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NCERT Solutions for Class 11 Miscellaneous Exercise of Chapter 7 Permutation and Combination
NCERT Solutions for Class 11 Miscellaneous Exercise of Chapter 7 Permutation and Combination are created here for the preparation of the class 11 CBSE exams.These NCERT Solutions are the best study material for doing your work and class test, unit test and semi-annual and annual exams of class 11 CBSE.NCERT Solutions for Class 11 Miscellaneous Exercise of Chapter 7 Permutation and Combination are created by maths expert of CBSE.
NCERT Solutions for Class 11 Maths Chapter 7 Permutation and Combination
Exercise 7.1- Permutations and Combinations
Exercise 7.2 -Permutation and Combination
Exercise 7.3-Permutation and Combination
Exercise 7.4-Permutation and Combination
Miscellaneous Exercise-Permutation and Combination
NCERT Solutions for Class 11 Miscellaneous Exercise of Chapter 7 Permutation and Combination
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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. We have to find out the number of ways to build 5 letters word in which there are 2 vowels and 3 consonants.
The vowels in the word DAUGHTER are A,U,E (i.e 3)
The consonant in the word DAUGHTER are D,G, H,T,R (i.e 5)
The number of ways selecting 2 vowels out of 3 vowels is = 3c2
The number of ways selecting 3 consonantss out of 5 vowels is = 5c3
Total number of ways selecting 2 vowels and 3 consonants is =3c2× 5c3
= 3 × 10 = 30
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.We have to find out the number of ways to form 5 letters word in which there are 2 vowels and 3 consonants.
The vowels in the word EQUATION are E,U,A,I,O (i.e 5)
The consonant in the word EQUATION are Q,T, N (i.e 3)
The number of ways selecting 5 vowels out of 5 vowels is = 5p5
The number of ways selecting 3 consonantss out of 3 consonant is = 3p3
The number of ways selecting vowels and consonants is =2
Total number of ways selecting 2 vowels and 3 consonants in the word of 5 letters word
= 5p5 ×3p3
= 5× 4× 3×2×3×2×2 = 1440
Therefore, the number of ways that letters of the word EQUATION are arranged at a time so that the vowels and consonants occur together is 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) The number of ways choosing three girls out of 4 girls is =4c3
The number of ways choosing rest of members (7-3=4) out of 9 boys is =9c4
The number of ways choosing 7 members commitee =4c3 × 9c4
= 4 ×9 ×2×7= 504
Hence the number of ways of selecting 7 members in forming a commitee is 504.
(ii) The number of ways selecting a commitee when at least 3 girls has to be taken is computed as follows
The mumber of ways selecting 3 girls out of 4 girls and selecting rest of members(7-3=4) boys out of 9 boys= 4c3 ×9c4
The mumber of ways selecting 4 girls out of 4 girls and selecting rest of members(7-4=3) boys out of 9 boys = 4c4 ×9c3
The total number of ways selecting at least 3 girls for a commitee is = 4c3 ×9c4+4c4 ×9c3
= 504 + 84 = 588
Hence total number of ways selecting at least 3 girls for a commitee of 7 members is 588
(iii) The number of ways selecting a commitee when at most 3 girls has to be taken is computed as follows
The mumber of ways selecting 3 girls out of 4 girls and selecting rest of members(7-3=4) boys out of 9 boys= 4c3 ×9c4
The mumber of ways selecting 2 girls out of 4 girls and selecting rest of members(7-2=5) boys out of 9 boys = 4c2 ×9c5
The mumber of ways selecting 1 girl out of 4 girls and selecting rest of members(7-1=6) boys out of 9 boys = 4c1 ×9c6
The mumber of ways selecting 0 girls out of 4 girls and selecting rest of members 7 boys out of 9 boys = 4c0 ×9c7
The total number of ways selecting at least 3 girls for a commitee is = 4c3 ×9c4+4c2 ×9c5
+4c1 ×9c6 +4c0 ×9c7
=504 + 6× 126 +4×84 + 36 = 504 + 756+ 336+ 36 = 1632
Hence the number of ways selecting at most 3 girls for a 7 member commitee is 1632
Q4.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?
Ans.The number of letters in the world EXAMINATION is = 11
The frequencies of the repeating letters are =2A, 2I,2N
Since we have to form all the words before the first word starting with E ,there is only one letter A before E,A is fixed at first place, therefore, the number of permutations of remaining letters is computed as following.
= 10× 9×8 ×7×6×5 ×3 ×2 = 907200
907200 words are there in this list before the first word starting with E
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 of selecting 6 digit numbers from the digits 0,1,3,5,7 and 9 which are divisible by 10 ,hence at the once place 0 is fixed,the permutation of remaining 5 digits is =5p5
Hence the required number of 6 digit numbers divisible by 6 are 120
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.The number of ways selecting 2 vowels from 5 vowels is =5c2
The number of ways selecting 2 consonants from 21 consonant is =21c2
The number of ways selecting 2 different vowels and 2 different consonants =5c2×21c2
= 5×2 × 21×10 =2100
Each of these letters 2 vowels and 2 consonant can be arranged by 4! ways
4 ! = 4 ×3×2 =24
Total number of the words with two different vowels and 2 different consonants can be formed = 2100 ×24 =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.Part I contains 5 questions and part II contains 7 questions,students are required to attempt 8 questions in all , students have to select at least 3 questions from each part
The number of ways selecting at least 3 questions from part I is computed as follows
The number of ways of selecting at least 3 questions from part I= Number of ways selecting 3 questions from part I and selecting 5 questions from part II + Number of ways selecting 4 questions from part I and selecting 4 questions from part II +Number of ways selecting 5 questions from part I and selecting 3 questions from part II
Number of ways selecting 3 questions from part I and selecting 5 questions from part II= 5c3 × 7c5
Number of ways selecting 4 questions from part I and selecting 4 questions from part II =5c4 × 7c4
Number of ways selecting 5 questions from part I and selecting 3 questions from part II=5c5 × 7c3
Hence the number of ways of selecting at least 3 questions from part I = 210+175 +35 =420
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.The number of ways selecting 1 king out of 4 kings is = 4c1
The number of ways choosing remaining 4 cards out of remaing (52-4 = 48) cards = 48c4
The number of 5 card combinations in which each combination has a king =4c1 × 48c4
Q9.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. 5 men and 4 women in a row can be arranged with the condition that women occupy even places as follows
MWMWMWMWM
The number of ways selecting 5 men in 5 odd places = 5p5
The number of ways selecting 4 women in 4 even places = 4p4
Total number of ways selecting 5 men 4 women with the condition that women occupy even places is = 120 ×24 = 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. There are 25 students in the class , among these students, 10 have to be chosen for the excursion party , in this selection following case arises.
Case 1. When 3 students join the excursion party
The number of ways selecting remaining (10 -3 = 7) students out of the total remaining students (25-3 =22) of the class =22p7
Case 2. When 3 students don’t join the excursion party
The number of ways selecting 10 students out of the total 25 students of the class =25p10
Total number of ways selecting 10 students in both cases is = 22p7+25p10
=10544 +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.Total letters in the given world ASSASSINATION are = 13
If all S’s occurs together they are supposed as a single letter,therefore let’s find the number of ways arranging the (13 -4+1 = 10 ) letters
Among 9 letters the frequencies of the letters are as follows
3A,2N,2I
Hence the number of ways to arrange the letters of ASSASSINATION so that all the S’s are together?
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| Chapter 1-Sets | Chapter 9-Sequences and Series |
| Chapter 2- Relations and functions | Chapter 10- Straight Lines |
| Chapter 3- Trigonometry | Chapter 11-Conic Sections |
| Chapter 4-Principle of mathematical induction | Chapter 12-Introduction to three Dimensional Geometry |
| Chapter 5-Complex numbers | Chapter 13- Limits and Derivatives |
| Chapter 6- Linear Inequalities | Chapter 14-Mathematical Reasoning |
| Chapter 7- Permutations and Combinations | Chapter 15- Statistics |
| Chapter 8- Binomial Theorem | Chapter 16- Probability |
CBSE Class 11-Question paper of maths 2015
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| Chapter 1-Relations and Functions | Chapter 9-Differential Equations |
| Chapter 2-Inverse Trigonometric Functions | Chapter 10-Vector Algebra |
| Chapter 3-Matrices | Chapter 11 – Three Dimensional Geometry |
| Chapter 4-Determinants | Chapter 12-Linear Programming |
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| Chapter 6- Application of Derivation | CBSE Class 12- Question paper of maths 2021 with solutions |
| Chapter 7- Integrals | |
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