Exercise Miscellaneous - Permutations and Combinations NCERT Solutions | Mathematics (Maths) Class 11 - Commerce PDF Download

Q1: 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: The word DAUGHTER has 3 vowels A, E, and 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.
∴ The number of choosing 2 vowels and 5 consonants would be ³C₂ ×⁵C₃

= 30
∴ The 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

Q2: 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: 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

Q3: 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?
Ans: (i) Given exactly 3 girls.
The total numbers of girls are 4.
Out of which, 3 are to be chosen.
∴ The 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 for making a 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

The total number 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

The number of choosing 3 girls and 4 boys has been done in (1)
= 504
The total number of ways in which a committee can have at most 3 girls are = 36 + 336 + 756 + 504 = 1632.

Q4: 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 starts with E?
Ans: In a dictionary, words are listed alphabetically, so to find the words
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 in 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 the remaining number of letters, p1 and p2 are the 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

Q5: 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?
Ans: 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 of the places can be filled are ⁵P₅


Q6: 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?
Ans: We know that there are 5 vowels and 21 consonants in the English alphabet.
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 to select 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

Q7: 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: 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


Q8: Determine the number of 5-card combinations out of a deck of 52 cards if each selection of 5 cards has exactly one king.
Ans: 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


Q9: 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?
Ans: 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 place where as men will be sitting in 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 number of ways in which these can be seated = ⁵P₅

The total numbers of sitting arrangements possible are
24 × 120 = 2880

Q10: 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: In this question, we get 2 options, which are
(i) Either all 3 will go
Then, the remaining students in the class are: 25 – 3 = 22
The number of students remained to be chosen for party = 7
Number of ways to choose the remaining 22 students = ²²C₇

(ii) None of them will go
The students going will be 10.
Remaining students eligible for going = 22
The number of ways in which these 10 students can be selected are ²²C₁₀

The total number of ways in which students can be chosen is
= 170544 + 646646 = 817190

Q11: In how many ways can the letters of the word ASSASSINATION be arranged so that all the S’s are together?
Ans: In the given word ASSASSINATION, there are 4 ‘S’. Since all the 4 ‘S’ have to be arranged together, let us take them as one unit.
The remaining letters are= 3 ‘A’, 2 ‘I’, 2 ‘N’, T
The number of letters to be arranged is 9 (including 4 ‘S’).
Using the formula
where n is the 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