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

$=\frac{3!}{2!\left ( 3-2 \right )! }\times \frac{5!}{3!\left ( 5-3 \right )!}$

$=\frac{3\times 2}{2 }\times \frac{5\times 4\times 3!}{3!\times 2}$

= 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

^{= 5}p₅×³p₃

$=\frac{5!}{\left ( 5-5 \right )!}\times \frac{3!}{\left ( 3-3 \right )!}\times 2$

= 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

$=\frac{4!}{3!\left ( 4-3 \right )!}\times \frac{9!}{4!\left ( 9-4 \right )!}$

$=\frac{4\times 3!}{3!}\times \frac{9\times 8\times 7\times 6\times 5!}{4\times 3\times 2\times 5!}$

= 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

$=504+\frac{4!}{4!\left ( 4-4 \right )!}\times \frac{9!}{3!\left ( 9-3 \right )!}$

$=504+ \frac{9\times 8\times 7\times 6!}{3\times 2\times 6!}$

= 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

$=504+\frac{4! }{2!\left (4-2 \right )!}\times \frac{9! }{5!\left ( 9-5 \right )!}+\frac{4!}{1!\left ( 4-1 \right )!}\times \frac{9!}{6!\left ( 9-6 \right )!}+\frac{9!}{7!\left ( 9-7 \right )!}$

$=504+\frac{4\times 3\times 2}{2\times 2}\times \frac{9\times 8\times 7\times 6\times 5!}{5!\times 4\times 3\times 2}+\frac{4\times 3!}{3!}\times \frac{9\times 8\times 7\times 6!}{6!\times 3\times 2}+\frac{9\times 8\times 7!}{7!\times 2}$

=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.

$=\frac{10!}{2!\times 2!}$

$=\frac{10\times 9\times 8\times 7\times 6\times 5\times 4\times 3\times 2}{2\times 2}$

= 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

$=\frac{5!}{\left ( 5-5 \right )!}=5!=5\times 4\times 3\times 2=120$

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

$=\frac{5!}{2!\left ( 5-2 \right )!}\times \frac{21!}{2!\left ( 21-2 \right )!}$

$=\frac{5\times 4\times 3\times 2}{2\times 3\times 2}\times \frac{21\times 20\times 19!}{2\times 19!}$

= 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

$=\frac{5!}{3!\times 2!}\times \frac{7!}{5!\times 2!}=\frac{5\times 4\times 3!}{3!\times 2}\times \frac{7\times 6\times 5!}{5!\times 2}=210$

Number of ways selecting 4 questions from part I and selecting 4 questions from part II =⁵c₄× ⁷c₄

$=\frac{5!}{4!\times 1!}\times \frac{7!}{4!\times 3!}=\frac{5\times 4!}{4!}\times \frac{7\times 6\times 5\times 4!}{4!\times 3\times 2}=5\times 7\times 5=175$

Number of ways selecting 5 questions from part I and selecting 3 questions from part II=⁵c₅× ⁷c₃

$=\frac{5!}{5!\times 0!}\times \frac{7!}{3!\times 4!}=1\times \frac{7\times 6\times 5\times 4!}{3\times 2\times 4!}=35$

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

$=\frac{4!}{1!\times 3!}\times \frac{48!}{4!\times 44!}$

$\frac{4\times 3!}{3!}\times \frac{48\times 47\times 46\times 45\times 44!}{4\times 3\times 2\times 44!}=4\times 2\times 47\times 46\times 45=778320$

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

$\frac{5!}{\left ( 5-5 \right )!}=5!=5\times 4\times 3\times 2=120$

The number of ways selecting 4 women in 4 even places = ⁴p₄

$\frac{4!}{\left ( 4-4 \right )!}=4!= 4\times 3\times 2=24$

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

$=\frac{22!}{7!\left ( 22-7 \right )!}+\frac{25!}{10!\left ( 25-10 \right )!}$

$=\frac{22\times 21\times 20\times 19\times 18\times 1\times 16\times 15!}{7\times 6\times 5\times 4\times 3\times 2\times 15!}+\frac{25\times 24\times 23\times 22\times 21\times 20\times 19\times 18\times 17\times 16\times 15!}{10\times 9\times 8\times 7\times 6\times 5\times 4\times 3\times 2\times 15!}$

=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?

$=\frac{10!}{3!\times 2!\times 2!}$

$=\frac{10\times 9\times 8\times 7\times 6\times 5\times 4\times 3!}{3!\times 2\times 2}=10\times 9\times 8\times 7\times 6\times 5=151200$

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

NCERT Solutions for Class 11 Maths Chapter 7 Permutation and Combination