NCERT Solutions Class 11th Maths Chapter – 7 Permutations and Combinations Miscellaneous Exercise

NCERT Solutions Class 11th Maths Chapter – 7 Permutations and Combinations Miscellaneous Exercise

NCERT Solutions Class 11th Maths Chapter – 7 Permutations and Combinations Miscellaneous Exercise

?Chapter – 7?

✍Permutations and Combinations✍

?Miscellaneous 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 ³C₂  as only two vowels are to be chosen.
Similarly, the five consonants can be chosen in ⁵C₃ ways.
∴ Number of choosing 2 vowels and 5 consonants would be ³C₂ ×⁵C₃

= 30
∴ Total number of ways of is 30
Each of these 5 letters can be arranged in 5 ways to form different words = ⁵P₅

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 ⁵C₅

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

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 × ⁵C₅ × ³C₃
∴ 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 = ⁴C₃
Numbers of boys are 9 out of which 4 are to be chosen which is given by ⁹C₄
Total ways of forming the committee with exactly three girls
= ⁴C₃ × ⁹C₄

(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 ⁴C₄ ways
And choosing 3 boys would be done in ⁹C₃
Total ways =  ⁴C₄ ×⁹C₃

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 = ⁴C₀ × ⁹C₇

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 p₁ and p₂ 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 ⁵P₅

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 ⁵C₂ ways
Choosing 2 consonants out of 21 can be done in ²¹C₂ ways
The total number of ways selecting 2 vowels and 2 consonants
= ⁵C₂ × ²¹C₂

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

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 = ⁴C₁
Ways of selecting the remaining 4 cards from 48 cards= ⁴⁸C₄
Total number of selecting the 5 cards having one king always
= ⁴C₁ × ⁴⁸C₄

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= ⁴P₄

5 men can occupy 5 seats in 5 ways
The numbers of ways in which these can be seated = ⁵P₅

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 =  ²²C₇
=

(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 ²²C₁₀

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 p₁, p₂ p₃ are the number of times the repeating letters repeat themselves.
Here  p₁= 3, p₂= 2, p₃ = 2
Putting the values in formula we get