NCERT Solutions for Miscellaneous Exercise Chapter 6 Class 11 - Permutations and Combinations

NCERT Solutions for Miscellaneous Exercise Chapter 6 Class 11 - Permutations and Combinations

Komal MiglaniUpdated on 06 May 2025, 02:29 PM IST

In our daily lives we often need to assess the possible number of arrangements in various situations related to the seating arrangement of people, arranging digits or making an arrangement of letters and numbers. When the sample size is small, it is easy to arrange by the conventional method, but when the sample is of large numbers, we need better advanced accounting techniques, which we study under the topic named permutations and combinations.

Miscellaneous Exercise of Chapter 6 in the NCERT discusses the questions related to the important principle of counting, along with the practical use of permutations and combinations in daily life. The exercise also involves solving questions such as arranging letters or digits, selecting teams, and solving real-life problems involving arrangements and selections. The NCERT solutions discuss step-by-step methodology to apply these techniques effectively.

This Story also Contains

  1. Class 11 Maths Chapter 1 Permutations and Combinations Miscellaneous Exercise Solutions - Download PDF
  2. NCERT Solutions Class 11 Maths Chapter 6 Miscellaneous Exercise
  3. Topics covered in Chapter 6 Permutations and Combinations Miscellaneous Exercise
  4. NCERT Solutions of Class 11 Subject Wise
  5. Subject-Wise NCERT Exemplar Solutions


Class 11 Maths Chapter 1 Permutations and Combinations Miscellaneous Exercise Solutions - Download PDF

Download PDF

NCERT Solutions Class 11 Maths Chapter 6 Miscellaneous Exercise

Question:1 How many words, with or without meaning, each of 2 vowels and 3 consonants can be formed from the letters of the word DAUGHTER?

Answer:

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 $= 5!\times 30=120\times 30=3600$

Question:2 How many words, with or without meaning, can be formed using all the letters of the word EQUATION at a time so that the vowels and consonants occur together?

Answer:

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 $= 2\times 5!\times 3!=2\times 120\times 6=1440$

Question:3(i) A committee of 7 has to be formed from 9 boys and 4 girls. In how many ways can this be done when the committee consists of:

exactly 3 girls?

Answer:

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 $=^4C_3.^9C_4$

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

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

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

Question:3(ii) A committee of 7 has to be formed from 9 boys and 4 girls. In how many ways can this be done when the committee consists of:

at least 3 girls?

Answer:

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

Question:3(iii) A committee of 7 has to be formed from 9 boys and 4 girls. In how many ways can this be done when the committee consists of:

atmost 3 girls?

Answer:

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

Question:4 If the different permutations of all the letter of the word EXAMINATION are listed as in a dictionary, how many words are there in this list before the first word starting with E?

Answer:

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

Question:5 How many 6-digit numbers can be formed from the digits 0, 1, 3, 5, 7 and 9 which are divisible by 10 and no digit is repeated?

Answer:

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$=5!=120$

Question:6 The English alphabet has 5 vowels and 21 consonants. How many words with two different vowels and 2 different consonants can be formed from the alphabet?

Answer:

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 $={ }^{21} C_2$

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

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 $=210\times 4!=50400$

Question:7 In an examination, a question paper consists of 12 questions divided into 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. In how many ways can a student select the questions?

Answer:

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 :

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

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

$=210+175+35$

$=420$

Question:8 Determine the number of 5-card combinations out of a deck of 52 cards if each selection of 5 cards has exactly one king.

Answer:

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 ${ }^{48} C_4$ ways.

Thus, the required number of 5 card combinations $={ }^4 C_1{ }^{48} C_4$

Question:9 It is required to seat 5 men and 4 women in a row so that the women occupy the even places. How many such arrangements are possible?

Answer:

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 $=5!\times 4!=120\times 24=2880$

Question:10 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. In how many ways can the excursion party be chosen ?

Answer:

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 ${ }^{22} C_7$ ways.

Case II : All 3 of them do not join.

Then,10 students can be chosen from 22 students in ${ }^{22} C_{10}$ ways.

Thus, the required number of ways for the excursion of party $={ }^{22} C_7+{ }^{22} C_{10}$

Question:11 In how many ways can the letters of the word ASSASSINATION be arranged so that all the S’s are together?

Answer:

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

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

Topics covered in Chapter 6 Permutations and Combinations Miscellaneous Exercise

1) Introduction to Permutations and Combinations

Permutations are arrangements of items where order matters in the arrangement, and combinations are selections of items where order does not matter. Permutations and combinations are used to count possible outcomes in different scenarios.

2) Fundamental Principle of Counting

If one event can occur in m ways and another in n ways, then both events together can occur in m × n ways. This fundamental principle of counting helps calculate total outcomes step by step.

Also Read

NCERT Solutions of Class 11 Subject Wise

Students can also access the NCERT solutions for other subjects and make their learning feasible.

Subject-Wise NCERT Exemplar Solutions

Use the links provided in the table below to get your hands on the NCERT exemplar solutions available for all the subjects.


Frequently Asked Questions (FAQs)

Q: What is the weightage of Statistics and Probability in CBSE Class 11 Maths ?
A:

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

Q: What is weightage of Straight Lines in the JEE Main Maths ?
A:

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

Q: Evaluate n!
A:

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

Q: Does 4! + 5! = 9! ?
A:

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!

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