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## Math Assignment  Class XI  Ch -07PERMUTATIONS & COMBINATIONS

Extra questions of chapter 07 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

Note: Solution hint to the difficult questions are given below the Assignment

Question 1

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

Ans: nr

Question 2

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

Ans: n = a + b

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

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

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

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

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

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

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

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

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

Question 16

Find the number of arrangements of the letters of the word ‘REPUBLIC’ in how many of these arrangements
Ans: 720

b) All the vowels occur together
Ans: 4320

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

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

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

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

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

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SOLUTION HINTS
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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

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)

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

Solution Hint: 10

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

Solution Hint: 11(i)

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

Solution Hint: 11(ii)  2! ✕ 3!

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

Solution Hint:13

ATQ   $^{n}C_{2}$  -n = 44  ⇒  n = 11

No of triangles formed =  $^{n}C_{3}$

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

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

Solution Hint: 16(a)

No. of vowels = 3

No. of consonants = 5

3 vowels can be arranged in 3! ways

5 consonants can be arranged in 5! 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

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

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

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

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