KSEEB SSLC Solutions for Class 10 Maths Chapter 4 Permutations and Combinations

In this chapter, we provide KSEEB Solutions for Class 10 Maths Chapter 4 Permutations and Combinations for English medium students, Which will very helpful for every student in their exams. Students can download the latest KSEEB Solutions for Class 10 Maths Chapter 4 Permutations and Combinations pdf, free KSEEB solutions for Class 10 Maths Chapter 4 Permutations and Combinations book pdf download. Now you will get step by step solution to each question.

KSEEB SSLC Solutions for Class 10 Maths – Permutations and Combinations (English Medium)

KSEEB SSLC Solutions for Class 10 Maths Chapter 4 Exercise 4.1

Question 1:

How many 3-digit numbers can be formed using the digits 1, 2, 3, 4, 5, 6 without repeating any digit?

Solution :

Digits are 1, 2, 3, 4, 5, and 6.
Now, the unit place can be filled in 6 ways using all the six digits.
As repetition of digits is not allowed, the tens place can be filled in 5 ways and the hundreds place can be filled in 4 ways.
Thus, total number of 3-digit numbers
= 6 × 5 × 4
= 120

Question 2:

How many 3 digit even numbers can be formed using the digits 3, 5, 7, 8, 9, if the digits are not repeated?

Solution :

Digits are 3, 5, 7, 8, and 9.
As the even number contains an even digit in the units place, 8 should be placed in the units place.
∴ The tens place can be filled in 4 ways by using 3, 5, 7, and 9; and the hundreds place can be filled in 3 ways.
Thus, the total number of ways = 4 × 3 = 12
Hence, 12 even numbers can be formed.

Question 3:

How many 3 letter code can be formed using the first 10 letters of English alphabet, if no letter can be repeated?

Solution :

The first 10 letters of the English alphabet are a, b, c, d, e, f, g, h, i, and j.
Using these 10 letters, a 3 letter code is to be formed.
∴ 1^st place can be filled in 10 ways,
2^nd place can be filled in 9 ways and
3^rd place can be filled in 8 ways.
Thus, the total number of 3 letter code which can be formed
= 10 × 9 × 8
= 720

Question 4:

How many 5 digit telephone numbers can be formed using the digits 0 to 9, if each number starts with 65 and no digit appears more than once?

Solution :

There are 10 numbers in all from 0 to 9.
If each telephone number starts with 65, then eight digits are available to fill in the next three places.
Thus, the telephone numbers are in the form 6 5 _ _ _
∴ 1^st place can be filled in 8 ways,
2^nd place can be filled in 7 ways and
3^rd place can be filled in 6 ways.
Thus, the total number of 5-digit telephone numbers which can be formed
= 8 × 7 × 6
= 336

Question 5:

If a fair coin is tossed 3 times, find the number of outcomes.

Solution :

Each time a fair coin is tossed, number of outcomes = 2
i.e., Head (H) or Tail (T)
Thus, when a coin is tossed 3 times, the total number of outcomes
= 2 × 2 × 2
= 8

Question 6:

Given 5 flags of different colours, how many different signals can be generated if each signal requires the use of 2 flags one below the other?

Solution :

5 flags of different colours are to be used to form a signal with 2 flags.
The upper portion of the flag can be filled with five colours in 5 ways.
And, the lower portion of the flag can be filled with four colours in 4 ways.
Thus, the total number of signals which can be generated
= 5 × 4
= 20

KSEEB SSLC Solutions for Class 10 Maths Chapter 4 Exercise 4.2

Question I:

Below are given situations for arrangements and selections. Classify them as examples of permutations and combinations.

1. A committee of 5 members to be chosen from a group of 12 people.
2. Five different subject books to be arranged on a shelf.
3. There are 8 chairs and 8 people to occupy them.
4. In a committee of 7 persons, a chairperson, a secretary and a treasurer are to be chosen.
5. The owner of children’s clothing shop has 10 designs of frocks and 3 of them have to be displayed in the front window.
6. Three-letter words to be formed from the letters in the word ‘ARITHMETIC’.
7. In a question paper having 12 questions, students must answer the first 2 questions but may select any 8 of the remaining ones.
8. Ajar contains 5 black and 7 white balls. 3 balls to be drawn in such a way that 2 are black and 1 is white.
9. Three-digit numbers are to be formed from the digits 1, 3, 5, 7, 9 where repetitions are not allowed.
10. Five keys are to be arranged in a circular key ring.
11. There are 7 points in a plane and no 3 of the points are collinear. Triangles are to be drawn by joining three non-collinear points.
12. A collection of 10 toys are to be divided equally between two children.

In each of the above examples give reason to explain why it is permutation or combination.

Solution :

1. Combination-as it is a mere selection of members.
2. Permutation-as it is an arrangement of five different books.
3. Permutation-as it requires an order.
4. Combination-as it is a mere selection of members.
5. Permutation-as it requires an order.
6. Permutation-as it requires an arrangement of letters.
7. Combination-as it requires the mere selection of questions.
8. Combination-as it requires only selection and order is not required.
9. Permutation-as it requires an order of numbers.
10. Permutation-as it is an arrangement.
11. Combination-as joining of points does not require an order.
12. Combination-as it requires mere selection and order is not required.

KSEEB SSLC Solutions for Class 10 Maths Chapter 4 Exercise 4.3

Question 1:

Convert the following products into factorials.

i. 1 × 2 × 3 × 4 × 5 × 6 × 7
ii. 18 × 17 × … × 3 × 2 × 1
iii. 6 × 7 × 8 × 9
iv. 2 × 4 × 6 × 8

Solution :

*Note: Answer of the subpart(iv) given in the book is incorrect.

Question 2:

Solution :

Question 4:

Find the L.C.M. of 4!, 5!, 6!

Solution :

4! = 4 × 3 × 2 × 1
5! = 5 × 4 × 3 × 2 × 1
6! = 6 × 5 × 4 × 3 × 2 × 1
Now, 4! and 5! are multiples of 6!
∴ L.C.M. of 4!, 5!, 6! = 6! = 720

Question 5:

Solution :

Question 6:

Solution :

Exercise 4.4:

Question 1(i):

Evaluate:
¹²P₄

Solution :

Question 1(ii):

Evaluate:
⁷⁵P₂

Solution :

Question 1(iii):

Evaluate:
⁸P₈

Solution :

Question 1(iv):

Evaluate:
¹⁵P₁

Solution :

Question 1(v):

Evaluate:
³⁸P₀

Solution :

Question 2(i):

If ⁿP₄ = 20ⁿP₂ find ‘n’.

Solution :

Question 2(ii):

If 16ⁿP₃ = 13ⁿ⁺¹P₃ find ‘n’.

Solution :

Question 2(iii):

If ⁵P_r = 2.⁶P_r-1 find ‘r’.

Solution :

Question 3(i):

If ⁿP₄ : ⁿP₅ = 1 : 2 find ‘n’.

Solution :

Question 3(ii):

If ^{n-1P₃ :n+1}P₃ = 5 : 12 find ‘n’.

Solution :

Question 4(i):

If ⁹P₅ +5 ⁹P₄ = ¹⁰P_r,find ‘r’.

Solution :

Question 4(ii):

If ¹⁰p_{r+1 :}¹¹P_r = 30 : 11, find ‘r’.

Solution :

Question 5(i):

Prove that ⁿ⁺¹P_r+1 = (n + 1) ⁿP_r

Solution :

Question 5(ii):

Show that ¹⁰P₃ = ⁹P₃ + 3⁹P₂

Solution :

KSEEB SSLC Solutions for Class 10 Maths Chapter 4 Exercise 4.5:

Question 1:

How many words with or without dictionary meaning can be formed using all the letters of the word ‘Joule’ using each letter exactly once?

Solution :

Number of letters in the word ‘Joule’ = 5
Number of words which can be formed using all 5 letters (each letter exactly once)
= ⁵P₅
= 5!
= 5 × 4 × 3 × 2 × 1
= 120
Thus, 120 words with or without meaning can be formed using all the letters of the word ‘Joule’.

Question 2:

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?

Solution :

Number of men = 5
Number of women = 4
4 women can occupy the even places in ⁴P₄ ways.
The remaining 5 places can be filled by 5 men in ⁵P₅ ways.
∴ Total number of arrangements,
= ⁴P₄ × ⁵P₅
= 4! × 5!
= (4 × 3 × 2 × 1) × (5 × 4 × 3 × 2 × 1)
= 24 × 120
= 2880
Thus, 2880 arrangements are possible.

Question 3:

In how many ways can 6 women draw water from 6 wells, if no well remains unused?

Solution :

Number of women = 6
Number of wells = 6
Thus, number of women = number of wells
∴ Number of ways of drawing water from all the wells
= ⁶P₆
= 6!
= 6 × 5 × 4 × 3 × 2 × 1
= 720
Hence, water can be drawn from the 6 wells by 6 women in 720 ways.

Question 4:

Seven students are contesting the election for the Presidentship of the student’s union. In how many ways can their names be listed on the ballot papers?

Solution :

Number of students contesting for Presidentship = 7
Therefore, the number ways in which their names can be listed on the ballot papers
= ⁷P₇
= 7!
= 7 × 6 × 5 × 4 × 3 × 2 × 1
= 5040
*Note: Answer given in the book is incorrect.

Question 5:

8 students are participating in a competition. In how many ways can the first three prizes be won?

Solution :

Question 6:

Find the total number of 2-digit numbers.

Solution :

The numbers required to form the 2-digit number are 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9.
∴ Total numbers = 10
The units place can be filled using all the ten numerals in ¹⁰P₁ ways.
As ‘0’ cannot be filled in the tens place, the tens place can be filled in ⁹P₁ ways.
Total number of 2-digits numerals,
= ¹⁰P₁ × ⁹P₁
= 10 × 9
= 90
Hence, the total number of 2-digit numbers are 90.

Question 7:

There are 5 stickers, 3 of them red and the two of them green. It is desired to make a design by arranging them in a row. How many such designs are possible?

Solution :

Total number of stickers = 5
Number of ways of arranging 5 stickers
= ⁵P₅
= 5!
= 5 × 4 × 3 × 2 × 1
= 120 ways
Thus, 120 designs are possible.

Question 8:

How many 4-digit numbers can be formed using the digits 1, 2, 3, 7, 8, 9 (repetitions not allowed)?

1. How many of these are less than 6000?
2. How many of these are even?
3. How many of these end with 7?

Solution :

Number of digits = 6
Therefore, the number of ways 4-digit numbers can be formed,
= ⁶P₄
= 6! /(6-4)!
= 6 × 5 × 4 × 3
= 360
Hence, 360 numbers can be formed.

a. For a number less than 6000,
We can fill the thousands place with 1 or 2 or 3 in
     ³P₁ ways and the remaining 3 places can be filled in ⁵P₃ ways.
Therefore, numbers less than 6000
= ³P₁ × ⁵P₃
    = 3 × 5 × 4 × 3
= 180

b. For a number to be even, the units place should contain 2 or 8.
∴ The units place can be filled in ²P₁ ways.
The remaining three places can be filled in ⁵P₃ ways.
Therefore, the total number of even numbers
= ⁵P₃ × ²P₁
    = 5 × 4 × 3 × 2
= 120

c. If a number is ending with 7, the units place should contain 7.
∴ The units place can be filled in ¹P₁ = 1 way
The remaining 3 places can be filled in ⁵P₃ ways.
Therefore, total numbers ending with 7,
= ⁵P₃ × 1
= 5 × 4 × 3 × 1
= 60

Question 9:

There are 15 buses running between two towns. In how many ways can a man go to one town and return by a different bus?

Solution :

Number of buses = 15
Therefore, the number of ways by which a man can go to one town and return by a different bus,
= ¹⁵P₂
= 15 × 14
= 210

KSEEB SSLC Solutions for Class 10 Maths Chapter 4 Exercise 4.6

Question 1:

Evaluate

^{i. 10}C₃
^{ii. 60}C₆₀
iii. ¹⁰⁰C₉₇

Solution :

Question 2(i):

If ⁿC₄ = ⁿC₇ find n.

Solution :

ⁿC₄ = ⁿC₇
∴ ⁿC₄ = ⁿC_{n – 7}
∴ 4 = n – 7
∴ n = 4 + 7
∴ n = 11

Question 2(ii):

If ⁿP_r = 840, ⁿC_r = 35, find n.

Solution :

Question 3:

If ²ⁿC₃ : ⁿC₃ = 11 : 1, find n.

Solution :

Question 4:

Verify that ⁸C₄ + ⁸C₅ = ⁹C₄

Solution :

Question 5:

Solution :

Question 6:

If ⁿC_r-1 : ⁿC_r: ⁿC_r+1 = 3 : 4 : 5, find r.

Solution :

KSEEB SSLC Solutions for Class 10 Maths Chapter 4 Exercise 4.7

Question 1:

Out of 7 Consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed?

Solution :

3 consonants out of 7 consonants can be selected in ⁷C₃ ways, and
2 vowels out of 4 vowels can be selected in ⁴C₂ ways.
∴ The total number of words which can be formed
= ⁷C₃ × ⁴C₂

Thus, 210 words can be formed.
*Note: Answer given in the book is incorrect

Question 2:

In how many ways can 5 sportsmen be selected from a group of 10?

Solution :

Question 3:

In how many ways a cricket team of eleven be selected from 17 players in which 5 players can bowlers? Each cricket team must included 2 bowlers?

Solution :

Question 4:

How many (i) lines (ii) triangles can be drawn through 8 points on a circle?

Solution :

Question 5:

How many diagonals can be drawn in a (i) decagon (ii) icosagon?

Solution :

Question 6:

A Polygon has 44 diagonals. Find the number of sides.

Solution :

Question 7:

There are 3 white and 4 red roses in a garden. In how many ways can 4 flowers of which 2 red be plucked?

Solution :

4 flowers are to be plucked, out of which 2 are red.
∴ The other two are white.
2 red roses out of 4 red roses can be plucked in ⁴C₂ ways.
2 white roses out of 3 white roses can be plucked in ³C₂ ways.
∴ Total number of ways

Question 8:

In how many ways can a student choose 5 courses out of 9 courses if 2 courses are compulsory for every student?

Solution :

Total number of courses = 9
Out of these 9 courses, 2 courses are compulsory.
Hence, the courses left for selection = 9 – 2 = 7
Hence,  the total number of ways by which the 3 courses can be selected from 7 courses
= ⁷C₃

Question 9:

There are 5 questions in a question paper. In how many ways can a boy solve one or more questions?

Solution :

Question 10:

In how many ways 4 cards from a pack of 52 playing cards can be chosen?

Solution :

Number of ways by which 4 cards can be chosen from 52 cards
= ⁵²C₄

Question 11:

At an election there are 7 candidates and three are to be elected. A voter is allowed to vote for any number of candidates not greater than the number to be elected. In how many ways can we vote?

Solution :

KSEEB SSLC Solutions for Class 10 Maths Chapter 4 Exercise 4.8

Question 1:

If there are 6 periods in each working day of a school, in how many ways can one arrange 5 subjects such that each subject is allowed at least one period?

Solution :

Total number of periods = 6
Total number of subjects = 5
5 unique subjects can be arranged in 6 periods in ⁶P₅ ways.
And the remaining one period can be allotted with any one of the 5 subjects in ⁵P₁ ways.
But that one subject will be repeated twice.
So the total numbers of ways of arranging the subjects

= 1800
*Note: Answer given in the book is incorrect

Question 2:

A committee of 5 is to be formed out of 6 men and 4 ladies. In how many ways can this be done when

1. at least 2 ladies are included.
2. at most 2 ladies are included.

Solution :

i. The committee should contain atleast 2 ladies,
i.e., it may contain 2, 3 or 4 ladies.
If 2 are ladies, the other 3 are men.
The committee can be formed in ⁴C₂ × ⁶C₃ ways.
If 3 are ladies, the other 2 are men.
The committee can be formed in ⁴C₃ × ⁶C₂ ways.
If 4 are ladies, the other 1 is man.
The committee can be formed in ⁴C₄ × ⁶C₁ ways.

ii. The committee should contain at most 2 ladies,
i.e., it may contain 2, 1 or 0 ladies.
If 2 are ladies, the other 3 are men.
The committee can be formed in ⁴C₂ × ⁶C₃ ways.
If 1 is lady, the other 4 are men.
The committee can be formed in ⁴C₁ × ⁶C₄ ways.
If no lady, all are men.
The committee can be formed in ⁴C₀ × ⁶C₅ ways.

Question 3:

A sports team of 11 students is to be constituted choosing at least 5 from class IX and at least 5 from class X. If there are 8 students in each of these classes, in how many ways can the team be constituted?

Solution :

A sports team of 11 students should contain at least 5 students each from class IX and X.
If 5 students are selected from class IX, then 6 students can be selected from class X.
It can be done in ⁸C₅ × ⁸C₆ ways.
If 6 students are selected from class IX, then 5 students can be selected from class X.
It can be done in ⁸C₆ × ⁸C₅ ways.

Question 4:

In a vegetable mela or fair, an artist wants to make mascot (a toy that represents an organization) with the following in the four sectors of a circle.
(i) Beans
(ii) Carrot
(iii) Peas
(iv) Tomato
In how many ways can the artist make the mascot?

Solution :

There are four vegetables and four sectors in a circle.
∴ The number of ways by which an artist can make a mascot
= ⁴P₄
= 4 × 3 × 2 × 1
= 24

Question 5:

From a group of 12 students, 8 are to be chosen for an excursion. There are 3 students who decide that either of them will join or none of them will join. In how many ways can the 8 be chosen?

Solution :

Question 6:

How many chords can be drawn through 20 points on a circle?

Solution :

Question 7:

The English alphabet has 5 vowels and 21 consonants. How many words with 2 different vowels and 2 different consonants can be formed from the alphabet?

Solution :

Question 8:

In how many ways 5 letters be posted in 3 post boxes, if any number of letters can be posted in all of the three post boxes?

Solution :

There are 5 letters and 3 post boxes.
As each letter can be posted to any of the three post boxes, each letter can be posted in 3 ways.
Since there are 5 letters,
the total number of ways of posting the 5 letters
= 3 × 3 ×3 × 3 ×3
= 243

Question 9:

In a group of 15 students there are 6 scouts. In how many ways can 12 students be selected, so as to include at least 4 scouts?

Solution :

Total number of students = 15
Total number of scouts = 6
∴ Remaining students = 15 – 6 = 9
A group of 12 students should contain at least 4 scouts,
i.e., it may contain 4, 5 or 6 scouts.
4 scouts and 8 other students can be selected in ⁶C₄ × ⁹C₈ ways.
5 scouts and 7 other students can be selected in ⁶C₅ × ⁹C₇ ways.
6 scouts and 6 other students can be selected in ⁶C₆ × ⁹C₆ ways.

All Chapter KSEEB Solutions For Class 10 Maths

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All Subject KSEEB Solutions For Class 10

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