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

Last updated date: 13th Aug 2024
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## NCERT Solutions for Class 11 Maths Permutations and Combinations Miscellaneous Exercise - Free PDF Download

NCERT Solutions for Class 11 Maths Chapter 6 Permutations And Combinations includes solutions to all Miscellaneous Exercise problems. The Miscellaneous Exercises NCERT Solutions for Class 11 Maths are based on the ideas presented in Maths Chapter 6. This activity is crucial for both the CBSE Board examinations and competitive tests. To perform well on the board exam, download the NCERT Solutions in PDF format and practice them offline. Download the CBSE Class 11 Maths Syllabus in PDF format and practise the topics offline regularly to do well in exams.

Table of Content
1. NCERT Solutions for Class 11 Maths Permutations and Combinations Miscellaneous Exercise - Free PDF Download
2. Access NCERT Solutions for Class 11 Maths Chapter 6 Permutations and Combinations
2.1Miscellaneous Exercise
3. Class 11 Maths Chapter 6: Exercises Breakdown
4. CBSE Class 11 Maths Chapter 6 Other Study Materials
5. Chapter-Specific NCERT Solutions for Class 11 Maths
FAQs
Competitive Exams after 12th Science

## Access NCERT Solutions for Class 11 Maths Chapter 6 Permutations and Combinations

### Miscellaneous Exercise

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?

Ans:  There are 3 vowels i.e. A, U and E and 5 consonants i.e. D, G, H, T and R in the given word.

Therefore, the number of ways of selecting 2 vowels out of 3 = $^{3}{{C}_{2}}$=3

Number of ways of selecting 3 consonants out of 5 = $^{5}{{C}_{3}}=10$

Thus, by multiplication principle, the number of combinations = $3\times 10=30$.

Each of these combinations can be arranged in 5! ways.

Hence, the number of different words = $30\times 5!=3600$.

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

Ans: There are 5 vowels i.e. A, E, I, O and U and 3 consonants i.e. Q, T and N.

Since, vowels and consonants occur together, both (AEIOU) and (QTN) can be considered as single objects.

Thus, there are 5! Permutations of 5 vowels taken all at a time and 3! permutations of 3 consonants taken all at a time.

Therefore, by multiplication principle, the number of words = $2!\, \times \,5!\, \times \,3!\, = \,1440$.

3. 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 :

(i) exactly 3 girls? (ii) at least 3 girls? (iii) at most 3 girls?

Ans: (i) Out of 9 boys and 4 girls, a committee of 7 has to be formed.

Given: exactly 3 girls should be there in a committee, hence, the number of boys = (7 - 3) = 4 boys only.

Therefore, many ways

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

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

= 504

(ii) Given: at least 3 girls are required in each committee. This can be done in 2 ways

(a) 3 girls and 4 boys or (b) 4 girls and 3 boys

3 girls and 4 boys can be selected in  $^{4}{{C}_{3}}{{\times }^{9}}{{C}_{4}}$ ways.

4 girls and 3 boys an be selected in  $^{4}{{C}_{4}}{{\times }^{9}}{{C}_{3}}$ ways.

Thus, the number of ways =

$^{4}{{C}_{3}}{{\times }^{9}}{{C}_{4}}{{+}^{4}}{{C}_{4}}{{\times }^{9}}{{C}_{3}}$

= 504+84

= 588

(iii) Given: at most 3 girls in every committee. This can be done in 4 ways

(a) 3 girls and 4 boys (b) 2 girls and 5 boys.

(c) 1 girl and 6 boys (d) No girl and 7 boys

3 girls and 4 boys can be selected in $^{4}{{C}_{3}}{{\times }^{9}}{{C}_{4}}$ ways

2 girls and 5 boys can be selected in $^{4}{{C}_{2}}{{\times }^{9}}{{C}_{5}}$ ways

1 girl and 6 boys can be selected in  $^{4}{{C}_{1}}{{\times }^{9}}{{C}_{6}}$ ways

No girl and 7 boys can be selected in $^{4}{{C}_{0}}{{\times }^{9}}{{C}_{7}}$ ways.

Thus, many ways

${{=}^{4}}{{C}_{3}}{{\times }^{9}}{{C}_{4}}{{+}^{4}}{{C}_{2}}{{\times }^{9}}{{C}_{5}}{{+}^{4}}{{C}_{1}}{{\times }^{9}}{{C}_{6}}{{+}^{4}}{{C}_{0}}{{\times }^{9}}{{C}_{7}}$

$=\frac{4!}{3!1!}\times \frac{9!}{4!5!}+\frac{4!}{2!2!}\times \frac{9!}{5!4!}+\frac{4!}{1!3!}\times \frac{9!}{6!3!}+\frac{4!}{0!4!}\times \frac{9!}{7!2!}$

$=504+756+336+36$

$=1632$

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

Ans: There are a total of 11 letters out of which A, I and N appear 2 times and other letters appear only once.

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

Thus, words starting with the letter A will have the letter A fixed at its extreme left end.

And remaining 10 letters are rearranged all at a time.

In the remaining 10 letters, there are 2 I’s and 2 N’s.

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

Therefore, the required number of words is 907200.

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?

Ans: A number can be divisible by 10 only if its unit digit is 0.

Hence, 0 is fixed at the unit place.

Therefore, the 5 vacant places can be filled by the remaining 5 digits.

These 5 empty places can be filled in 5! ways.

Therefore, number of 6-digit numbers = 5! = 120.

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?

Ans: Given: 2 vowels and 2 consonants should be selected from the English alphabet.

We know that there are 5 vowels.

Thus, the number of ways of selecting 2 vowels out of 5 = $^{5}{{C}_{2}}=\frac{5!}{2!3!}=10$

Also, we know that there are 21 consonants.

Thus, the number of ways of selecting 2 consonants out of 21 = $^{21}{{C}_{2}}=\frac{21!}{2!19!}=210$

Thus, the total number of combinations of selecting 2 vowels and 2 consonants is

= $10\, \times \,210\, = \,2100$.

Every 2100 combination consists of 4 letters, which can be arranged in 4! ways.

Therefore, number of words = $2100\, \times \,4!\, = \,50400$

7. In an examination, a paper consists of 12 Questions divided into 2 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?

Ans:  Given: 12 Qs are divided into 2 parts – Part I and Part II consisting of 5 and 7 Qs respectively. A student must attempt 8 Qs with at least 3 from each part. This can be done as:

1. 3 Qs from part I and 5 Qs from part II.

2. 4 Qs from part I and 4 Qs from part II.

3. 5 Qs from part I and 3 Qs from part II.

The first case can be selected in $^{5}{{C}_{3}}{{\times }^{7}}{{C}_{5}}$ ways.

The second case can be selected in $^{5}{{C}_{4}}{{\times }^{7}}{{C}_{4}}$ ways.

The third case can be selected in $^{5}{{C}_{5}}{{\times }^{7}}{{C}_{3}}$ ways.

Thus, number of ways of selecting Q’s

${{=}^{5}}{{C}_{3}}{{\times }^{7}}{{C}_{5}}{{+}^{5}}{{C}_{4}}{{\times }^{7}}{{C}_{4}}{{+}^{5}}{{C}_{5}}{{\times }^{7}}{{C}_{3}}$

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

$=210+175+35$

$=420$

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.

Ans:  Out of 52 cards, 5-card combinations have to be made in such a way that in each selection of 5 cards, there is exactly one king. We know that in 52 cards, there are 4 kings.

Out of 4 kings, 1 king can be selected in $^{4}{{C}_{1}}$ ways.

Out of the remaining 48 cards, 4 cards can be selected in $^{48}{{C}_{4}}$ ways.

Therefore, number of 5-card combinations =  $^{4}{{C}_{1}}{{\times }^{48}}{{C}_{4}}$ ways.

9. It is required to seat 5 men and 4 women in a row so that the women can occupy the even places.

How many such arrangements are possible?

Ans: Given 5 men and 4 women should be seated so that women always occupy the even places.

Thus, the men can be seated in 5! ways.

For each arrangement, the women can be seated only in even places.

Thus, the women can be seated in 4! ways.

Therefore, possible number of arrangements = $4!\, \times \,5!\, = \,24\, \times \,120\, = \,2880$

10. From the 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?

Ans:  Given: 10 are chosen for an excursion party out of 25 students.

There are 2 cases where 3 students decided either all or one of them would join.

Case 1: All the 3 students join.

The remaining 7 students can be chosen out of 22 students in $^{22}{{C}_{7}}$ ways.

Case 2: None of the 3 students join.

Thus, 10 students can be chosen in $^{22}{{C}_{10}}$ ways.

Hence, number of ways of choosing excursion party = $^{22}{{C}_{7}}{{+}^{22}}{{C}_{10}}$

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

Ans: There are 3 A’s, 4 S’s, 2 I’s and all other letters appear only once in the word ASSASSINATION.

The given word should be arranged such that all the S’s are together.

The 4 S’s can be treated as a single object for the time being. This single object with the remaining objects will be 10 objects together.

These 10 objects with 3 A’s, 2 I’s and 2 N’s can be arranged in $\dfrac{{10!}}{{2!3!2!}}$ways.

Therefore, number of ways of arranging the given word = $\dfrac{{10!}}{{2!3!2!}}\, = \,15120$.

## Conclusion

Miscellaneous Exercise in Class 11 Maths Chapter 6 is crucial for understanding various concepts thoroughly. It covers diverse problems that require the application of multiple formulas and techniques. It's important to focus on understanding the underlying principles behind each question rather than just memorising solutions. Remember to understand the theory behind each concept, practice regularly, and refer to solved examples to master this exercise effectively.

## Class 11 Maths Chapter 6: Exercises Breakdown

 Exercise Number of Questions Exercise 6.1 6 Questions & Solutions Exercise 6.2 5 Questions & Solutions Exercise 6.3 11 Questions & Solutions Exercise 6.4 9 Questions & Solutions

## Chapter-Specific NCERT Solutions for Class 11 Maths

Given below are the chapter-wise NCERT Solutions for Class 11 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

## FAQs on NCERT Solutions for Class 11 Maths Miscellaneous Exercise Chapter 6 - Permutations and Combinations

1. What are permutations and combinations in NCERT solutions of Class 11 Maths Chapter 6 Miscellaneous Exercise Solutions?

In NCERT Solutions of Class 11 Maths Chapter 6 Miscellaneous Exercise Solutions, permutations involve arranging objects in a specific order, while combinations involve selecting objects without considering the order. Both concepts are essential for solving counting problems.

2. How do I distinguish between permutations and combinations in a problem in NCERT Class 11 Maths Chapter 6 Miscellaneous Exercise Solutions?

In NCERT Class 11 Maths Chapter 6 Miscellaneous Exercise Solutions, if the order matters, it's a permutation; if it doesn't, it's a combination. Look for keywords like "arrange" for permutations and "select" for combinations to identify the type of problem.

3. What formulas are important for permutations in NCERT Permutation And Combination Class 11 Miscellaneous Exercise?

In NCERT Permutation And Combination Class 11 Miscellaneous Exercise, the main formula is nPr = n! / (n-r)! where n is the total number of items and r is the number of items to arrange. Understanding and applying factorial notation is crucial.

4. What formulas are crucial for combinations in NCERT Permutation And Combination Class 11 Miscellaneous Exercise?

The key formula is nCr = n! / [r! * (n-r)!]. This formula helps in problems where selection is required without regard to order.

5. What types of questions are typically asked in the NCERT solutions of class 11 maths chapter 6 miscellaneous exercise?

In NCERT solutions of class 11 maths chapter 6 miscellaneous exercise questions often involve applying formulas directly, solving word problems, and dealing with real-life scenarios. Practising these will build a strong understanding of the concepts.

6. What should I focus on while solving problems in NCERT Permutation And Combination Miscellaneous Exercise Class 11 Solutions?

In NCERT Permutation And Combination Miscellaneous Exercise Class 11 Solutions focuses on understanding the problem, determining whether it involves permutations or combinations, and applying the correct formula accurately. Double-check your calculations to avoid errors.

7. What topics are covered in the Permutation And Combination Miscellaneous Exercise Class 11 Solutions?

The Permutation And Combination Miscellaneous Exercise Class 11 Solutions includes a variety of problems related to the concepts of complex numbers and quadratic equations.

8. How can NCERT Solutions help me with the Miscellaneous Exercise Class 11 Chapter 6?

NCERT Solutions provides clear, step-by-step answers to each question. This helps you understand how to solve each problem and learn the correct methods.

9. Why is practicing the Miscellaneous Exercise Class 11 Chapter 6 important?

Practicing the Miscellaneous Exercise Class 11 Chapter 6 is important because it helps you apply what you’ve learned in the chapter. It covers different types of questions that can appear in exams, giving you good practice.

10. What is a good way to approach solving these problems of Miscellaneous Exercise Class 11 Chapter 6?

A good way to solve these problems is to start by trying them on your own. Then, use NCERT Solutions to check your work and understand any mistakes. This helps reinforce your learning.

11. Are the questions in the Miscellaneous Exercise Class 11 Chapter 6 likely to appear in exams?

Yes, the questions from the Miscellaneous Exercise are often similar to those asked in exams. Practising these problems can help you prepare effectively for your tests.