NCERT Solutions for last exercise of the seventh chapter of class 11 Maths have been provided here for students to prepare well for the exam. These solutions are designed by the subject experts at BYJUâ€™S and are in accordance with the NCERT syllabus and guidelines. The Miscellaneous Exercise of NCERT Solutions for Class 11 Maths Chapter 7- Permutations and Combinations is based on the following topics:

- Introduction to Permutations and Combinations
- Fundamental Principle of Counting

Understanding proper methods to solve the NCERT solutions for Class 11 Maths will help the students in analysing the different types of questions that are likely to be asked in the examination. Once the students become proficient in these solutions, their problem solving speed will increase, helping them in boosting their self confidence.

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

Â

### Solutions for Class 11 Maths Chapter 7 â€“ 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?**

**Solution:**

The word DAUGHTER has 3 vowels A, E, U and 5 consonants D, G, H, T and R.

The three vowels can be chosen inÂ ^{3}C_{2}Â as only two vowels are to be chosen.

Similarly, the five consonants can be chosen in ^{5}C_{3}Â ways.

âˆ´Â Number of choosing 2 vowels and 5 consonants would beÂ ^{3}C_{2}Â Ã—^{5}C_{3}

= 30

âˆ´Â Total number of ways of is 30

Each of these 5 letters can be arranged in 5 ways to form different words =Â ^{5}P_{5}

Total number of words formed would be = 30 Ã— 120 = 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 the vowels and consonants occur together?**

**Solution:**

In the word EQUATION there are 5 vowels (A, E, I, O, U) and 3 consonants (Q, T, N)

The numbers of ways in which 5 vowels can be arranged areÂ ^{5}C_{5}

â€¦â€¦â€¦â€¦â€¦ (i)

Similarly, the numbers of ways in which 3 consonants can be arranged areÂ ^{3}P_{3}

â€¦â€¦â€¦â€¦â€¦.. (ii)

There are two ways in which vowels and consonants can appear together

(AEIOU) (QTN) or (QTN) (AEIOU)

âˆ´Â The total number of ways in which vowel and consonant can appear together are 2 Ã—Â ^{5}C_{5}Â Ã—Â ^{3}C_{3}

âˆ´ 2 Ã— 120 Ã— 6 = 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?**

**Solution:**

(i) Given exactly 3 girls

Total numbers of girls are 4

Out of which 3 are to be chosen

âˆ´Â Number of ways in which choice would be made =Â ^{4}C_{3}

Numbers of boys are 9 out of which 4 are to be chosen which is given byÂ ^{9}C_{4}

Total ways of forming the committee with exactly three girls

=Â ^{4}C_{3}Â Ã—Â ^{9}C_{4}

=**Â **

(ii) Given at least 3 girls

There are two possibilities of making committee choosing at least 3 girls

There are 3 girls and 4 boys or there are 4 girls and 3 boys

Choosing three girls we have done in (i)

Choosing four girls and 3 boys would be done inÂ ^{4}C_{4}Â ways

And choosing 3 boys would be done inÂ ^{9}C_{3}

Total ways =Â ^{4}C_{4}Â Ã—^{9}C_{3}

Total numbers of ways of making the committee are

504 + 84 = 588

(iii) Given at most 3 girls

In this case the numbers of possibilities are

0 girl and 7 boys

1 girl and 6 boys

2 girls and 5 boys

3 girls and 4 boys

Number of ways to choose 0 girl and 7 boys =Â ^{4}C_{0}Â Ã—Â ^{9}C_{7}

Number of choosing 3 girls and 4 boys has been done in (1)

= 504

Total number of ways in which committee can have at most 3 girls are = 36 + 336 + 756 + 504 = 1632

**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?**

**Solution:**

In dictionary words are listed alphabetically, so to find the words

Listed before E should start with letter either A, B, C or D

But the word EXAMINATION doesn`t have B, C or D

Hence the words should start with letter A

The remaining 10 places are to be filled by the remaining letters of the word EXAMINATION which are E, X, A, M, 2N, T, 2I, 0

Since the letters are repeating the formula used would be

Where n is remaining number of letters p_{1}Â and p_{2}Â are number of times the repeated terms occurs.

The number of words in the list before the word starting with E

= words starting with letter A = 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?**

**Solution:**

The number is divisible by 10 if the unit place has 0 in it.

The 6-digit number is to be formed out of which unit place is fixed as 0

The remaining 5 places can be filled by 1, 3, 5, 7 and 9

Here n = 5

And the numbers of choice available are 5

So, the total ways in which the rest the places can be filled areÂ ^{5}P_{5}

**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?**

**Solution:**

We know that there are 5 vowels and 21 consonants in English alphabets.

Choosing two vowels out of 5 would be done inÂ ^{5}C_{2}Â ways

Choosing 2 consonants out of 21 can be done inÂ ^{21}C_{2}Â ways

The total number of ways selecting 2 vowels and 2 consonants

=Â ^{5}C_{2}Â Ã—Â ^{21}C_{2}

Each of these four letters can be arranged in four waysÂ ^{4}P_{4}

Total numbers of words that can be formed are

24 Ã— 2100 = 50400

**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?**

**Solution:**

The student can choose 3 questions from part I and 5 from part II

Or

4 questions from part I and 4 from part II

5 questions from part 1 and 3 from part II

**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.**

**Solution:**

We have a deck of cards has 4 kings.

The numbers of remaining cards are 52.

Ways of selecting a king from the deck =Â ^{4}C_{1}

Ways of selecting the remaining 4 cards from 48 cards=Â ^{48}C_{4}

Total number of selecting the 5 cards having one king always

=Â ^{4}C_{1}Â Ã—Â ^{48}C_{4}

**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?**

**Solution:**

Given there are total 9 people

Women occupies even places that means they will be sitting on 2^{nd}, 4^{th}, 6^{th}Â and 8^{th}Â place where as men will be sitting on 1^{st}, 3^{rd}, 5^{th},7^{th}Â and 9^{th}Â place.

4 women can sit in four places and ways they can be seated=Â ^{4}P_{4}

5 men can occupy 5 seats in 5 ways

The numbers of ways in which these can be seated =Â ^{5}P_{5}

The total numbers of sitting arrangements possible are

24 Ã— 120 = 2880

**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?**

**Solution:**

In this question we get 2 options that is

(i) Either all 3 will go

Then remaining students in class are: 25 â€“ 3 = 22

Number of students remained to be chosen for party = 7

Number of ways choosing the remaining 22 students = ^{22}C_{7}

=

(ii) None of them will go

The students going will be 10

Remaining students eligible for going = 22

Number of ways in which these 10 students can be selected are ^{22}C_{10}

Total numbers of ways in which students can be chosen are

= 170544 + 646646 = 817190

**11. In how many ways can the letters of the word ASSASSINATION be arranged so that all the Sâ€™s are together?**

**Solution:**

In the given word ASSASSINATION, there are 4 â€˜Sâ€™. Since all the 4 â€˜Sâ€™ have to be arranged together so let as take them as one unit.

The remaining letters are= 3 â€˜Aâ€™, 2 â€˜Iâ€™, 2 â€˜Nâ€™, T

The number of letters to be arranged are 9 (including 4 â€˜Sâ€™)

Using the formula

where n is number of terms and p_{1}, p_{2}Â p_{3}Â are the number of times the repeating letters repeat themselves.

HereÂ p_{1}= 3, p_{2}= 2, p_{3}Â = 2

Putting the values in formula we get

### Access other exercise solutions of Class 11 Maths Chapter 7- Permutations and Combinations

Exercise 7.1 Solutions 6 Questions

Exercise 7.2 Solutions 5 Questions

Exercise 7.3 Solutions 11 Questions

Exercise 7.4 Solutions 9 Questions