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**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?**

**Ans. **There are 8 letters in the word DAUGHTER. In this word 3 vowels and 5 consonants. 2 vowels and 3 consonants are to be selected.

Number of ways of selection = = = = 30

Now, each word contains 5 letters which can be arranged among themselves 5! Ways.

Therefore, total number of words = = 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?**

**Ans. **There are 8 letters in the word EQUATION. In this word 5 vowels and 3 consonants. 2 vowels and 3 consonants are to be selected.

Now, 5 vowels can be arranged in 5! Ways and 3 consonants can be arranged in 3! Ways.

Also the groups of vowels and consonants can be arranged in 2! Ways.

Total number of permutations = = = 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) almost 3 girls?**

**Ans. (i)** There are 9 boys and 4 girls. 3 girls and 4 boys have to be selected.

Number of ways of selection = =

= = 504

**(ii)** We have to select at least 3 girls. So the committee consists of 3 girls and 4 boys or 4 girls and 3 boys.

Number of ways of selection =

=

=

= 504 + 84 = 588

**(iii)** We have to select at most 3 girls. So the committee consists of no girls and 7 boys or 1 girl and 6 boys or 2 girls and 5 boys or 3 girls and 4 boys.

Number of ways of selection =

=

=

= 36 + 336 + 756 + 504 = 1632

**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?**

**Ans. **In the word EXAMINATION, there are two I’s and two N’s and all other letters are different.

Number of ways of arrangement =

=

= 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 digits are repeated?**

**Ans. **A number divisible by 10 have unit place digit 0. So digit 0 is fixed at unit place and the remaining 5 placed filled with remaining five digits in ways.

Required numbers = = = 120

**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 alphabets?**

**Ans. **2 vowels out of 5 vowels and 2 consonants out of 21 consonants have to be selected and these 4 letters in 4 ways.

Required number of words = =

= = = 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?**

**Ans. **Here, we have to select 8 questions at least 3 questions from each section. Therefore, we have required selections are 3 from part I and 5 from part II or 4 from part I and 4 from part II or 5 from part I and 3 from part II.

Number of ways of selection =

=

=

=

= 210 + 175 + 35 = 420

**8. Determine the number of 5-card combinations out of a deck of 52 cards is each selection of 5 cards has exactly one king.**

**Ans. **Here, we have to select 5 cards containing 1 king and 4 other cards i.e., we have to select 1 king out of 4 kings and 4 other cards out of 48 other cards.

Number of ways of selection =

=

=

= = 778320

**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?**

**Ans. **Given: women occupy the even places.

1 2 3 4 5 6 7 8 9

M W M W M W M W M

Therefore, we can arrange four women in 4! Ways and 5 men in 5! Ways.

Number of ways of selection = = 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?**

**Ans. **According to the question, Number of ways of selection =

=

=

= 170544 + 646646 = 817190

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

**Ans. **In the word ASSASSINATION, A appears 3 times, S appears 4 times, I appears in 2 times and N appears in 2 times. Now, 4 S’ taken together become a single letter and other remaining letters taken with this single letter.

Number of arrangements =

= 151200

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