Commerce Exam  >  Commerce Notes  >  Mathematics (Maths) Class 11  >  NCERT Exemplar: Permutations and Combinations- 2

NCERT Exemplar: Permutations and Combinations- 2 | Mathematics (Maths) Class 11 - Commerce PDF Download

Fill in the Blanks
Q.41. If nPr = 840, nCr = 35, then r = ______.
Ans.
Given that nPr = 840 and nCr = 35
NCERT Exemplar: Permutations and Combinations- 2 | Mathematics (Maths) Class 11 - Commerce
Dividing eq. (i) by eq. (ii) we get
NCERT Exemplar: Permutations and Combinations- 2 | Mathematics (Maths) Class 11 - Commerce
⇒ r! = 24 ⇒ r! = 4 x 3 x 2 x 1
⇒ r! = 4!  ∴ r = 4
Hence the value of the filler is 4.

Q.42. 15C8 + 15C915C615C7 = ______.
Ans.
15C+ 15C915C6 15C= 15C15– 8 + 15C15– 9 15C615C7 
[∴ nCr = nCn – r]
= 15C7 + 15C615C615C7 = 0
Hence, the value of the filler is 0.

Q.43. The number of permutations of n different objects, taken r at a line, when repetitions are allowed,  is ______.
Ans.
Number of permutation of n different objects, taken r at a time is nr.

Q.44. The number of different words that can be formed from the letters of the word INTERMEDIATE such that two vowels never come together is ______. [Hint: Number of ways of arranging 6 consonants of which two are alike isNCERT Exemplar: Permutations and Combinations- 2 | Mathematics (Maths) Class 11 - Commerce
and number of ways of arranging vowelsNCERT Exemplar: Permutations and Combinations- 2 | Mathematics (Maths) Class 11 - Commerce
Ans.
Total number of words is INTERMEDIATE = 12
which have 6 vowels and 6 consonants
If two vowels never come together then we can arrange as under
V C V C V C V C V C V C V
Here, vowels are IEEIAE where 2 I’s and 3 E’s are there.
∴ Number of ways of arranging vowels =NCERT Exemplar: Permutations and Combinations- 2 | Mathematics (Maths) Class 11 - Commerce
Consonants are NTRMDT where 2T’s are there
∴ Number of ways arranging consonants =NCERT Exemplar: Permutations and Combinations- 2 | Mathematics (Maths) Class 11 - Commerce
NCERT Exemplar: Permutations and Combinations- 2 | Mathematics (Maths) Class 11 - Commerce
So, the total number of words are
= 420 x 360 = 151200
Hence, the value of the filler is 151200.

Q.45. Three balls are drawn from a bag containing 5 red, 4 white and 3 black balls. The number of ways in which this can be done if at least 2 are red is ______.
Ans.
We have 5 red, 4 white and 3 black balls out of which atleast 2 red  balls are to be drawn
∴ Number of ways = 5C2 x 7C1 + 5C3 
= 10 x 7 + 10 = 70 + 10 = 80
Hence, the value of the filler = 80.

Q.46. The number of six-digit numbers, all digits of which are odd is ______.
Ans.
Out of the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 the odd digits are 1, 3, 5, 7, 9.
Therefore number of 6 digit numbers = (5)6 
Hence, the value of the filler is (5)6.

Q.47. In a football championship, 153 matches were played . Every two teams played one match with each other. The number of teams, participating in the championship is ______.
Ans.
Let the number of participating teams be n
Given that every two teams played one match with each other.
∴ Total number of matches played = nC2 
So nC2 = 153
NCERT Exemplar: Permutations and Combinations- 2 | Mathematics (Maths) Class 11 - Commerce
⇒ n2 – n – 306 = 0 ⇒ n2 – 18n + 17n – 306 = 0
⇒ n(n – 18) + 17 (n – 18) = 0 ⇒ (n - 18) (n + 17) = 0
⇒ n – 18 = 0 and n + 17 = 0
⇒ n = 18, n ≠ - 17
Hence, the value of the filler is 18.

Q.48. The total number of ways in which six ‘+’ and four ‘–’ signs can be arranged in a line such that no two signs ‘–’ occur together is ______.
Ans.
The  following may be the arrangement of (–) and (+)
(–) (+) (–) (+) (–) (+) (–) (+) (–) (+) (–) (+) (–)
Therefore, ‘+’ sign can be arranged only is 1 way because all are identical.
and 4(–) signs can be arranged at 7 places in 7C4 ways
∴ Total number of ways = 7C4 x 1 =NCERT Exemplar: Permutations and Combinations- 2 | Mathematics (Maths) Class 11 - Commerce
Hence, the value of the filler is 35.

Q.49. A committee of 6 is to be chosen from 10 men and 7 women so as to contain atleast 3 men and 2 women. In how many different ways can this be done if two particular women refuse to serve on the same committee.
[Hint:At least 3 men and 2 women: The number of ways = 10C3 × 7C3 + 10C4 × 7C2For 2 particular women to be always there: the number of ways = 10C4 + 10C3 × 5C1The total number of committees when two particular women are never together = Total – together.]
Ans.
We have 10 men and 7 women out of which a committee of 6 is to be formed which contain atleast 3 men and 2 women
Therefore, Number of ways = 10C3 x 7C3 + 10C4 x 7C2
NCERT Exemplar: Permutations and Combinations- 2 | Mathematics (Maths) Class 11 - Commerce
= 120 x 35 + 210 x 21 = 4200 + 4410 = 8610
If 2 particular women to be always present, then the number
of ways = 10C4 x 5C0 + 10C3 x 5C1
NCERT Exemplar: Permutations and Combinations- 2 | Mathematics (Maths) Class 11 - Commerce
= 210 + 600 = 810
∴ Total number of committee = 8610 - 810 = 7800
Hence, the value of the filler is 7800.

Q.50. A box contains 2 white balls, 3 black balls and 4 red balls. The number of ways three balls be drawn from the box if at least one black ball is to be included in the draw is ______.
Ans.
We have 2 white, 3 black and 4 red balls It is given that atleast 1 black ball is to be included.
∴ Required number of ways = 3C1 x 6C2 + 3C2 x 6C1 + 3C3
= 3 x 15 + 3 x 6 + 1 = 45 + 18 + 1 = 64
Hence, the value of the filler is 64.

TRUE/FALSE STATEMENT
Q.51. There are 12 points in a plane of which 5 points are collinear, then the number of lines obtained by joining these points in pairs is 12C25C2.
Ans.
Required number of lines = 12C25C2 + 1
Hence, the given statement is ‘False’

Q.52. Three letters can be posted in five letterboxes in 35 ways.
Ans.
Given that 3 letters are to be posted in 5 letter boxes
∴ Required number of ways = 53 = 125
Hence, the given statement is ‘False’

Q.53. In the permutations of n things, r taken together, the number of permutations in which m particular things occur together is n–mPr–m × rPm.

Ans.
Arrangement of n things, r taken at a time in which m things occur together.
So, number of object excluding m object = (r – m)
Here, we first arrange (r – m + 1) object
∴ Number of arrangements = (r - m + 1)!
m objects can be arranged in m! ways
So, the required number of arrangements  = (r - m + 1)! x m!
Hence, the given statement is ‘False’.

Q.54. In a steamer there are stalls for 12 animals, and there are horses, cows and calves (not less than 12 each) ready to be shipped. They can be loaded in 312 ways.
Ans.
There are 3 types of animals horses, cows and calves not less than 12 each.
So, number of ways of loading = 312
Hence, the given statement is ’True‘.

Q.55. If some or all of n objects are taken at a time, the number of combinations is 2n–1.
Ans.
When some or all objects, taken at a time, then the number of selection will be
nC1 + nC2 + nC3 + … + nCn = 2n –1
[∴ nC0 + nC1 + nC2 + … + nCn = 2n]
Hence, the given statement is ‘True’.

Q.56. There will be only 24 selections containing at least one red ball out of a bag containing 4 red and 5 black balls. It is being given that the balls of the same colour are identical.
Ans.
We have 4 red and 5 black balls in a box and atleast one red ball is to be drawn
∴ Number of selection = [(4 + 1) (5 + 1) - 1] - 5 = [5 x 6 - 1] - 5
29 - 5 = 24
Hence, the given statement is ‘True’.

Q.57. Eighteen guests are to be seated, half on each side of a long table. Four particular guests desire to sit on one particular side and three others on other side of the table. The number of ways in which the seating arrangements can be made isNCERT Exemplar: Permutations and Combinations- 2 | Mathematics (Maths) Class 11 - Commerce
[Hint: After sending 4 on one side and 3 on the other side, we have to select out of 11; 5 on one side and 6 on the other. Now there are 9 on each side of the long table and each can be arranged in 9! ways.]
Ans.
When 4 guests sit can one side and 3 on the other side, we have to select out of 11. 5 sit one one side and 6 sit on the other side.
Now, remaining selecting on one half side = 18 – 4 – 3C5 = 11C5
and the other half side = (11 – 5)C6 = 6C6
So, the total arrangements = 11C5 x 9! x 6C6 x 9!
NCERT Exemplar: Permutations and Combinations- 2 | Mathematics (Maths) Class 11 - Commerce
Hence, the given statement is ’True‘.

Q.58. A candidate is required to answer 7 questions out of 12 questions which are divided into two groups, each containing 6 questions. He is not permitted to attempt more than 5 questions from either group. He can choose the seven questions in 650 ways.

Ans.
The candidate may attempt in following manner
NCERT Exemplar: Permutations and Combinations- 2 | Mathematics (Maths) Class 11 - Commerce
So, the number of attempts of 7 questions
= 6C2 x 6C5 + 6C3 x 6C4 + 6C4 x 6C3 + 6C5 x 6C2
= 2[6C2 x 6C5 + 6C3 x 6C4]
= 2 [15 x 6 + 20 x 15] = 2[90 + 300] = 2 x 390 = 780.

Q.59. To fill 12 vacancies there are 25 candidates of which 5 are from scheduled castes. If 3 of the vacancies are reserved for scheduled caste candidates while the rest are open to all, the number of ways in which the selection can be made is 5C3 × 20C9.
Ans.
Number of ways to select 3 scheduled caste candidate out of 5 = 5C3
We have to select 9 other candidates out of 22.
So the number of ways = 22C9
Required number of selection = 5C3 x 22C9
Hence, the given statement is False.

MATCH THE COLUMNS
Q.60. There are 3 books on Mathematics, 4 on Physics and 5 on English. How many different collections can be made such that each collection consists of :
NCERT Exemplar: Permutations and Combinations- 2 | Mathematics (Maths) Class 11 - Commerce
Ans.
We have 3 books of Mathematics, 4 of Physics and 5 on English
(a) One book of each subject = 3C1 x 4C1 x 5C1 = 3 x 4 x 5 = 60
(b) Atleast one book of each subject = (23 - 1) x (24 - 1) x (25 - 1)
= 7 x 15 x 31 = 3255
(c) Atleast one book of English = (25 – 1) x 27 = 31 x 128 = 3968.
Hence the required matching is
(a) ↔ (ii), (b) ↔ (iii) and (c) ↔ (i)

Q.61. Five boys and five girls form a line. Find the number of ways of making the seating arrangement under the following condition:
NCERT Exemplar: Permutations and Combinations- 2 | Mathematics (Maths) Class 11 - Commerce
Ans.
(a) Total number of arrangement when boys and girls alternate: = (5!)2 + (5!)2
(b) No two girls sit together: = 5! 6!
(c) All the girls sit together = 2! 5! 5!
(d) All the girls sit never together = 10! – 5! 6!
Hence, the required matching is
(a) ↔ (iii), (b) ↔ (i), (c) ↔ (iv), (d) ↔ (ii)

Q.62. There are 10 professors and 20 lecturers out of whom a committee of 2 professors and 3 lecturer is to be formed. Find :
NCERT Exemplar: Permutations and Combinations- 2 | Mathematics (Maths) Class 11 - Commerce
Ans.
(a) We have to select 2 professor out of 10 and 3 lecturers out of 20
∴ Number of ways of selection = 10C2 x 20C3 
(b) When a particular professor is included taken the number of ways = 10-1C1 x 20C3 = 9C1 x 20C3
(c) When a particular lecturer is included then number of ways = 10C2 x 19C2 
(d) When a particular lecturer is excluded, then number of ways = 10C2 x 19C3 
Hence the required matching is
(a) ↔ (iv), (b) ↔ (iii), (c) ↔ (ii), (d) ↔ (i)

Q.63. Using the digits 1, 2, 3, 4, 5, 6, 7, a number of 4 different digits is formed. Find
NCERT Exemplar: Permutations and Combinations- 2 | Mathematics (Maths) Class 11 - Commerce
Ans.
(a) Total of 4 digit number formed with 1, 2, 3, 4, 5, 6, 7
NCERT Exemplar: Permutations and Combinations- 2 | Mathematics (Maths) Class 11 - Commerce
(b) When a number is divisible by 2 = 4 x 5 x 6 x 3 = 360
(c) Total numbers which are divisible by 25 = 40
(d) Total numbers which are divisible by  4 (last two digits is divisible by 4) = 200
Hence, the required matching is
(a) ↔ (i), (b) ↔ (iii), (c) ↔ (iv), (d) ↔ (ii)

Q.64. How many words (with or without dictionary meaning) can be made from the letters of the word MONDAY, assuming that no letter is repeated, if
NCERT Exemplar: Permutations and Combinations- 2 | Mathematics (Maths) Class 11 - Commerce
Ans.
 (a) 4 letters are used at a time =NCERT Exemplar: Permutations and Combinations- 2 | Mathematics (Maths) Class 11 - Commerce
(b) All letters are used  at a time  = 6P6 = 6! = 720
(c) All letters are used but first letter is vowel = 2 x 5! = 2 x 120 = 240
Hence, the required matching is
(a) ↔ (iii), (b) ↔ (i), (c) ↔ (ii)

The document NCERT Exemplar: Permutations and Combinations- 2 | Mathematics (Maths) Class 11 - Commerce is a part of the Commerce Course Mathematics (Maths) Class 11.
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FAQs on NCERT Exemplar: Permutations and Combinations- 2 - Mathematics (Maths) Class 11 - Commerce

1. What is the difference between permutations and combinations in the context of the NCERT Exemplar: Permutations and Combinations-2?
Ans. Permutations and combinations are both concepts in mathematics that deal with the arrangement or selection of objects. The main difference between the two is that permutations refer to the arrangement of objects in a specific order, while combinations refer to the selection of objects without considering the order. In the context of the NCERT Exemplar: Permutations and Combinations-2, these concepts are likely to be explained in detail with relevant examples.
2. How can permutations be calculated in the NCERT Exemplar: Permutations and Combinations-2?
Ans. In the NCERT Exemplar: Permutations and Combinations-2, permutations can be calculated using the formula nPr = n! / (n-r)!, where n represents the total number of objects and r represents the number of objects selected for arrangement. This formula helps in finding the number of permutations possible for a given set of objects.
3. What is the significance of combinations in the NCERT Exemplar: Permutations and Combinations-2?
Ans. Combinations are significant in the NCERT Exemplar: Permutations and Combinations-2 as they allow us to determine the number of ways to select objects from a given set without considering the order. Combinations help in solving problems related to choosing a committee, forming teams, or selecting a group of objects where the order doesn't matter.
4. How can the concept of permutations and combinations be applied in real-life scenarios?
Ans. The concept of permutations and combinations can be applied in various real-life scenarios. For example, in a lottery system, the calculation of the number of possible combinations helps determine the odds of winning. In sports fixtures, the arrangement of teams in a league or tournament can be understood through permutations. Additionally, the selection of a committee from a group of individuals is a practical application of combinations.
5. Are there any practical tips or tricks to solve permutation and combination problems in the NCERT Exemplar: Permutations and Combinations-2?
Ans. While solving permutation and combination problems in the NCERT Exemplar: Permutations and Combinations-2, it is helpful to understand the concepts thoroughly and practice solving different types of problems. Additionally, using visual aids such as diagrams or charts can aid in understanding the problem better. It is also important to carefully read the problem statement and identify the given information to determine whether it requires the use of permutations or combinations.
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