NCERT Solutions for Class 11 Maths chapter-7 Permutation And Combination Miscellaneous Exercise

NCERT Solutions for Class-11 Maths Chapter-7 Permutation and Combination

NCERT Solutions For Class-11 Maths Chapter-7 Permutation and Combination Miscellaneous Exercise prepared by expert of pw score more with pw NCERT Class-11 maths solutions. You can download solution of all chapters from Physics Wallah NCERT solutions of class 11.

NCERT Solutions for Class-11 Maths Chapter-7 Miscellaneous Exercise

Question1. 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 :There are 3 vowels i.e. A, U and E and 5 consonants i.e. D, G, H, T and R in the given word.

Therefore, number of ways of selecting 2 vowels out of 3 = ³C₂ =3

Number of ways of selecting 3 consonants out of 5 = ⁵C₃ = 10

Thus, by multiplication principle, number of combinations = 3×10=30

.

Each of these combinations can be arranged in 5! ways.

Hence, number of different words = 30×5!=3600
.

Question2. 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 :
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 = 2!×5!×3!=1440= 1440

Question3. 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?

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

⇒ Number of ways of selection = ⁴C₃ × ⁹C₄ =

= = 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 =  ⁴C₃ × ⁹C₄ + ⁴C₄ × ⁹C₃

= ⁴C₃×⁹C₄+⁴C₄×⁹C₃

 

= 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 = ⁴C₃  ×  ⁹C₄  +  ⁴C₂ × ⁹C₅ + ⁴C₁ × ⁹C₆ + ⁴C₀ × ⁹C₇

 

= 504 + 756 + 336 + 36

= 36 + 336 + 756 + 504 = 1632

Question4. 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 :
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

Question5. 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?

Solution :
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 5! ways.

∴ Required numbers = 5! = 120

Question6. 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?

Solution :
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 =

 

Question7. 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 :
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 ⁵C₃ × ⁷C⁵  ways.

4 questions from part I and 4 questions from part II can be selected in ⁵C₄ × ⁷C₄ ways.

5 questions from part I and 3 questions from part II can be selected in ⁵C₅ × ⁷C₃ ways.

Thus, required number of ways of selecting questions

=⁵C₃ × ⁷C₅ + ⁵C₄ × ⁷C₄ + ⁵C₅ × ⁷C₃

=

= 210 + 175 + 35 = 420

Question8. Determine the number of 5-card combinations out of a deck of 52 cards is each selection of 5 cards has exactly one king.

Solution :
From a deck of 52 cards, 5-card combinations have to be made in such a way that in each selection of 5 cards, there is exactly one king.

In a deck of 52 cards, there are 4 kings.

1 king can be selected out of 4 kings in ⁴C₁  ways.

4 cards out of the remaining 48 cards can be selected in ⁴⁸C₄ ways.

Thus, the required number of 5-card combinations is ⁴C₁ × ⁴⁸C₄.

Question9. 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 :
5 men and 4 women are to be seated in a row such that the women occupy the even places.

The 5 men can be seated in 5! ways. For each arrangement, the 4 women can be seated only at the cross marked places (so that women occupy the even places).

M x M x M x M x M

Therefore, the women can be seated in 4! ways.

Thus, possible number of arrangements = 4! × 5! = 24 × 120 = 2880 

Question10. 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 :
 

10 are chosen for an excursion party out of 25 students.


There are 2 cases since 3 students decide either all or one of them will join.


Case 1: All the 3 students join.


The remaining 7 students can be chosen out of 22 students in ²²C₇

 ways.


Case 2: None of the 3 students join.


Thus, 10 students can be chosen in ²²C₁₀

 ways.

Hence, number of ways of choosing excursion party = ²²C₇+²²C₁₀

= 170544 +  646646  = 817190

Question11. In how many ways can be letters of the word ASSASSINATION be arranged so that all the S’s are together?

Solution :
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 = 10! / 2!3!2!

= 151200