### NCERT Solutions For Class 11 Math Chapter – 7 Miscelleneous Exercise

**NCERT Solutions For Class 11 Math Chapter – 7 Miscellneous 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 in ^{3}C_{2} as only two vowels are to be chosen.

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

∴ Number of choosing 2 vowels and 5 consonants would be ^{3}C_{2} ×^{5}C_{3}

=

∴ Total number of ways of is 30

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

Total number of words formed would be = 30 × 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 are ^{5}C_{5}

…………… (i)

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

…………….. (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 are 2 × ^{5}C_{5} × ^{3}C_{3}

∴ 2 × 120 × 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

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

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

Total ways of forming the committee with exactly three girls

= ^{4}C_{3} × ^{9}C_{4}

=** **

(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 in ^{4}C_{4} ways

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

Total ways = ^{4}C_{4} ×^{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 = ^{4}C_{0} × ^{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} and p_{2} 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 are ^{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 in ^{5}C_{2} ways

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

The total number of ways selecting 2 vowels and 2 consonants

= ^{5}C_{2} × ^{21}C_{2}

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

Total numbers of words that can be formed are

24 × 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 = ^{4}C_{1}

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

Total number of selecting the 5 cards having one king always

= ^{4}C_{1} × ^{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} and 8^{th} place where as men will be sitting on 1^{st}, 3^{rd}, 5^{th},7^{th} and 9^{th} place.

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

5 men can occupy 5 seats in 5 ways

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

The total numbers of sitting arrangements possible are

24 × 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 – 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 S’s are together?**

**Solution:**

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

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

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

Using the formula

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

Here p_{1}= 3, p_{2}= 2, p_{3} = 2

Putting the values in formula we get