Chapter 7 Miscellaneous Ncert math solution class 11

 Chapter 7 ( Miscellaneous Exercise)

Chapter 7 Miscellaneous Ncert math solution class 11

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?

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

Number of way of selecting 2 vowels from 3  = 3C2

Similarly, Number of way of selecting 3 consonants from 5 =  5C3

∴ The number of choosing 2 vowels and 5 consonants=  3C2 ×5C3

=\dfrac{3!}{2!(3-2)!}\times \dfrac{5!}{3!(5-3)!}

=\dfrac{3!}{2!1!}\times \dfrac{5!}{3!2!}

=\dfrac{5\times 4\times 3\times 2\times 1}{2\times 1\times2\times1}

= 30

∴ The total number of ways is 30.

Each of these 5 letters can be arranged in 5 ways to form different words = 5P5

=\dfrac{5!}{(5-5)!}=\dfrac{5!}{0!}

=\dfrac{5\times 4\times 3\times 2\times 1}{1}

=120

The total number of words formed would be = 30 × 120 = 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?

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 =  5P5

=\dfrac{5!}{(5-5)!}=\dfrac{5!}{0!}

=\dfrac{5\times 4\times 3\times 2\times 1}{1}

=120  …………… (i)

Similarly, the numbers of ways in which 3 consonants can be arranged = 3P3

=\dfrac{3!}{(3-3)!}=\dfrac{3!}{0!}

=\dfrac{3\times 2\times 1}{1}

=6  …………….. (ii)

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

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

∴ The total number of ways in which vowel and consonant can appear together = 2 ×5P5 × 3P3

= 2 × 120 × 6 = 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?

(ii) At least 3 girls?

(iii) At most 3 girls?

Solution: (i) Given, exactly 3 girls.

The total numbers of girls = 4.

∴ The number of ways of selection of 3 girls from 4  = 4C3

The number of boys is 9, out of which 4 are to be chosen = 9C4

Total ways of forming the committee with exactly three girls

4C3 × 9C4

= \frac{4!}{3!(4-3)!}\times \frac{9!}{4!(9-4)!}

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

= \frac{9\times 8\tmes 7\times 6 \times 5\times 4 \times 3\times 2\times 1}{\times 3\times 2\times 1\times 5\times 4 \times 3\times 2\times 1}

=  504

(ii) Given, at least 3 girls.

There are two possibilities for making a committee choosing at least 3 girls  = 3 girls and 4 boys + 4 girls and 3 boys

={}^4C_3 \times {}^9C_4 + {}^4C_4 \times {}^9C_3

= {}^4C_3 \times {}^9C_4</span>  + <span style="font-size: 16px;">{}^4C_4\times {}^9C_3

= \frac{4!}{3!(4-3)!}\times \frac{9!}{4!(9-4)!} + \frac{4!}{4!(4-4)!}\times \frac{9!}{3!(9-3)!}

= \frac{4!}{3!1!}\times \frac{9!}{4!5!}+ \frac{4!}{4!}\times \frac{9\times 8\tmes 7\times 6!}{3\times 2\times 1\times 6!}

= 504 + 84 = 588

The total number of ways of making the committee are =- 588

(iii) Given, at most 3 girls.

In this case, the numbers of possibilities are = 0 girls and 7 boys + 1 girl and 6 boys + 2 girls and 5 boys + 3 girls and 4 boys

= {}^4C_0\times {}^9C_7 + {}^4C_1\times {}^9C_6 + {}^4C_2\times {}^9C_5+ {}^4C_3\times {}^9C_4

=  1\times \frac{9\times 8}{2!}+4\times \frac{9\times 8 \times 7}{3!}+\frac{4\times 3}{2!}\times \frac{9\times 8\times 7\times 6}{4!} + 504

= 36+336+756+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 letters 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:   Listed before E should start with the letter either A, B, C or D.

But the word EXAMINATION doesn’t have B, C or D.

Hence, the words should start with the 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

=\frac{n!}{P_1!P_2!P_3!}

Where n is the remaining number of letters, pand p2 are the number of times the repeated terms occur.

=\frac{10!}{2!2!} = 907200

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

= words starting with letter A = 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?

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

So, the total ways in which the rest of the places can be filled = 5P5

=\frac{5!}{(5-5)!}

= \frac{5!}{1!}

= 5 × 4 × 3 × 2 × 1

= 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?

Solution: We know that there are 5 vowels and 21 consonants in the English alphabet.

Choosing two vowels out of 5  =  5C2 ways.

Choosing 2 consonants out of 21  = 21C2 ways.

The total number of ways to select 2 vowels and 2 consonants

5C2 × 21C2

=\frac{5!}{2!3!}\times \frac{21!}{2!19!}

= \frac{5\times 4\times 3!}{2!3!}\times \frac{21 \times 20\times 19!}{2\times 1\times 19!}

= 2100

Each of these four letters can be arranged in four ways =4P4

=\frac{4!}{0!}

= 4\times 3\times 2 \times 1 = 24

The total numbers of words that can be formed are

=24 × 2100 = 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?

Solution: Number of ways of selection can a student select the questions

=  3 questions from part I and 5 from part II + 4 questions from part I and 4 from part II +  5 questions from parts 1 and 3 from part II.

= {}^5C_3\times {}^7C_5 + {}^5C_4 \times {}^7C_4 + {}^5C_5\times {}^7C_3

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

Solution: 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 = 4C1

Ways of selecting the remaining 4 cards from 48 cards= 48C4

The total number of selecting the 5 cards having one king always

4C1 × 48C4

= \frac{4!}{1!3!}\times \frac{48!}{4!44!}

= \frac{4\times 3!}{3!}\times \frac{48\times 47\times 46\times 45\times 44!}{4\times 3\times 2\times 1\times 44!}

= 778320

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

Solution: Given, there is a total of 9 people.

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

4 women can sit in four places and ways they can be seated= 4P4

=\frac{4!}{(4-4)!}

=4! = 24

5 men can occupy 5 seats in 5 ways.

The number of ways in which these can be seated = 5P5

=\frac{5!}{(5-5)!}

= 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?

Solution:  In this question, we get 2 options that are

(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 = 22C7

= \frac{22!}{7!15!}

\frac{22\time 21\times 20 \times 19 \times 18 \times 17 \times 16 \times 15!}{7!15!}

= 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 =  22C10

=\frac{22!}{10!12!}

= 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 are together?

Solution: In the given word ASSASSINATION, there are 4 ‘S’. Since all the 4 ‘S’ have to be arranged together, let us take them as one unit.

The remaining letters are= 3 ‘A’, 2 ‘I’, 2 ‘N’, T

The number of letters to be arranged is 9 (including 4 ‘S’)

Using the formula =\frac{n!}{P_1!P_2!P_3!}

where n is the number of terms and p1, p2 p3 are the number of times the repeating letters repeat themselves.

Here, p1= 3, p2= 2, p3 = 2

Putting the values in the formula, we get

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

= 151200

Chapter 7: permutation and combination Class 11

Exercise 7.1 ncert solutions maths class-11

Exercise 7.2 ncert solutions maths class 11

Exercise 7.3 ncert solutions maths class 11

Exercise 7.4 ncert solutions maths class 11

Chapter 7 Miscellaneous Ncert math solution class 11

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