NCERT Solutions for Class 11 Miscellaneous Exercise of Chapter 7 Permutation and Combination - Future Study Point

NCERT Solutions for Class 11 Miscellaneous Exercise of Chapter 7 Permutation and Combination

class 11 maths ex miscellaneous permutation

NCERT Solutions for Class 11 Miscellaneous Exercise of Chapter 7 Permutation and Combination

class 11 maths ex miscellaneous permutation

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

= 5p×3p

= 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 =4c

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 =4c× 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   = 4c×9c

The total number of ways selecting at least 3 girls for a commitee is = 4c3 ×9c4+4c×9c

= 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 ×9c

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 ×9c

The mumber of ways selecting 0 girls out of 4 girls and selecting rest of members 7 boys out of 9 boys   = 4c0 ×9c

The total number of ways selecting at least 3 girls for a commitee is = 4c3 ×9c4+4c2 ×9c

+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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