# 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 4Exercise 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.
∴ 1st place can be filled in 10 ways,
2nd place can be filled in 9 ways and
3rd 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 _ _ _
∴ 1st place can be filled in 8 ways,
2nd place can be filled in 7 ways and
3rd 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 4Exercise 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 4Exercise 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 5Pr = 2.6Pr-1 find ‘r’.

Solution :

Question 3(i):

If nP4 : nP5 = 1 : 2 find ‘n’.

Solution :

Question 3(ii):

If n-1P3 :n+1P3 = 5 : 12 find ‘n’.

Solution :

Question 4(i):

If 9P5 +5 9P4 = 10Pr,find ‘r’.

Solution :

Question 4(ii):

If 10pr+1 :11Pr = 30 : 11, find ‘r’.

Solution :

Question 5(i):

Prove that n+1Pr+1 = (n + 1) nPr

Solution :

Question 5(ii):

Show that 10P3 = 9P3 + 39P2

Solution :

#### KSEEB SSLC Solutions for Class 10 Maths Chapter 4Exercise 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. 10C3
ii. 60C60
iii. 100C97

Solution :

Question 2(i):

If nC4 = nC7 find n.

Solution :

nC4 = nC7
nC4 = nCn – 7
∴ 4 = n – 7
∴ n = 4 + 7
∴ n = 11

Question 2(ii):

If nPr = 840, nCr = 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 nCr-1 : nCr: nCr+1 = 3 : 4 : 5, find r.

Solution :

#### KSEEB SSLC Solutions for Class 10 Maths Chapter 4Exercise 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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