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

**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

**Exercise 4.2:**

**Question I:**

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

- A committee of 5 members to be chosen from a group of 12 people.
- Five different subject books to be arranged on a shelf.
- There are 8 chairs and 8 people to occupy them.
- In a committee of 7 persons, a chairperson, a secretary and a treasurer are to be chosen.
- The owner of children’s clothing shop has 10 designs of frocks and 3 of them have to be displayed in the front window.
- Three-letter words to be formed from the letters in the word ‘ARITHMETIC’.
- In a question paper having 12 questions, students must answer the first 2 questions but may select any 8 of the remaining ones.
- 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.
- Three-digit numbers are to be formed from the digits 1, 3, 5, 7, 9 where repetitions are not allowed.
- Five keys are to be arranged in a circular key ring.
- 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.
- 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 :**

- Combination-as it is a mere selection of members.
- Permutation-as it is an arrangement of five different books.
- Permutation-as it requires an order.
- Combination-as it is a mere selection of members.
- Permutation-as it requires an order.
- Permutation-as it requires an arrangement of letters.
- Combination-as it requires the mere selection of questions.
- Combination-as it requires only selection and order is not required.
- Permutation-as it requires an order of numbers.
- Permutation-as it is an arrangement.
- Combination-as joining of points does not require an order.
- Combination-as it requires mere selection and order is not required.

**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 3:**

**Solution :**

*Note: Answer given in the book is incorrect.

**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:

^{12}P_{4}

**Solution :**

**Question 1(ii):**

Evaluate:

^{75}P_{2}

**Solution :**

**Question 1(iii):**

Evaluate:

^{8}P_{8}

**Solution :**

**Question 1(iv):**

Evaluate:

^{15}P_{1}

**Solution :**

**Question 1(v):**

Evaluate:

^{38}P_{0}

**Solution :**

**Question 2(i):**

If ^{n}P_{4} = 20^{n}P_{2} find ‘n’.

**Solution :**

**Question 2(ii):**

If 16^{n}P_{3} = 13^{n+1}P_{3} find ‘n’.

**Solution :**

**Question 2(iii):**

If ^{5}P_{r} = 2.^{6}P_{r-1} find ‘r’.

**Solution :**

**Question 3(i):**

If ^{n}P_{4} : ^{n}P_{5} = 1 : 2 find ‘n’.

**Solution :**

**Question 3(ii):**

If ^{n-1P3 :}^{n+1}P_{3} = 5 : 12 find ‘n’.

**Solution :**

**Question 4(i):**

If ^{9}P_{5} +5 ^{9}P_{4} = ^{10}P_{r},find ‘r’.

**Solution :**

**Question 4(ii):**

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

**Solution :**

**Question 5(i):**

Prove that ^{n+1}P_{r+1} = (n + 1) ^{n}P_{r}

**Solution :**

**Question 5(ii):**

Show that ^{10}P_{3} = ^{9}P_{3} + 3^{9}P_{2}

**Solution :**

**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)

= ^{5}P_{5
}= 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 ^{4}P_{4} ways.

The remaining 5 places can be filled by 5 men in ^{5}P_{5} ways.

∴ Total number of arrangements,

= ^{4}P_{4} × ^{5}P_{5
}= 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

= ^{6}P_{6
}= 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

= ^{7}P_{7
}= 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 ^{10}P_{1} ways.

As ‘0’ cannot be filled in the tens place, the tens place can be filled in ^{9}P_{1} ways.

Total number of 2-digits numerals,

= ^{10}P_{1} × ^{9}P_{1
}= 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

= ^{5}P_{5
}= 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)?

- How many of these are less than 6000?
- How many of these are even?
- How many of these end with 7?

**Solution :**

Number of digits = 6

Therefore, the number of ways 4-digit numbers can be formed,

= ^{6}P_{4
}= 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

^{ 3}P_{1} ways and the remaining 3 places can be filled in ^{5}P_{3} ways.

Therefore, numbers less than 6000

= ^{3}P_{1} × ^{5}P_{3
} = 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 ^{2}P_{1} ways.

The remaining three places can be filled in ^{5}P_{3} ways.

Therefore, the total number of even numbers

= ^{5}P_{3} × ^{2}P_{1
} = 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 ^{1}P_{1} = 1 way

The remaining 3 places can be filled in ^{5}P_{3} ways.

Therefore, total numbers ending with 7,

= ^{5}P_{3 }× 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,

= ^{15}P_{2
}= 15 × 14

= 210

**Exercise 4.6:**

**Question 1:**

Evaluate

^{i. 10}C_{3
}^{ii. 60}C_{60
}iii. ^{100}C_{97}

**Solution :**

**Question 2(i):**

If ^{n}C_{4} = ^{n}C_{7} find n.

**Solution :**

^{n}C_{4} = ^{n}C_{7
}∴ ^{n}C_{4} = ^{n}C_{n}_{ – 7
}∴ 4 = n – 7

∴ n = 4 + 7

∴ n = 11

**Question 2(ii):**

If ^{n}P_{r} = 840, ^{n}C_{r} = 35, find n.

**Solution :**

**Question 3:**

If ^{2n}C_{3 }: ^{n}C_{3} = 11 : 1, find n.

**Solution :**

**Question 4:**

Verify that ^{8}C_{4} + ^{8}C_{5} = ^{9}C_{4}

**Solution :**

**Question 5:**

**Solution :**

**Question 6:**

If ^{n}C_{r-1 }: ^{n}C_{r}: ^{n}C_{r+1} = 3 : 4 : 5, find r.

**Solution :**

**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 ^{7}C_{3} ways, and

2 vowels out of 4 vowels can be selected in ^{4}C_{2} ways.

∴ The total number of words which can be formed

= ^{7}C_{3} × ^{4}C_{2}

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 ^{4}C_{2} ways.

2 white roses out of 3 white roses can be plucked in ^{3}C_{2} 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

= ^{7}C_{3}

**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

= ^{52}C_{4}

**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 :**

**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 ^{6}P_{5} ways.

And the remaining one period can be allotted with any one of the 5 subjects in ^{5}P_{1} 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

- at least 2 ladies are included.
- 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 ^{4}C_{2} × ^{6}C_{3} ways.

If 3 are ladies, the other 2 are men.

The committee can be formed in ^{4}C_{3} × ^{6}C_{2} ways.

If 4 are ladies, the other 1 is man.

The committee can be formed in ^{4}C_{4} × ^{6}C_{1} 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 ^{4}C_{2} × ^{6}C_{3} ways.

If 1 is lady, the other 4 are men.

The committee can be formed in ^{4}C_{1} × ^{6}C_{4} ways.

If no lady, all are men.

The committee can be formed in ^{4}C_{0} × ^{6}C_{5} 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 ^{8}C_{5} × ^{8}C_{6} ways.

If 6 students are selected from class IX, then 5 students can be selected from class X.

It can be done in ^{8}C_{6} × ^{8}C_{5} 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

= ^{4}P_{4
}= 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 ^{6}C_{4} × ^{9}C_{8} ways.

5 scouts and 7 other students can be selected in ^{6}C_{5} × ^{9}C_{7} ways.

6 scouts and 6 other students can be selected in ^{6}C_{6} × ^{9}C_{6} ways.