

$= 9\times 8\times 7=504$

(b) Girls =4, so boys in committee= 7-4=3

Thus, the required number of ways $=^4C_4.^9C_3$

$=\frac{4!}{4!0!}\times \frac{9!}{3!6!}$

$= \frac{9\times 8\times 7\times 6!}{ 3\times 2\times 6!}$

$=84$

Hence, in this case, the number of ways = 504+84=588

atmost 3 girls?

There are 9 boys and 4 girls. A committee of 7 has to be formed.

(ii) atmost 3 girls, there can be 4 cases :

(a) Girls =0, so boys in committee= 7-0=7

Thus, the required number of ways $=^4C_0.^9C_7$

$=\frac{4!}{4!0!}\times \frac{9!}{2!7!}$

$=9\times 4=36$

(b) Girls =1, so boys in committee= 7-1=6

Thus, the required number of ways $=^4C_1.^9C_6$

$=\frac{4!}{3!1!}\times \frac{9!}{3!6!}$

$=336$

(c) Girls =2, so boys in committee= 7-2=5

Thus, the required number of ways $=^4C_2.^9C_5$

$=\frac{4!}{2!2!}\times \frac{9!}{4!5!}$

$=756$

(d) Girls =3, so boys in committee= 7-3=4

Thus, the required number of ways $=^4C_3.^9C_4$

$=\frac{4!}{3!1!}\times \frac{9!}{4!5!}$

$=504$

Hence, in this case, the number of ways = 36+336+756+504=1632

In the word EXAMINATION, we have 11 letters out of which A,I, N appear twice and all other letters appear once.

The word that will be listed before the first word starting with E will be words starting with A.

Therefore, to get the number of words starting with A, letter A is fixed at extreme left position, the remaining 10 letters can be arranged.

Since there are 2 I's and 2 N's in the remaining 10 letters.

Number of words starting with A $=\frac{10!}{2!2!}=907200$

Thus, the required number of different words = 907200

For the number to be divisible by 10, unit digit should be 0.

Thus, 0 is fixed at a unit place.

Therefore, the remaining 5 places should be filled with 1,3,5,7,9.

The remaining 5 vacant places can be filled in $5!$ ways.

Hence, the required number of 6 digit numbers which are divisible by 10$=5!=120$

Two different vowels and 2 different consonants are to be selected from the English alphabets.

Since there are 5 different vowels so the number of ways of selecting two different vowels = $^5C_2$

$=\frac{5!}{2!3!}=10$

Since there are 21 different consonants so the number of ways of selecting two different consonants = $^2^1C_2$

$=\frac{21!}{2!19!}=210$

Therefore, the number of combinations of 2 vowels and 2 consonants $=10\times 210=2100$

Each of these 2100 combinations has 4 letters and these 4 letters arrange among themselves in $4!$ ways.

Hence, the required number of words $=210\times 4!=50400$

It is given that a question paper consists of 12 questions divided in two parts i.e., Part I and Part II, containing 5 and 7 questions, respectively.

A student is required to attempt 8 questions in all, selecting at least 3 from each part.

This can be done as follows:

(i) 3 questions from part I and 5 questions from part II

(ii) 4 questions from part I and 4 questions from part II

(iii) 5 questions from part I and 3 questions from part II

3 questions from part I and 5 questions from part II can be selected in $^5C_3.^7C_5$ ways.

4 questions from part I and 4 questions from part II can be selected in $^5C_4.^7C_4$ ways.

5 questions from part I and 3 questions from part II can be selected in $^5C_5.^7C_3$ ways.

Hence, required number of ways of selecting questions :

$=^5C_3.^7C_5+^5C_4.^7C_4+^5C_5.^7C_3$

$=\frac{5!}{2!3!}\times \frac{7}{2!5!}+\frac{5!}{1!4!}\times \frac{7}{4!3!}+\frac{5!}{5!0!}\times \frac{7}{3!4!}$

$=210+175+35$

$=420$

From a deck of 52 cards, 5 cards combinations have to be made in such a way that in each selection of 5 cards there is exactly 1 king.

Number of kings =4

Number of ways of selecting 1 king $=^4C_1$

4 cards from the remaining 48 cards are selected in $^4^8C_4$ ways.

Thus, the required number of 5 card combinations $=^4C_1 .^4^8C_4$

It is required to seat 5 men and 4 women in a row so that the women occupy the even places.
The 5 men can be seated in $5!$ ways.

4 women can be seated at cross marked places (so that women occupy even places)

Therefore, women can be seated in $4!$ ways.

Thus, the possible arrangements $=5!\times 4!=120\times 24=2880$

From a class of 25 students, 10 are to be chosen for an excursion party.

There are 3 students who decide that either all of them will join or none of them will join, there are two cases :

The case I: All 3 off them join.

Then, the remaining 7 students can be chosen from 22 students in $^2^2C_7$ ways.

Case II : All 3 of them do not join.

Then,10 students can be chosen from 22 students in $^2^2C_1_0$ ways.

Thus, the required number of ways for the excursion of party $=^2^2C_7+^2^2C_1_0$

In the word ASSASSINATION, we have

number of S =4

number of A =3

number of I= 2

number of N =2

Rest of letters appear at once.

Since all words have to be arranged in such a way that all the S are together so we can assume SSSS as an object.

The single object SSSS with other 9 objects is counted as 10.

These 10 objects can be arranged in (we have 3 A's,2 I's,2 N's)

$=\frac{10!}{3!2!2!}$ ways.

Hence, requires the number of ways of arranging letters

$=\frac{10!}{3!2!2!}=151200$

## More About NCERT Solutions for Class 11 Maths Chapter 7 Miscellaneous Exercise:-

Miscellaneous exercise chapter 7 Class 11 consists of questions related to practical applications of permutations and combinations. There are four solved examples related to real-life examples of counting given before the of and miscellaneous exercise chapter 7 Class 11. You can go through these solved examples in order to get conceptual clarity. You must try to solve the 11 questions given in this exercise after completing the previous exercises of this chapter.

## Topic Based on Class 11 Maths ch 7 Miscellaneous Exercise Solutions

The Miscellaneous Exercise in NCERT Solutions for Class 11 Maths Chapter 7 - Permutations and Combinations covers the following topics:

1. Introduction to Permutations and Combinations
2. Fundamental Principle of Counting
JEE Main Highest Scoring Chapters & Topics
Just Study 40% Syllabus and Score upto 100%

Proficiency in solving problems using proper methods, as outlined in these NCERT Class 11 Maths solutions, assists students in analyzing various question types that may appear in board examinations. Mastering these solutions enhances problem-solving speed, ultimately boosting students' self-confidence.

## Benefits of NCERT Solutions for Class 11 Maths Chapter 7 Miscellaneous Exercise:-

• Miscellaneous exercises are not considered to be important for the CBSE exams but these exercises are very important for engineering competitive exams.
• If you are not able to solve miscellaneous exercise problems by yourself, Class 11 Maths chapter 7 miscellaneous exercise solutions are here to help you.
• Class 11 Maths chapter 7 miscellaneous solutions could be used for help in the homework for the students.

## Key Features of Class 11 Chapter 7 Maths Miscellaneous Solutions

• Comprehensive Coverage: The miscellaneous exercise class 11 chapter 7 Solutions offer a comprehensive explanation of the diverse problems presented in the chapter, ensuring a thorough understanding of permutations and combinations.
• Step-by-Step Solutions: Each problem in the class 11 chapter 7 maths miscellaneous solutions is meticulously solved with a step-by-step approach, guiding students through the logical progression of concepts and problem-solving techniques.
• Clear and Concise Language: The class 11 maths miscellaneous exercise chapter 7 solutions are articulated in clear and concise language, making intricate mathematical concepts related to permutations and combinations more accessible to students.
• Board Examination Relevance: The class 11 maths ch 7 miscellaneous exercise solutions are designed with relevance to the expected question types in board examinations, aiding students in effective exam preparation.
• Free PDF: The class 11 chapter 7 miscellaneous exercise solutions are provided in a downloadable PDF format, allowing students the flexibility to access and study the material conveniently at their own pace.

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## Subject Wise NCERT Exampler Solutions

Happy learning!!!

1. Evaluate n!

n! = 1 x 2 x 3 x 4 ........ x n

2. Does 4! + 5! = 9! ?

4! = 1 x 2 x 3 x 4 = 24

5! = 1 x 2 x 3 x 4 x 5 = 120

9! = 1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 x 9 = 362880

Clearly 4! + 5! is not equal to 9!

3. What is the weightage of Statistics and Probability in CBSE Class 11 Maths ?

Statistics and Probability have 10 marks weightage in the CBSE Class 11 Maths final exams.

4. What is weightage of Straight Lines in the JEE Main Maths ?

Generally, two questions from Straight Lines are asked in the JEE Main Maths. It has 6.6 % weightage in the JEE Main Maths paper.

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