Mathematics Class 10 Lab Manual | 21 Lab Activities

Mathematics Lab Manual Class X   lab activities for class 10 with complete observation Tables strictly according to the CBSE syllabus also very useful & helpful for the students and teachers.

Math Assignment  Class XI  Ch -06

Extra questions of chapter 06 Permutations & Combinations class 11  with answers and  hints to the difficult questions, strictly according to the cbse syllabus. Important and useful math. assignment for the students of class 11

PERMUTATIONS & COMBINATIONS (XI)

Strictly according to the CBSE and DAV Board

Question 1

Find the number of permutations of n different objects taken r at a time when repetition is allowed.

Ans: nr

Question 2

(i) If  $^{n}C_{a}=^{n}C_{b}$ , then find 'n'

Ans: n = a + b

(ii)  If  $^{n^{2}-n}C_{2}=\:^{n^{2}-n}C_{4}$ , then find the value of n

(iii) If nCr-1 = 36, nCr = 84, nCr+1 = 126, then find the value of  rC

Question 3

What is the relation between  $^{n}P_{r}$  and $^{n}C_{r}$

Ans:  $^{n}P_{r}$ = r!  $^{n}C_{r}$

Question 4

If  $^{22}P_{r+1}$ : $^{20}P_{r+2}$ = 11 : 52, then find r

Ans: r = 7

Solution Hint: 4

Solution Hint

Solving these two ratios we get the result

(21 - r)(20 - r)(19 - r) = 14✕13✕12

Comparing these corresponding factors we get r = 7

Question 5

Write the number of diagonals in a polygon with 7 sides.

Ans: 14

Question 6

If  $^{n}C_{12}$ = $^{n}C_{8}$ , then find the value of  $^{n}C_{3}$

Ans: n = 20,  $^{n}C_{3}$ = 1140

Question 7

n points are marked on a circle such that number of chords that can be formed by joining these points is 35 more than the number of points. Find the value of n?

Ans: 10
Solution Hint: 7

No. of chord can be formed by joining n points

= $^{n}C_{2}$ =  $\frac{n(n-1)}{2}$

ATQ :  $\frac{n(n-1)}{2}$  - n = 35

⇒ n- 3n - 70 = 0  ⇒ n = 10 and n = - 7(rejected)

Question 8

How many words with or without meaning can we formed by using all the letters of the word mathematics so that
a) Each word begin with M and end with S.
Ans: 90720

b) Their are always 6 letters between M and M
Ans: 362880

Solution Hint: 8(b)

 M * * * * * * M

Here M can take 4 positions

Leaving M and M we left with 9 letters with 2T tand 2A

So remaining 9 letters can be arranged in  $\frac{9!}{2!2!}$

Total word formed = $\frac{9!}{2!2!}$ ✕ 4 = 362880

Question 9

How many natural numbers less than 1000 can be formed with digits 1, 2, 3, 4, 5 if
(i) No digit is repeated ?
Ans: 60+20+5 = 85

(ii) Repetition of digits is allowed ?
Ans: 125+25+5 = 155

Question 10

A box contains 5 red and 6 white balls. In how many ways can 6 balls be selected so that there are at least 2 balls of each color ?
Ans: 425

Solution Hint: 10

Solution Hint:  $^{5}C_{2}\times ^{6}C_{4}+ ^{5}C_{3}\times ^{6}C_{3}+^{5}C_{4}\times ^{6}C_{2}$

Question 11

A student has 4 library tickets and, in the library, there are 2 language books, 4 subject specific books and 3 fictional books of his interest. Of these 9 books, he chooses exactly 2 subject specific books and 2 other books.
Based on the above information, answer the following questions:
(i) In how many ways can he borrow the four books?
Ans: 60

(ii) Once selected, in how many ways, can he now arrange the borrowed books in his bookshelf so that the subject specific books are always kept together?
Ans: 12

Solution Hint: 11(i)

Solution Hint: $^{4}C_{2}\times ^{5}C_{2}$

Solution Hint: 11(ii)  2! ✕ 3!
Question 12

How many words with or without meaning can be formed using the letters of the word DAUGHTER if
a) All vowels are never together
Ans: 36000

b) Vowels occupy odd places.
Ans: 2880

Solution Hint: 12(a)

Total number of words formed = 8! = 40320

No. of words when all vowels are together = 6!  3! = 4320

No of words when vowels are not together = 40320 - 4320 = 36000

Solution Hint : 12(b)
 Total Places in DAUGHTER * * * * * * * * Odd & Even Places E O E O E O E O Odd Places 4 3 2 1

Total letters = 8

No. of consonants = 5

5 vowels can be arranged in  5! ways

No. of vowels = 3

Odd places can be filled with 3 vowels in 4P3 ways

Total words formed with vowels at odd places = 4P3 X 5! = 2880

Question 13

A polygon has 44 diagonals if n denotes the number of vertices of a polygon, find the value of n. Hence find the number of triangles that can be formed by joining these n points
Ans: 165

Solution Hint:13

ATQ   $^{n}C_{2}$  -n = 44  ⇒  n = 11

No of triangles formed =  $^{n}C_{3}$
Question 14

How many words with or without meaning, containing 3 vowels and 2 consonants can be formed using the letters of the word EQUATION ?
Ans: 3600

Solution Hint: 14

3 vowels can be selected from 5 in $^{5}C_{3}$ ways

2 consonants can be selected from 3 in  $^{3}C_{2}$ ways

5 letters can be arranged in 5! ways

Total words formed: $^{5}C_{3}\times ^{3}C_{2}\times 5!$

Question 15

How many words with or without meaning, can be formed using the letters of the word ALLAHABAD so that vowels are never together ?
Ans: 7200

Solution Hint: 15

Total letters in ALLAHABAD = 9 with 4A, 2L

Total words formed =  $\frac{9!}{4!2!}$ = 7560

No. of vowels = 4 (AAAA)

No. of Consonant = 5 with 2L

When all the vowels in one block then

Total letters = 9 - 4 + 1 = 6

Total words with all vowels together = $\frac{6!}{2!}\times \frac{4!}{4!}$ = 360

No. of words so that vowels are not together = 7560 - 360 = 7200

Question 16

Find the number of arrangements of the letters of the word ‘REPUBLIC’ in how many of these arrangements
Ans: 15120

b) All the vowels occur together
Ans: 4320

c) What is the significance of republic day in our life

Solution Hint: 16(a)

No. of vowels = 3

No. of consonants = 5

First letter  (vowels) can be arranged in 3 ways

Remaining 7 letters can be arranged in 7! ways.

Solution Hint: 16(b)

Let all the vowels are in one block then

Total letters = 8 - 3 + 1 = 6

6 letters can be arranged in 6! ways

3 vowels can be arranged in 3! ways

Total words = 6! ✕ 3! = 4320

Question 17
How many words with or without meaning can be made using the letters of the word ‘FESTIVAL’ taken all at a time. In how many of them vowels occupy odd places.

Ans: Total words formed = 40320
No. of words with vowels at odd places = 2880
Solution Hint: 17

Total letters in FESTIVAL = 8

Total words formed = 8! = 40320

No. of consonants = 5

Consonants can be arranged in 5! ways

No. of vowels = 3

When vowels placed at odd places then vowels can occupies only 4 positions

 Total Places in FESTIVAL * * * * * * * * Odd & Even Places E O E O E O E O Odd Places 4 3 2 1

So 3 vowels can be placed at 4 places (odd places) in $^{4}P_{3}$ ways

Total words formed with vowels at odd places = 5! ✕ $^{4}P_{3}$ = 2880

Question 18

A group consists of 5 girls and 6 boys in how many ways can a team of 4 members be selected if the team has
a) At least one boy and one girl
Ans: 310 ways

b) At most two boys
Ans: 215

Question 19
Find the number words formed using all the letters of the word 'MATHEMATICS' such that
a) All vowels are together.
Ans: 120960

b) No two vowels are together.
Ans: 1058400
Solution Hint 19(a)

Given word is MATHEMATICS

Total letters = 11

No. of vowels 4 (AEAI) with 2A

No. of consonant = 7 (M, T, H, M, T, C, S) with 2M and 2T

When all vowels are together then

Total letters = 11-4+1 = 8 with 2M and 2T

Total words with all vowels together = $\frac{8!}{2!2!}\times \frac{4!}{2!}$ = 120960

Solution Hint 19(b)

7 consonants can be arranged in  $\frac{7!}{2!2!}$ ways

Vowels can be placed at the * marked places: *M*T*H*M*T*C*S*

⇒ Vowels can be placed at 8 places

So 8 places can be filled with 4 vowels in  $^{8}P_{4}$  ways

But vowel 'A' is repeated

So vowels can be arranged in  $\frac{^{8}P_{4}}{2!}$  ways

Hence total words formed with no vowel together:

= $\frac{7!}{2!2!}$ ✕ $\frac{^{8}P_{4}}{2!}$ = 1058400

Question 20

A student wants to make words with or without meaning by using all letters of the word "ORIGIN"

(i) How many maximum number of words, student can make?

Ans: 360

(ii) If these words are written as in dictionary. What will be the 181th word?

Ans: NGIIOR

Solution Hint: 20(ii)

Dictionary order of ORIGIN = GIINOR

No of words start with G = $\frac{5!}{2!}$ = 60

Total 60+120 = 180

So 181th word is = NGIIOR

Question 21
There are 12 points in a plane, no three of which are in the same straight line except 5 which are in the same line. Find :-
(i) Number of lines.

Ans: 57

(ii) Number of triangles. which can be formed by joining them.

Ans: 210

Solution Hint 21(i)

No. of lines formed by joining the 12 points, taking 2 at a time = 12C2
No of lines formed by joining the 5 points taking 2 at a time = 5C2
But 5 collinear points when joined pairwise, give only one line
The required number of lines = 12C2 − 5C+ 1
= 66 – 10 + 1
= 57

Solution Hint 21(ii)

No of triangles formed by joining 12 points by taking 3 points at a time  = 12C3
No. of triangles formed by joining 5 points by taking 3 points at a time        = 5C3
But there is no triangle formed by joining 3 points out of 5 collinear points
The required number of triangles formed

12C3 − 5C3
= 220 – 10 = 210

Question 22

How many four letter words can be formed out of the letters of the word ' MATHEMATICS'

Solution Hint

The word MATHEMATICS has 2M, 2T, 2A and 1 each of H, E, I, C and S.

2M, 2T, 2A are three groups of repeated letters.

Now 4 letters can be chosen in 3 ways.

Case 1:  2 alike of one kind and 2 alike of the second kind from three groups of repeated letters.

Number of words = $^{3}C_{2}\times\frac{4!}{2!2!}=18$

Case II:  2 alike of one kind from 3 groups and 2 different from remaining seven

Number of words = $^{3}C_{1}\times^{7}C_{2}\times\frac{4!}{2!}=756$

Case III: All different letters

Number of words =  $^{8}C_{4}\times4!=1680$

Total number of words = 18 + 756 + 1680 = 2454

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