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

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

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

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

$= 504 + 84 = 588$

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, p₁and 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 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= 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

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