NCERT Solutions for Class 11 Maths Miscellaneous Exercise Chapter 6 Permutations and Combinations help students revise and apply the complete concepts of permutations and combinations in a clear and exam-oriented manner. This exercise includes a variety of questions that test both arrangement and selection-based problems.
Prepared according to the latest CBSE Class 11 Maths syllabus, this exercise focuses on applying formulas of permutations (nPr) and combinations (nCr) in different scenarios. It also helps students understand when to use permutation (order matters) and combination (order does not matter).
NCERT Solutions for Class 11 Maths Miscellaneous Exercise Chapter 6 Permutations and Combinations
Q.
How many 3-digit numbers can be formed from the digits 1, 2, 3, 4 and 5 assuming that
(i) repetition of the digits is allowed?
(ii) repetition of the digits is not allowed?
Q.
How many words, with or without meaning, each of 2 vowels and 3 consonants can be formed from the letters of the word DAUGHTER?
Q.
How many chords can be drawn through 21 points on a circle?
Q.
In how many ways can a team of 3 boys and 3 girls be selected from 5 boys and 4 girls?
Q.
Find the number of ways of selecting 9 balls from 6 red balls, 5 white balls and 5 blue balls if each selection consists of 3 balls of each colour.
Q.
Determine the number of 5 card combinations out of a deck of 52 cards if there is exactly one ace in each combination.
Q.
In how many ways can one select a cricket team of eleven from 17 players in which only 5 players can bowl if each cricket team of 11 must include exactly 4 bowlers?
Q.
A bag contains 5 black and 6 red balls. Determine the number of ways in which 2 black and 3 red balls can be selected.
Q.
In how many ways can a student choose a programme of 5 courses if 9 courses are available and 2 specific courses are compulsory for every student?
Q.
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?
Q.
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) atleast 3 girls ?
(iii) atmost 3 girls ?
Q.
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?
Q.
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?
Q.
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?
Q.
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?
Q.
Determine the number of 5-card combinations out of a deck of 52 cards if each selection of 5 cards has exactly one king.
Q.
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?
Q.
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?
Q.
Determine n if
(i) 2nC3 : nC3 = 12 : 1
(ii) 2nC3 : nC3 = 11 : 1
Q.
In how many ways can the letters of the word PERMUTATIONS be arranged if the
(i) words start with P and end with S,
(ii) vowels are all together,
(iii) there are always 4 letters between P and S?
Q.
How many 3-digit even numbers can be formed from the digits 1, 2, 3, 4, 5, 6 if the digits can be repeated?
Q.
Q.
How many 4-letter codes can be formed using the first 10 letters of the English alphabet, if no letter can be repeated?
Q.
How many 5-digit telephone numbers can be constructed using the digits 0 to 9 if each number starts with 67 and no digit appears more than once?
Q.
A coin is tossed 3 times and the outcomes are recorded. How many possible outcomes are there?
Q.
Given 5 flags of different colours, how many different signals can be generated if each signal requires the use of 2 flags, one below the other?
Q.
Evaluate: (i) 8! (ii) 4! – 3!
Q.
How many 3-digit numbers can be formed by using the digits 1 to 9 if no digit is repeated?
Q.
In how many of the distinct permutations of the letters in MISSISSIPPI do the four I’s not come together?
Q.
How many 4-digit numbers are there with no digit repeated?
Q.
How many 3-digit even numbers can be made using the digits 1, 2, 3, 4, 6, 7, if no digit is repeated?
Q.
Find the number of 4-digit numbers that can be formed using the digits 1, 2, 3, 4, 5 if no digit is repeated. How many of these will be even?
Q.
From a committee of 8 persons, in how many ways can we choose a chairman and a vice chairman assuming one person can not hold more than one position?
Q.
Findnifn−1P3:nP4=1:9
Q.
Q.
How many words, with or without meaning, can be formed using all the letters of the word EQUATION, using each letter exactly once?
Q.
How many words, with or without meaning can be made from the letters of the word MONDAY, assuming that no letter is repeated, if.
(i) 4 letters are used at a time,
(ii) all letters are used at a time,
(iii) all letters are used but first letter is a vowel?
Q.
In how many ways can the letters of the word ASSASSINATION be arranged so that all the S’s are together?
NCERT Solutions for Class 11 Maths Miscellaneous Exercise Chapter 6 Permutations and Combinations
The solutions are explained in a structured, step-by-step format so students can easily identify the correct approach and solve problems accurately in exams.
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 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.
2. How many words, with or without meaning, can be formed using all the letters of the word EQUATION at a time so that vowels and consonants occur together?
Ans: There are 5 vowels i.e. A, E, I, O and U and 3 consonants i.e. Q, T and N.
Since, vowels and consonants occur together, both (AEIOU) and (QTN) can be considered as single objects.
Thus, there are 5! permutations of 5 vowels taken all at a time and 3! permutations of 3 consonants taken all at a time.
Therefore, by multiplication principle, the number of words =
2! × 5! × 3! = 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?
Ans:
(i)
Out of 9 boys and 4 girls, a committee of 7 has to be formed.
Given: exactly 3 girls should be there in a committee, hence, number of boys = (7 - 3) = 4 boys only.
Therefore, number of ways
⁴C₃ × ⁹C₄ = (4! / 3!1!) × (9! / 4!5!)
= 4 × (9 × 8 × 7 × 6 × 5!) / (4 × 3 × 2 × 5!)
= 504
(ii)
Given: at least 3 girls are required in each committee. This can be done in 2 ways
(a) 3 girls and 4 boys or
(b) 4 girls and 3 boys
3 girls and 4 boys can be selected in
⁴C₃ × ⁹C₄ ways.
4 girls and 3 boys can be selected in
⁴C₄ × ⁹C₃ ways.
Thus, number of ways =
⁴C₃ × ⁹C₄ + ⁴C₄ × ⁹C₃
= 504 + 84
= 588
(iii)
Given: at most 3 girls in every committee. This can be done in 4 ways
(a) 3 girls and 4 boys
(b) 2 girls and 5 boys
(c) 1 girl and 6 boys
(d) No girl and 7 boys
3 girls and 4 boys can be selected in
⁴C₃ × ⁹C₄ ways
2 girls and 5 boys can be selected in
⁴C₂ × ⁹C₅ ways
1 girl and 6 boys can be selected in
⁴C₁ × ⁹C₆ ways
No girl and 7 boys can be selected in
⁴C₀ × ⁹C₇ ways
Thus, number of ways
= ⁴C₃ × ⁹C₄ + ⁴C₂ × ⁹C₅ + ⁴C₁ × ⁹C₆ + ⁴C₀ × ⁹C₇
= (4! / 3!1!) × (9! / 4!5!) + (4! / 2!2!) × (9! / 5!4!) + (4! / 1!3!) × (9! / 6!3!) + (4! / 0!4!) × (9! / 7!2!)
= 504 + 756 + 336 + 36
= 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 the list before the first word starting with E?
Ans: There are a total of 11 letters out of which A, I and N appears 2 times and other letters appear only once.
The words starting with A will be the words listed before the words starting with E.
Thus, words starting with letter A will have letter A fixed at its extreme left end.
And remaining 10 letters are rearranged all at a time.
In the remaining 10 letters, there are 2 I’s and 2 N’s.
Number of words starting with A =
10! / 2!2! = 907200
Therefore, required number of words is 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?
Ans: A number can be divisible by 10 only if its units digit is 0.
Hence, 0 is fixed at units place.
Therefore, the 5 vacant places can be filled by remaining 5 digits.
These 5 empty places can be filled in 5! ways.
Therefore, number of 6-digit numbers = 5! = 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 alphabet?
Ans: Given: 2 vowels and 2 consonants should be selected from the English alphabet.
We know that, there are 5 vowels.
Thus, number of ways of selecting 2 vowels out of 5 =
⁵C₂ = 5! / 2!3! = 10
Also we know that, there are 21 consonants.
Thus, number of ways of selecting 2 consonants out of 21 =
²¹C₂ = 21! / 2!19! = 210
Thus, total number of combinations of selecting 2 vowels and 2 consonants is
= 10 × 210 = 2100.
Every 2100 combination consists of 4 letters, which can be arranged in 4! ways.
Therefore, number of words =
2100 × 4! = 50400
7. In an examination, a paper consists of 12 Qs divided into 2 parts i.e., Part I and Part II, containing 5 and 7 Qs, respectively. A student is required to attempt 8 Qs in all, selecting at least 3 from each part. In how many ways can a student select the Qs?
Ans: Given: 12 Qs are divided into 2 parts – Part I and Part II consisting of 5 and 7 Qs respectively. A student must attempt 8 Qs with atleast 3 from each part. This can be done as:
-
3 Qs from part I and 5 Qs from part II.
-
4 Qs from part I and 4 Qs from part II.
-
5 Qs from part I and 3 Qs from part II.
The first case can be selected in
⁵C₃ × ⁷C₅ ways.
The second case can be selected in
⁵C₄ × ⁷C₄ ways.
The third case can be selected in
⁵C₅ × ⁷C₃ ways.
Thus, number of ways of selecting Q’s
= ⁵C₃ × ⁷C₅ + ⁵C₄ × ⁷C₄ + ⁵C₅ × ⁷C₃
= (5! / 2!3!) × (7! / 2!5!) + (5! / 4!1!) × (7! / 4!3!) + (5! / 5!0!) × (7! / 3!4!)
= 210 + 175 + 35 = 420
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.
Ans: Out 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. We know that in 52 cards, there are 4 kings.
Out of 4 kings, 1 king can be selected in
⁴C₁ ways.
Out of the remaining 48 cards, 4 cards can be selected in
⁴⁸C₄ ways.
Therefore, number of 5-card combinations =
⁴C₁ × ⁴⁸C₄ ways.
9. It is required to seat 5 men and 4 women in a row so that the women can occupy the even places.
How many such arrangements are possible?
Ans: Given: 5 men and 4 women should be seated so that women always occupy the even places.
Thus, the men can be seated in 5! ways.
For each arrangement the women can be seated only in even places.
Thus, the women can be seated in 4! ways.
Therefore, possible number of arrangements =
4! × 5! = 24 × 120 = 2880
10. From the 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: Given: 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
11. In how many words can the letters of the word ASSASSINATION be arranged so that all the S’s are together?
Ans: There are 3 A’s, 4 S’s, 2 I’s and all other letters appear only once in the word ASSASSINATION.
The given word should be arranged such that all the S’s are together.
The 4 S’s can be treated as a single object for time being. This single object with the remaining objects will be 10 objects together.
These 10 objects with 3 A’s, 2 I’s and 2 N’s can be arranged in
10! / 2!3!2! ways.
Therefore, number of ways of arranging the given word =
10! / 2!3!2! = 15120.
FAQs – Class 11 Maths Miscellaneous Exercise Chapter 6 Permutations and Combinations
Q1. What is the focus of the Miscellaneous Exercise in Chapter 6?
The Miscellaneous Exercise focuses on applying both permutations and combinations concepts in mixed and advanced problems.
Q2. What topics are covered in this exercise?
This exercise includes problems based on arrangements, selections, factorials, permutations, and combinations.
Q3. Why is this exercise important for exams?
This exercise helps in revising the entire chapter and includes different types of questions that are commonly asked in exams.
Q4. How can students prepare effectively for this exercise?
Students should revise all formulas, understand the difference between permutation and combination, and practice a variety of problems to improve accuracy and speed.