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
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 , then find 'n'
Ans: n = a + b
(ii) If
Answer: n = 3
(iii) If nCr-1 = 36, nCr = 84, nCr+1 = 126, then find the value of rC2 Answer:
What is the relation between and
Ans: = r! ✕
If :
= 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
Question 7n 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?
No. of chord can be formed by joining n points
= =
ATQ : - n = 35
⇒ n2 - 3n - 70 = 0 ⇒ n = 10 and n = - 7(rejected)
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 |
Total word formed = ✕ 4 = 362880
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
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:
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:
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
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
-n = 44 ⇒ n = 11
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
3 vowels can be selected from 5 in ways
2 consonants can be selected from 3 in ways
5 letters can be arranged in 5! ways
Total words formed:
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
Total letters in ALLAHABAD = 9 with 4A, 2L
Total words formed = = 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 = = 360
No. of words so that vowels are not together = 7560 - 360 = 7200
Question 16Find the number of arrangements of the letters of the word ‘REPUBLIC’ in how many of these arrangements
a) Does the word start with the vowel.
Ans: 15120
b) All the vowels occur together
Ans: 4320
c) What is the significance of republic day in our life
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.
Total words start with vowels = 3 ✕ 7! = 15120
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
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 ways
Total words formed with vowels at odd places = 5! ✕ = 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
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 = = 120960
Solution Hint 19(b)
7 consonants can be arranged in 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 ways
But vowel 'A' is repeated
So vowels can be arranged in ways
Hence total words formed with no vowel together:
= ✕
= 1058400
Question 20
Solution Hint: 20(ii)
Dictionary order of ORIGIN = GIINOR
No of words start with G = = 60
No. of words start with I = 5! = 120
Total 60+120 = 180
181'th word is the first word start with N
So 181th word is = NGIIOR
Ans: 57
(ii) Number of triangles. which can be formed by joining them.
Ans: 210
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 − 5C2 + 1
= 66 – 10 + 1
= 57
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 =
Case II: 2 alike of one kind from 3 groups and 2 different from remaining seven
Number of words =
Case III: All different letters
Number of words =
Total number of words = 18 + 756 + 1680 = 2454
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