**Miscellaneous Exercise**

Permutations and combinations

**Question and Answers**

**Class 11 – Maths**

Permutations and combinations

Class | Class 11 |

Subject | Mathematics |

Chapter Name | Permutations and Combinations |

Chapter No. | Chapter 6 |

Exercise | Miscellaneous Exercise |

Category | Class 11 Maths NCERT Solutions |

**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, there are 3 vowels namely, A, U, and E, and 5 consonants namely, D, G, H, T, and R.

Number of ways of selecting 2 vowels out of 3 vowels = ^{3}C_{2} = 3

Number of ways of selecting 3 consonants out of 5 consonants = ^{5}C_{3} = 10

Therefore, number of combinations of 2 vowels and 3 consonants = 3 × 10 = 30

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

Hence, required number of different words = 30 × 5! = 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, there are 5 vowels, namely, A, E, I, O, and U, and 3 consonants, namely, Q, T, and N.

Since all the vowels and consonants have to occur together, both (AEIOU) and (QTN) can be assumed as single objects. Then, the permutations of these 2 objects taken all at a time are counted. This number would be ^{2}C_{2} = 2!

Corresponding to each of these permutations, there are 5! permutations of the five vowels taken all at a time and 3! permutations of the 3 consonants taken all at a time.

Hence, by multiplication principle, required number of words = 2! × 5! × 3! = 1440

**Question 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 ?**

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

Exactly 3 girls should be there in a committee, hence, number of boys = (7 – 3) = 4 boys only.

Total ways of forming the committee with exactly three girls = ^{4}C_{3} × ^{9}C_{4}

** (ii) atleast 3 girls ?**

Since at least 3 girls are to be there in every committee, the committee can consist of

(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} × ^{9}C_{4} ways.

4 girls and 3 boys can be selected in ^{4}C_{3} × ^{9}C_{4} ways.

Therefore, in this case, required number of ways = ^{4}C_{3} × ^{9}C_{4} + ^{4}C_{3} × ^{9}C_{4}

= 504 + 84 = 588

** (iii) atmost 3 girls ?**

Since at most 3 girls are to be there in every committee, the committee can consist of

(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} × ^{9}C_{4} ways.

2 girls and 5 boys can be selected in ^{4}C_{2} × ^{9}C_{5} ways.

1 girl and 6 boys can be selected in ^{4}C_{1} × ^{9}C_{6} ways .

No girl and 7 boys can be selected in ^{4}C_{0} × ^{9}C_{7} ways .

Therefore, in this case, required number of ways .

The number of choosing 3 girls and 4 boys = 504

The total number of ways in which a committee can have at most 3 girls are = 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** There are a total of 11 letters out of which A, I and N appears 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 letter A will have 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.

**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** The number is divisible by 10 if the unit place has 0 in it.Therefore, 0 is fixed at the units place.

The 6-digit number is to be formed

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

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

Therefore, number of 6-digit numbers = 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 **We know that there are 5 vowels and 21 consonants in the English alphabet.

Choosing two vowels out of 5 would be done in ^{5}C_{2} ways.

Choosing 2 consonants out of 21 can be done in ^{21}C_{2} ways.

The total number of ways to select 2 vowels and 2 consonants = ^{5}C_{2} × ^{21}C_{2}

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

Therefore, number of words = 2100 × 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 the question paper consists of 12 questions divided into two parts – Part I and Part II, containing 5 and 7 questions, respectively.

A student has to attempt 8 questions, selecting at least 3 from each part.

This can be done as follows.

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

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

(c) 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 ^{5}C_{3} × ^{7}C_{5} ways.

4 questions from part I and 4 questions from part II can be selected in ^{5}C_{4} × ^{7}C_{4} ways.

5 questions from part I and 3 questions from part II can be selected in ^{5}C_{5} × ^{7}C_{3} ways.

Now the total number of ways in which a student can choose the questions are

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 **We have a deck of cards that has 4 kings.

The numbers of remaining cards are 52.

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

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

The total number of selecting the 5 cards having one king always = ^{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 ** 5 men and 4 women should be seated so that women always occupy the even places.

Women occupy even places, which means they will be sitting in 2^{nd}, 4^{th}, 6^{th} and 8^{th} places, whereas men will be sitting in 1^{st}, 3^{rd}, 5^{th},7^{th} and 9^{th} places.

4 women can sit in four places and ways they can be seated= ^{4}P_{4 }= 24

5 men can occupy 5 seats in 5 ways and the number of ways in which these can be seated = ^{5}P_{5 }= 120

The total number of sitting arrangements possible = 24 × 120 = 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 the class of 25 students, 10 are to be chosen for an excursion party.

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

**(i)** Either all 3 will go

The remaining students in the class are: 25 – 3 = 22

The number of students that remain to be chosen for the party = 7

The number of ways to choose the remaining 22 students = ^{22}C_{7 }= 170544

**(ii)** None of them will go.

The students going will be 10.

Remaining students eligible for going = 22

The number of ways in which these 10 students can be selected is ^{22}C_{10 }= 646646

The total number of ways in which students can be chosen = 170544 + 646646 = 817190

**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 given word ASSASSINATION, the letter A appears 3 times, S appears 4 times, I appears 2 times, N appears 2 times, and all the other letters appear only once.

Since all the words have to be arranged in such a way that all the S are together, SSSS is treated as a single object for the time being. This single object together with the remaining 9 objects will account for 10 objects.

These 10 objects in which there are 3 As, 2 Is, and 2 Ns can be arranged in

## Leave a Reply