# NCERT Solutions Class 11 Maths Chapter 7

## NCERT Solutions Class 11 Mathematics Chapter 7- Permutations and Combinations

Mathematics can be a complex subject in which students tend to find many concepts difficult to comprehend. Hence,  it is essential for students to understand the crux of the problem to solve it correctly. The NCERT Solutions Class 11 Mathematics Chapter 7 is a reliable and convenient learning resource. It will help to perform strategic learning and enable regular practice.

Through explicit answers to the textual problems NCERT Solutions Class 11 Mathematics Chapter 7 explains all fundamental concepts based on Permutations and Combinations. Its basic applications involve arranging numbers, alphabets, etc., in a specific order. They also have various applications in the fields of computer science and physics.

Having a deep knowledge of Class 11 Mathematics Chapter 7 is crucial. With the help of NCERT Solutions Class 11 Mathematics Chapter 7, students will build their core fundamentals and get well-versed with this topic.

Students are encouraged to refer to NCERT Solutions Class 10, NCERT Solutions Class 9 and NCERT Solutions Class 8 which will help them score better in their examinations.

### Key Topics Covered in NCERT Solutions Class 11 Mathematics Chapter 7

The Chapter 7 Mathematics Class 11 is very important for each student to attain high scores in the examinations. With the help of NCERT Solutions Class 11 Mathematics Chapter 7 provided by the Extramarks, students can attain the required knowledge and solve unlimited questions and exercises provided in these solutions.

To master the Class 11 Mathematics NCERT solutions Chapter 7, students will have to study the following topics thoroughly:

 Exercise Topic 7.1 Introduction 7.2 Fundamental Principle of Counting 7.3 Permutations 7.4 Combinations

7.1 Introduction:

In the NCERT Solutions Class 11 Mathematics Chapter 7, students will learn about Permutations and Combinations. The topics included in the solutions are factorial notation, principles of counting permutations, derivation of formulas and theorems. In the Class 11 Mathematics Chapter 7, students will learn basic techniques and different ways to arrange and select objects.

Students are advised to study using the NCERT Solutions Class 11 Mathematics Chapter 7 to avail accurate answers to the textbook exercises of this  chapter.

7.2 Fundamental Principle of Counting:

This principle states that “For two events if an event occurs in ‘m’ ways. Another event occurs in ‘n’ ways, then the total number of occurrences of the events in the given order  is m × n.”

For instance; If a person has 3 pants and 2 shirts. Then, the possible number of pairs that person can choose from will be 3×2=  6

For 3 events, the principle is as follows: “If an event occurs in m ways, and another event occurs in n ways, after which a third event occurs in p ways, then a total number of occurrences in the given order will be m × n × p.”

There are two types of questions that are included in the NCERT Solutions Class 11 Mathematics Chapter 7.

1. The number of ways the following events occur in succession:

(i) Choosing a pant

(ii) Choosing a shirt

1. The required number of ways is the number of ways an event occurs in succession:

(i) Choosing a school bag

(ii) Choosing a tiffin box

(iii) Choosing a water bottle.

In both the above cases, the events may occur in various orders. But, we must choose any one order and count the number of different ways the event occurs.

7.3 Permutations:

In this section of NCERT Solutions Class 11 Mathematics Chapter 7, students will learn about permutations. It is defined as the arrangement in a definite order of a number of objects chosen some or all at a single time.

Theorem 1: When objects are distinct, say n, the number of permutations taken r at a time, as 0 < r ≤ n then the objects do not repeat. It is n ( n – 1) ( n – 2). . .( n – r + 1) and is denoted by Prn

Factorial notation:

Under this section of permutation, students will learn to denote the product of several numbers such as 1 x 2 x 3 x 4 x 5 x …. X (n-1) x n = n!

The formula for Permutation, Prn = n!(n-r)! , where 0 < r ≤ n is very important. Students will also learn the derivation of this formula. Theorem 1 which states that the number of permutations of distinct objects, say n, taken r at a time with repetition is given as nr

When the number of objects are not distinct, p objects are the same kind, and the rest are different is given as n!p!.

7.4 Combinations:

In this section, students will learn the way to select items from the collection in any order. Several concepts related to this topic are explained here. In the NCERT Solutions Class 11 Mathematics Chapter 7, students will learn several new formulas to solve the problems easily. The formulas are mentioned below:

1. Prn = Crn r!, where 0 < r ≤ n
2. Crn = n!r! (n-r)!
3. Cnn = 1, where r = n
4. C0n = 1
5. Cn-rn = Crn = n!r! (n-r)!
6. C1n = Cbn, then a = b and n = a + b
7. Crn + Cr+1n = Crn+1

There are several solved examples that students must study in order to get an in-depth understanding of the NCERT Solutions Class 11 Mathematics Chapter 7 Permutations and Combinations.

### NCERT Solutions Class 11 Mathematics Chapter 7: Exercise & Solutions

Students can study with the help of the NCERT Solutions Class 11 Mathematics Chapter 7 from the extramarks platform. These solutions are prepared by experts and professionals in the field of Mathematics. It comprehensively explains each and every answer of the textual exercises along with   several other sums for practice. Students can depend on the Class 11 Mathematics NCERT  Solutions Chapter 7 for last minute revision. It includes all important formulas, theorems, derivations and key points.

The NCERT Solutions Class 1, NCERT Solutions Class 2 and NCERT Solutions Class 3 provides a helping hand for the primary section students in the studies.

Visit the links mentioned below to get access to the Extramarks NCERT Solutions Class 11 Mathematics Chapter 7:

Extramarks also provides study materials such as CBSE sample papers, mock tests, and revision notes along with several past years’ question papers. Students are advised to refer to the best NCERT reference books and NCERT Solutions Class 12 provided by the platform of Extramarks to get a deeper understanding of all concepts. Refer to the Class 11 Mathematics Chapter 7 for efficient preparation.

• NCERT Solutions Class 1
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### NCERT Exemplar Class 11 Mathematics

The NCERT Exemplar notes are important for annual examinations and competitive exam preparations. Students must refer to these notes to understand all complicated topics easily. The NCERT Exemplar and the NCERT Solutions Class 11 Mathematics Chapter 7 will help students to enhance their thinking abilities and problem-solving skills. It provides unlimited aptitude-based, higher levels and several twisted questions for students to practise. Students are advised to use the NCERT Exemplars provided on the Extramarks’ website to ace their examinations.

The platform of Extramarks aims to make learning better and easier through high-quality and dependable study notes. The NCERT Solutions Class 11 Mathematics Chapter 7 are prepared by subject matter enthusiasts through thorough research and analysis. The team also provides guidance and support in the student’s journey to pursue their dream careers.

### Key Features of NCERT Solutions Class 11 Mathematics Chapter 7

• All concepts in the NCERT Solutions Class 11 Mathematics Chapter 7 are explained in a detailed and well-structured manner for students to understand easily.
• Students will be able to tackle tricky questions with the help of the tricks and shortcut methods included in the notes.
• The NCERT Solutions for Class 11 Mathematics Chapter 7 provided by the Extramarks platform follows the CBSE books. It also adheres to all guidelines issued by the board.
• The solutions include all the answers to the textual questions having the concepts which student needs to study, learn and revise to gain top scores in the annual examinations.
• By practising the questions of Chapter 7 Mathematics Class 11 over and over again, students can improve their time-management skills by solving all the questions in the allotted time.
• The notes aim to clear all queries and doubts of the students.
• Students preparing for entrance examinations can also refer to these NCERT Solutions Class 11 Mathematics Chapter 7.

Q.1 How many 3-digit numbers can be formed from the digits 1, 2, 3, 4 and 5 assuming that

(i) repetition of the digits is allowed?
(ii) repetition of the digits is not allowed?

Ans

The given three digits are 1, 2, 3, 4 and 5.
Number of digits in number to be formed = 3
(i) If repetition of digits is allowed.
Number of ways to fill one’s digit place = 5
Number of ways to fill ten’s digit place = 5
No. of ways to fill hundred’s digit place = 5
So, total formed 3-digit numbers = 5 × 5 × 5
= 125.
Thus, total 3-digit numbers are 125.

(ii) If repetition of digits is not allowed.
Number of ways to fill one’s digit place = 5
Number of ways to fill ten’s digit place = 4
No. of ways to fill hundred’s digit place = 3
So, total formed 3-digit numbers = 5 × 4 × 3
= 60
Thus, total 3-digit numbers are 60.

Q.2 How many 3-digit even numbers can be formed from the digits 1, 2, 3, 4, 5, 6 if the digits can be repeated?

Ans

The given three digits are 1, 2, 3, 4, 5 and 6.
Number of digits in number to be formed = 3
If repetition of digits is allowed.
Number of ways to fill one’s digit place = 3
Number of ways to fill ten’s digit place = 6
No. of ways to fill hundred’s digit place = 6
So, total formed 3-digit numbers = 6 x 6 x 3 = 108.
Thus, total 3-digit even numbers are 108.

Q.3 How many 4-letter codes can be formed using the first 10 letters of the English alphabet, if no letter can be repeated?

Ans

Number of letters of the English alphabet taken = 10
Number of letter used in making letter codes = 4
Since, no letter can be repeated in these codes.

 I II III IV

So, number of ways to fill first place = 10
Number of ways to fill second place = 9
Number of ways to fill third place = 8
Number of ways to fill fourth place = 7
Total number of 4-letter codes = 10 x 9 x 8 x 7
= 5040.
Thus, total 4-letter codes are 5040.

Q.4 How many 5-digit telephone numbers can be constructed using the digits 0 to 9 if each number starts with 67 and no digit appears more than once?

Ans

Total given digits from 0 to 9 = 10
Number of digits in telephone number = 5
Since, each number starts with 67. So, number of
vacant places in telephone number = 3

 6 7 III IV V

Number of ways to fill third place = 8
Number of ways to fill fourth place = 7
Number of ways to fill fifth place = 6
So, number of 5-digit telephone numbers
= 8 x 7 x 6
= 336
Thus, total number of 5-digit telephone numbers is 336.

Q.5 A coin is tossed 3 times and the outcomes are recorded. How many possible outcomes are there?

Ans

Number of possible outcomes in first toss of coin = 2
Number of possible outcomes in 2nd toss of coin = 2
Number of possible outcomes in 3rd toss of coin = 2
Total number of possible outcomes in 3 tosses of coin
= 2 × 2 × 2
= 8
Thus, total possible outcomes are 8.

Q.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?

Ans

Number of flags of different colours = 5
Flags required for a signal = 2
Number of ways to take first flag for a signal
= 5
Number of ways to take second flag for a signal
= 4
Number of different signals generated by 5 flags
= 5 × 4
= 20
Thus, total number of signals generated by using 5 flags is 20.

Q.7 Evaluate: (i) 8! (ii) 4! – 3!

Ans

(i) 8! = 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1
= 40320
(ii) 4! – 3! = 4 × 3 × 2 × 1 – 3 × 2 × 1
= 24 – 6
= 18

Q.8 Is 3! + 4! = 7! ?

Ans

3! + 4! = 3 × 2 × 1 + 4 × 3 × 2 × 1
= 6 + 24
= 30
7! = 7 × 6 × 5 × 4 × 3 × 2 × 1
= 5040
No, 3! + 4! ≠ 7!

Q.9

$\mathrm{Compute}:\frac{8!}{6!×2!}$

Ans

$\begin{array}{l}\frac{8!}{6!\text{ }×\text{ }2!}=\frac{\text{8}×\text{7}×\text{6}×\text{5}×\text{4}×\text{3}×\text{2}×\text{1}}{\text{6}×\text{5}×\text{4}×\text{3}×\text{2}×\text{1}×\text{2}×\text{1}}\\ \text{ }=\frac{\text{8}×\text{7}}{\text{2}×\text{1}}\\ \text{ }=28\end{array}$

Q.10

$\mathrm{If}\text{\hspace{0.17em}}\frac{1}{6!}+\frac{1}{7!}=\frac{\mathrm{x}}{8!},\mathrm{find}\mathrm{x}.$

Ans

$\begin{array}{l}\mathrm{We}\text{ are given}\\ \text{ }\frac{1}{6!}+\frac{1}{7!}=\frac{\text{x}}{8!}\\ \text{ }\frac{1}{6!}+\frac{1}{7.6!}=\frac{\text{x}}{8!}\\ \text{ }\frac{1}{6!}\left(1+\frac{1}{7}\right)=\frac{\text{x}}{8!}\\ \text{ }\frac{1}{6!}\left(\frac{7+1}{7}\right)=\frac{\text{x}}{8!}\\ \text{ }\frac{1}{6!}\left(\frac{8}{7}\right)=\frac{\text{x}}{8!}\\ \text{ }\frac{8}{7!}=\frac{\text{x}}{8.\left(7!\right)}\\ \text{ }8=\frac{\text{x}}{8}\\ ⇒\text{ x}=64\end{array}$

Q.11

$\begin{array}{l}\mathbf{Evaluate}\frac{\mathbf{n}\mathbf{!}}{\mathbf{\left(}\mathbf{n}\mathbf{–}\mathbf{r}\mathbf{\right)}\mathbf{!}}\mathbf{,}\mathbf{\text{​​}}\mathbf{when}\\ \mathbf{\left(}\mathbf{i}\mathbf{\right)}\mathbf{\text{\hspace{0.17em}}}\mathbf{n}\mathbf{=}\mathbf{6}\mathbf{,}\mathbf{\text{\hspace{0.17em}\hspace{0.17em}}}\mathbf{r}\mathbf{=}\mathbf{2}\mathbf{}\mathbf{\left(}\mathbf{ii}\mathbf{\right)}\mathbf{}\mathbf{n}\mathbf{=}\mathbf{9}\mathbf{,}\mathbf{\text{\hspace{0.17em}\hspace{0.17em}}}\mathbf{r}\mathbf{=}\mathbf{5}\mathbf{.}\end{array}$

Ans

$\begin{array}{l}\left(\mathrm{i}\right)\text{\hspace{0.17em}}\mathrm{n}=6,\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{r}=2\\ \frac{\mathrm{n}!}{\left(\mathrm{n}-\mathrm{r}\right)!}=\frac{6!}{\left(6-2\right)!}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=\frac{6×5×4!}{4!}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}=30\\ \left(\mathrm{ii}\right)\mathrm{n}=9,\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{r}=5\\ \frac{\mathrm{n}!}{\left(\mathrm{n}-\mathrm{r}\right)!}=\frac{9!}{\left(9-5\right)!}\\ \begin{array}{l}\text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{9×8×7×6×5×4!}{4!}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=15120\end{array}\end{array}$

Q.12 How many 3-digit numbers can be formed by using the digits 1 to 9 if no digit is repeated?

Ans

Number of digits from 1 to 9 = 9
Number of digits in each number = 3
If no digit is repeated, then
Number of 3-digit numbers = 9P3
= 9! /(9 – 3)!
= 9 × 8 × 7
= 504
Thus, there are 504, 3-digit numbers which can be formed by using the digits 1 to 9.

Q.13 How many 4-digit numbers are there with no digit repeated?

Ans

Number of digits from 0 to 9 = 10
Number of digits in each number = 4
If no digit is repeated, then
Number of 4-digit numbers = 9 × 9P3
= 9 × 9 × 8 × 7
= 4536
Thus, there are 4536, 4-digit numbers which can be formed by using the digits 0 to 9.

Q.14 How many 3-digit even numbers can be made using the digits 1, 2, 3, 4, 6, 7, if no digit is repeated?

Ans

Number of given digits = 6
Number of digits in each number = 3
Since, each number is even, so last digit may be 2, 4 or 6.
If no digit is repeated, then Number of ways to fill remaining two places of 3-digit numbers = 5P2
= 5! /(5 – 2)!
= 5! / 3!
= 20
Number of ways to arrange 3 even numbers at one’s place = 3
Total formed 3-digit even numbers = 20 × 3 = 60
Thus, there are 60, 3-digit even numbers.

Q.15 Find the number of 4-digit numbers that can be formed using the digits 1, 2, 3, 4, 5 if no digit is repeated. How many of these will be even?

Ans

Number of given digits = 5
Number of digits in each number = 4
Number of 4-digit numbers = 5P4
= 5!
= 120
Now, we find 4 digit even numbers.
Only 2 numbers are possible at units place (2, 4) as we need even number.
If no digit is repeated, then number of ways to fill remaining three places of 4-digit numbers = 4P3
= 4! /(4 – 3)!
= 4! / 1!
= 24
Number of ways to arrange 2 even numbers at one’s place = 2
Total formed 4-digit even numbers = 24 x 2
= 48
Thus, there are 48, 4 digit even numbers.

Q.16 From a committee of 8 persons, in how many ways can we choose a chairman and a vice chairman assuming one person can not hold more than one position?

Ans

$\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Number}\text{of persons in committee}=8\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Number}\text{of chairman to be selected}=\text{1}\\ \mathrm{Number}\text{of vice chairman to be selected}=\text{1}\\ \text{Number of persons to be selected for both posts}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\text{2}\\ \text{Number of ways to select a chairman and vice chairman}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\text{\hspace{0.17em}}}^{8}{\mathrm{P}}_{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{8!}{\left(8-2\right)!}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{8×7×6!}{6!}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=56\\ \mathrm{Thus},\text{​ the number of ways to choose both persons is 56.}\end{array}$

Q.17

$\mathbf{Find}\mathbf{}\mathbf{n}\mathbf{}\mathbf{if}{\mathbf{}}^{\mathbf{n}\mathbf{–}\mathbf{1}}{\mathbf{P}}_{\mathbf{3}}\mathbf{}\mathbf{:}{\mathbf{}}^{\mathbf{n}}{\mathbf{P}}_{\mathbf{4}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{1}\mathbf{}\mathbf{:}\mathbf{}\mathbf{9}$

Ans

$\begin{array}{l}\mathrm{We}\text{ are given:}\\ {\text{ }}^{\text{n}-1}{\text{P}}_{3}\text{}:{}^{\text{n}}{\text{P}}_{4}=1:9\\ ⇒\text{ }\frac{{}^{\text{n}-1}{\text{P}}_{3}}{{}^{\text{n}}{\text{P}}_{4}}=\frac{1}{9}\\ ⇒\text{ }\frac{\left\{\frac{\left(\text{n}-1\right)!}{\left(\text{n}-1-3\right)!}\right\}}{\left\{\frac{\text{n}!}{\left(\text{n}-4\right)!}\right\}}=\frac{1}{9}\\ ⇒\text{ }\frac{\left(\text{n}-1\right)!}{\left(\text{n}-1-3\right)!}×\frac{\left(\text{n}-4\right)!}{\text{n}!}=\frac{1}{9}\\ ⇒\text{ }\frac{\left(\text{n}-1\right)!}{\left(\text{n}-4\right)!}×\frac{\left(\text{n}-4\right)!}{\text{n}\left(\text{n}-1\right)!}=\frac{1}{9}\\ ⇒\text{ }\frac{1}{\text{n}}=\frac{1}{9}\\ ⇒\text{ n}=9\end{array}$

Q.18

$\begin{array}{l}\mathbf{\text{Find r if}}\\ \left(\text{i}\right){\mathbf{\text{ }}}^{\mathbf{\text{5}}}{\mathbf{\text{P}}}_{\mathbf{\text{r}}}{\mathbf{\text{= 2 }}}^{\mathbf{\text{6}}}{\mathbf{\text{P}}}_{\mathbf{\text{r}}\mathbf{–}\mathbf{\text{1}}}\\ \left(\text{ii}\right){\mathbf{\text{ }}}^{\mathbf{\text{5}}}{\mathbf{\text{P}}}_{\mathbf{\text{r}}}{\mathbf{\text{= }}}^{\mathbf{\text{6}}}{\mathbf{\text{P}}}_{\mathbf{\text{r–1}}}\end{array}$

Ans

$\begin{array}{l}\left(\mathrm{i}\right){\text{ }}^{5}{\mathrm{P}}_{\mathrm{r}}={2}^{6}{\mathrm{P}}_{\mathrm{r}-1}\\ ⇒\text{ }\frac{5!}{\left(5-\mathrm{r}\right)!}=2\cdot \frac{6!}{\left(6-\mathrm{r}+1\right)!}\\ ⇒\text{ }\frac{5!}{\left(5-\mathrm{r}\right)!}=2\cdot \frac{6!}{\left(7-\mathrm{r}\right)!}\\ ⇒\text{ }\frac{5!}{\left(5-\mathrm{r}\right)!}=2\cdot \frac{6.5!}{\left(7-\mathrm{r}\right)\left(6-\mathrm{r}\right)\left(5-\mathrm{r}\right)!}\\ ⇒\text{ }\frac{1}{1}=2\cdot \frac{6}{\left(7-\mathrm{r}\right)\left(6-\mathrm{r}\right)}\\ ⇒\text{ }\left(7-\mathrm{r}\right)\left(6-\mathrm{r}\right)=12\\ ⇒\text{ }42-13\mathrm{r}+{\mathrm{r}}^{2}=12\\ ⇒\text{ }30-13\mathrm{r}+{\mathrm{r}}^{2}=0\\ ⇒\text{ }\left(10-\mathrm{r}\right)\left(3-\mathrm{r}\right)=0\\ ⇒\text{ }\left(10-\mathrm{r}\right)\left(3-\mathrm{r}\right)=0\\ ⇒\text{ }\mathrm{r}=10,3\\ ⇒\text{ }\mathrm{r}=3\left[\mathrm{Neglecting}\text{ }10,\mathrm{because}\text{ }0<\mathrm{r}\le \mathrm{n}.\right]\\ \left(\mathrm{ii}\right){\text{ }}^{5}{\mathrm{P}}_{\mathrm{r}}{=}^{6}{\mathrm{P}}_{\mathrm{r}-1}\\ ⇒\text{ }\frac{5!}{\left(5-\mathrm{r}\right)!}=\frac{6!}{\left(6-\mathrm{r}+1\right)!}\\ ⇒\text{ }\frac{5!}{\left(5-\mathrm{r}\right)!}=\frac{6!}{\left(7-\mathrm{r}\right)!}\\ ⇒\text{ }\frac{5!}{\left(5-\mathrm{r}\right)!}=\frac{6.5!}{\left(7-\mathrm{r}\right)\cdot \left(6-\mathrm{r}\right)\cdot \left(5-\mathrm{r}\right)!}\\ ⇒\text{ }1=\frac{6}{\left(7-\mathrm{r}\right)\left(6-\mathrm{r}\right)}\\ ⇒\text{ }\left(7-\mathrm{r}\right)\left(6-\mathrm{r}\right)=6\\ ⇒\text{ }42-13\mathrm{r}+{\mathrm{r}}^{2}=6\\ ⇒\text{ }36-13\mathrm{r}+{\mathrm{r}}^{2}=0\\ ⇒\text{ }\left(9-\mathrm{r}\right)\left(4-\mathrm{r}\right)=0\\ ⇒\text{ }\mathrm{r}=9,4\\ ⇒\text{ }\mathrm{r}=4\text{ }\left[\text{Neglecting}9,\text{ because}0<\mathrm{r}\le \mathrm{n}.\right]\end{array}$

Q.19 How many words, with or without meaning, can be formed using all the letters of the word EQUATION, using each letter exactly once?

Ans

$\begin{array}{l}\text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Number}\text{of letters in ‘EQUATION’}=8\\ \mathrm{Number}\text{\hspace{0.17em}of words using each letter once}={\text{\hspace{0.17em}}}^{8}{\mathrm{P}}_{8}\end{array}$ $\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{8!}{\left(8-8\right)!}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{8!}{0!}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=8×7×6×5×4×3×2×1\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=40320\\ \mathrm{Thus},\text{number of words formed by using letters of ‘Equation’}\\ \text{once is 40320.}\end{array}$

Q.20 How many words, with or without meaning can be made from the letters of the word MONDAY, assuming that no letter is repeated, if.

(i) 4 letters are used at a time,
(ii) all letters are used at a time,
(iii) all letters are used but first letter is a vowel?

Ans

$\begin{array}{l}\mathrm{Number}\text{of letters in ‘MONDAY’}=6\\ \left(\mathrm{i}\right)\text{If 4 letters are used at a time.}\\ \mathrm{Number}\text{of 4-letter words that can be formed from the letters}\\ \text{of ‘MONDAY’, then}\\ \text{Number of words}={\text{\hspace{0.17em}}}^{6}{\mathrm{P}}_{4}\\ =\frac{6!}{\left(6-4\right)!}\\ =\frac{6!}{2!}\\ =6×5×4×3\\ =360\end{array}$ $\begin{array}{l}\mathrm{Thus},\text{number of words formed by using 4 letters is 360.}\\ \left(\mathrm{ii}\right)\text{\hspace{0.17em}}\mathrm{all}\text{letters are used at a time.}\\ \text{Number of words}={\text{\hspace{0.17em}}}^{6}{\mathrm{P}}_{6}\\ \text{\hspace{0.17em}}=\frac{6!}{\left(6-6\right)!}\\ \text{\hspace{0.17em}}=\frac{6!}{0!}\\ \text{\hspace{0.17em}}=6×5×4×3×2×1\\ \text{\hspace{0.17em}}=720\\ \left(\mathrm{iii}\right)\mathrm{Number}\text{of vowels}=2\\ \mathrm{Number}\text{of ways to select first letter}\\ ={\text{\hspace{0.17em}}}^{2}{\mathrm{P}}_{1}\\ =\frac{2!}{\left(2-1\right)!}\\ =2\\ \mathrm{Number}\text{​ of ways to select remaining 5 letters}\\ ={\text{\hspace{0.17em}}}^{5}{\mathrm{P}}_{5}\\ =\frac{5!}{\left(5-5\right)!}\\ =5×4×3×2×1\\ =120\\ \mathrm{So},\text{\hspace{0.17em}}\mathrm{total}\text{number of words}=2×120\\ =240\end{array}$

Q.21 In how many of the distinct permutations of the letters in MISSISSIPPI do the four I’s not come together?

Ans

$\begin{array}{l}\mathrm{Number}\text{of letters in ‘MISSISSIPPI’}=11\\ \mathrm{Number}\text{of I’s}=4\\ \mathrm{Number}\text{of S’s}=4\\ \mathrm{Number}\text{of P’s}=2\end{array}$ $\begin{array}{l}\mathrm{Number}\text{of permutations}=\frac{11!}{4!4!2!}\\ =\frac{11×10×9×8×7×6×5×4!}{4!4!2!}\\ =\frac{11×10×9×8×7×6×5}{4×3×2×1×2×1}\\ =34650\\ \mathrm{When}\text{4 I’s come together.}\\ \mathrm{Then},\text{\hspace{0.17em}number of words}=\frac{8!}{4!2!}\\ =\frac{8×7×6×5×4!}{4!×2×1}\\ =840\\ \mathrm{So},\text{required distinct permutations}=34650-840\\ =33810\end{array}$

Q.22 In how many ways can the letters of the word PERMUTATIONS be arranged if the

(ii) vowels are all together,
(iii) there are always 4 letters between P and S?

Ans

$\begin{array}{l}\mathrm{Number}\text{of letters in ‘PERMUTATIONS’}=12\\ \text{ }\mathrm{Number}\text{of T’s}=\text{2}\\ \left(\text{i}\right)\text{ }\mathrm{When}\text{words start with P and end with S,}\\ \text{ then number of the words formed}=\frac{10!}{2!}\\ \text{ }=1814400\\ \left(\mathrm{ii}\right)\text{ }\mathrm{Vowels}\text{}\mathrm{are}\text{all together,}\\ \text{ Number of vowels in PERMUTATIONS}=5\\ \text{ }\mathrm{If}\text{all vowels are considered as a single letter, then}\\ \text{ number of letters to be arranged}=12-5+1\\ \text{ }=8\\ \text{ then number of the words formed}=\frac{8!}{2!}×{\text{ }}^{5}{\text{P}}_{5}\\ \text{ }=20160×120\\ \text{ }=2419200\\ \left(\mathrm{iii}\right)\text{ If there are always 4 letters between P and S, then}\\ \text{ }\mathrm{number}\text{of letters to be arranged}=12-2\\ \text{ }=10\\ \text{ }\mathrm{Number}\text{repeated letter T}=\text{2}\\ \text{ Number of ways to arrange 10 letters}=\frac{10!}{2!}\\ \text{ }=1814400\\ \text{ }\begin{array}{cccccccccccc}\text{P}& & & & & \text{S}& & & & & & \end{array}\\ \text{ }\begin{array}{ccccccc}1& 2& 3& 4& 5& 6& 7\end{array}\\ \text{ }\mathrm{Total}\text{number of places to shift P and S}=7\\ \text{ P and S can be interchange, so total number}\\ \text{ of ways of arrangement of P and S}=2×7\\ \therefore \text{ }\mathrm{Number}\text{of ways to arrange letters}=14×1814400\\ \text{ }=25401600\end{array}$

Q.23 If nC8 = nC2, find nC2.

Ans

$\mathrm{Given}{,}^{\text{n}}{\text{C}}_{\text{8}}={\text{\hspace{0.17em}}}^{\text{n}}{\text{C}}_{\text{2}}$ $\begin{array}{l}{⇒}^{\text{n}}{\text{C}}_{\text{8}}={\text{\hspace{0.17em}}}^{\text{n}}{\text{C}}_{\text{n-2}}\text{}\left[\because {\text{\hspace{0.17em}}}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}}={\text{\hspace{0.17em}}}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{n}-\mathrm{r}}\right]\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}8=\mathrm{n}-2\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}10=\mathrm{n}\\ \mathrm{So},{\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}}^{\mathrm{n}}{\mathrm{C}}_{2}={\text{\hspace{0.17em}}}^{10}{\mathrm{C}}_{2}\\ =\frac{10!}{2!\left(10-2\right)!}\\ =\frac{10×9×8!}{2!\text{\hspace{0.17em}}8!}\\ =45\end{array}$

Q.24 Determine n if
(i) 2nC3 : nC3 = 12 : 1
(ii) 2nC3 : nC3 = 11 : 1

Ans

$\begin{array}{l}\left(\text{i}\right)\text{ }\frac{{}^{\text{2n}}{\text{C}}_{\text{3}}}{{}^{\text{n}}{\text{C}}_{\text{3}}}=\frac{12}{1}\\ ⇒\text{ }\frac{\frac{2\text{n}!}{3!\left(2\text{n}-3\right)!}}{\frac{\text{n}!}{3!\left(\text{n}-3\right)!}}=12\\ ⇒\text{ }\frac{2\text{n}!}{3!\left(2\text{n}-3\right)!}×\frac{3!\left(\text{n}-3\right)!}{\text{n}\left(\text{n}-1\right)\left(\text{n}-2\right)\left(\text{n}-3\right)!}=12\\ ⇒\text{ }\frac{2\text{n}\left(2\text{n}-1\right)\left(2\text{n}-2\right)\left(2\text{n}-3\right)!}{\left(2\text{n}-3\right)!}×\frac{1}{\text{n}\left(\text{n}-1\right)\left(\text{n}-2\right)}=12\\ ⇒\text{ }2\text{n}\left(2\text{n}-1\right)2\left(\text{n}-1\right)×\frac{1}{\text{n}\left(\text{n}-1\right)\left(\text{n}-2\right)}=12\\ ⇒\text{ }\frac{\left(2\text{n}-1\right)}{\text{n}-2}=3\\ ⇒\text{ }2\text{n}-1=3\text{n}-6\\ ⇒\text{ }6-1=3\text{n}-2\text{n}\\ ⇒\text{ }5=\text{n}\\ ⇒\text{ n}=5.\\ \left(\text{ii}\right){\text{ }}^{\text{2n}}{\text{C}}_{\text{3}}:{\text{ }}^{\text{n}}{\text{C}}_{\text{3}}=\text{11}:\text{1}\\ \text{ }\frac{{}^{\text{2n}}{\text{C}}_{\text{3}}}{{}^{\text{n}}{\text{C}}_{\text{3}}}=\frac{11}{1}\\ ⇒\text{ }\frac{\frac{2\text{n}!}{3!\left(2\text{n}-3\right)!}}{\frac{\text{n}!}{3!\left(\text{n}-3\right)!}}=11\\ ⇒\text{ }\frac{2\text{n}!}{3!\left(2\text{n}-3\right)!}×\frac{3!\left(\text{n}-3\right)!}{\text{n}\left(\text{n}-1\right)\left(\text{n}-2\right)\left(\text{n}-3\right)!}=11\\ ⇒\text{ }\frac{2\text{n}\left(2\text{n}-1\right)\left(2\text{n}-2\right)\left(2\text{n}-3\right)!}{\left(2\text{n}-3\right)!}×\frac{1}{\text{n}\left(\text{n}-1\right)\left(\text{n}-2\right)}=11\\ ⇒\text{ }2\text{n}\left(2\text{n}-1\right)2\left(\text{n}-1\right)×\frac{1}{\text{n}\left(\text{n}-1\right)\left(\text{n}-2\right)}=11\\ \text{ }\frac{4\left(2\text{n}-1\right)}{\text{n}-2}=11\\ \text{ }8\text{n}-4=11\text{n}-22\\ \text{ }22-4=11\text{n}-8\text{n}\\ \text{ }18=3\text{n}\\ ⇒\text{ n}=6\end{array}$

Q.25 How many chords can be drawn through 21 points on a circle?

Ans

$\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Number}\text{of points on a circle}=\text{21}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Number of points used for a chord}=\text{2}\\ \mathrm{Number}\text{of chords made by 21 points}={\text{\hspace{0.17em}}}^{21}{\mathrm{C}}_{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{21!}{2!\left(21-2\right)!}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{21×20×19!}{2×1×19!}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=210\\ \mathrm{Thus},\text{​​ number of chords in a circle is 210.}\end{array}$

Q.26 In how many ways can a team of 3 boys and 3 girls be selected from 5 boys and 4 girls?

Ans

$\begin{array}{l}\mathrm{Total}\text{​ number of boys}=\text{5}\\ \text{Total number of girls}=4\\ \mathrm{Number}\text{​ of boys in a team}=\text{3}\\ \text{Number of girls in a team}=\text{3}\end{array}$ $\begin{array}{l}\text{Number of ways to select 3 boys}={\text{\hspace{0.17em}}}^{5}{\mathrm{C}}_{3}\\ =\frac{5!}{3!\left(5-3\right)!}\\ =\frac{5×4×3!}{3!\text{\hspace{0.17em}}×2!}\\ =10\\ \text{Number of ways to select 3 girls}={\text{\hspace{0.17em}}}^{4}{\mathrm{C}}_{3}\end{array}$ $\begin{array}{l}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\end{array}$ $\begin{array}{l}=\frac{4!}{3!\left(4-3\right)!}\\ =\frac{4×3!}{3!\text{\hspace{0.17em}}×1!}\\ =4\\ \mathrm{Total}\text{​\hspace{0.17em}number of ways to select 3}\\ \text{boys and 3 girls}=10×4\\ =40\end{array}$

Q.27 Find the number of ways of selecting 9 balls from 6 red balls, 5 white balls and 5 blue balls if each selection consists of 3 balls of each colour.

Ans

$\begin{array}{l}\mathrm{Number}\text{of red balls}=\text{6}\\ \mathrm{Number}\text{of white balls}=\text{5}\\ \mathrm{Number}\text{of blue balls}=\text{5}\\ \mathrm{Number}\text{of ways to select 3 red balls}={\text{\hspace{0.17em}}}^{6}{\mathrm{C}}_{3}\\ =\frac{6!}{3!\left(6-3\right)!}\end{array}$ $\begin{array}{l}=\frac{6×5×4×3!}{3!.3!}\\ =20\\ \mathrm{Number}\text{of ways to select 3 white balls}={\text{\hspace{0.17em}}}^{5}{\mathrm{C}}_{3}\\ =\frac{5!}{3!\left(5-3\right)!}\\ =\frac{5×4×3!}{3!.2!}\\ =10\\ \mathrm{Number}\text{of ways to select 3 blue balls}={\text{\hspace{0.17em}}}^{5}{\mathrm{C}}_{3}\\ =\frac{5!}{3!\left(5-3\right)!}\\ =\frac{5×4×3!}{3!.2!}\\ =10\\ \mathrm{Number}\text{\hspace{0.17em}of ways of selecting 9 balls}=20×10×10\\ =2000\end{array}$

Q.28 Determine the number of 5 card combinations out of a deck of 52 cards if there is exactly one ace in each combination.

Ans

$\begin{array}{l}\mathrm{Number}\text{of cards in a deck}=52\\ \mathrm{Number}\text{of cards in each combination}=\text{5}\\ \text{Number of aces in a deck}=4\\ \text{Number of ace in each combination}=\text{1}\\ \text{Number of ways to select 1 ace out of 4}={\text{\hspace{0.17em}}}^{4}{\mathrm{C}}_{1}\end{array}$ $\begin{array}{l}=\frac{4!}{1!\left(4-1\right)!}\\ =\frac{4!}{1!3!}\\ =4\\ \text{\hspace{0.17em}No. of ways to select 4 cards out of 48}={\text{\hspace{0.17em}}}^{48}{\mathrm{C}}_{4}\\ =\frac{48!}{4!\left(48-4\right)!}\\ =\frac{48!}{4!\text{\hspace{0.17em}}44!}\\ =\frac{48×47×46×45×44!}{4×3×2×1×44!}\\ =2×47×46×45\\ =194580\\ \text{Number of 5 cards combination}=\text{\hspace{0.17em}}2×47×46×45×4\\ =778320\end{array}$

Q.29 In how many ways can one select a cricket team of eleven from 17 players in which only 5 players can bowl if each cricket team of 11 must include exactly 4 bowlers?

Ans

$\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Total}\text{​​ number of players}=17\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Number}\text{of players who can bowl}=\text{5}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Number of players in cricket team}=11\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Number}\text{of bowlers in cricket team}=4\end{array}$ $\begin{array}{l}\mathrm{Number}\text{\hspace{0.17em}of players who can not bowl}=17-5\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=12\\ \text{\hspace{0.17em}}\mathrm{Number}\text{​ of ways to select 11 players}={\text{\hspace{0.17em}}}^{12}{\mathrm{C}}_{7}×{\text{\hspace{0.17em}}}^{5}{\mathrm{C}}_{4}\\ =\frac{12!}{7!\left(12-7\right)!}×\frac{5!}{4!\left(5-4\right)!}\\ =\frac{12!}{7!\text{\hspace{0.17em}}×5!}×\frac{5!}{4!1!}\\ =\frac{12×11×10×9×8×\overline{)7!}}{\overline{)7!\text{\hspace{0.17em}}}×\overline{)5!}}×\frac{\overline{)5!}}{4!}\\ =3960\end{array}$

Q.30 A bag contains 5 black and 6 red balls. Determine the number of ways in which 2 black and 3 red balls can be selected.

Ans

$\begin{array}{l}\mathrm{Number}\text{of black balls in a bag}=5\\ \mathrm{Number}\text{of red balls in a bag}=6\\ \mathrm{Number}\text{of ways to select 2 black and 3 red balls}\\ ={\text{ }}^{5}{\text{C}}_{2}\text{ }×{\text{ }}^{6}{\text{C}}_{3}\\ =\frac{5!}{2!\left(5-2\right)!}×\frac{6!}{3!\left(6-3\right)!}\\ =\frac{5!}{2!\text{ }×\text{ }3!}×\frac{6!}{3!\text{ }×\text{ }3!}\\ =\frac{5×4×\overline{)3!}}{2!\text{ }×\text{ }\overline{)3!}}×\frac{\overline{)6}×5×4×\overline{)3!}}{\overline{)3!}\text{ }×\text{ }\overline{)3!}}\\ =10×5×4\\ =200\end{array}$

Q.31 In how many ways can a student choose a programme of 5 courses if 9 courses are available and 2 specific courses are compulsory for every student?

Ans

$\begin{array}{l}\mathrm{Available}\text{courses for students}=9\\ \mathrm{Number}\text{of specific courses}=2\\ \mathrm{Courses}\text{to be chosen by students}=5\\ \mathrm{Not}\text{specific courses for students}=5-2\\ =3\\ \mathrm{Number}\text{of ways to select 3 remaining courses}\\ ={\text{​\hspace{0.17em}\hspace{0.17em}}}^{7}{\mathrm{C}}_{3}\\ =\frac{7!}{3!\text{\hspace{0.17em}}×\text{\hspace{0.17em}}\left(7-3\right)!}\left[\because {\text{\hspace{0.17em}}}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}}=\frac{\mathrm{n}!}{\mathrm{r}!\left(\mathrm{n}-\mathrm{r}\right)!}\right]\\ =\frac{7!}{3!\text{\hspace{0.17em}}×\text{\hspace{0.17em}}4!}\\ =\frac{7×6×5×4!}{3!\text{\hspace{0.17em}}×\text{\hspace{0.17em}}4!}\\ =35\\ \mathrm{Thus},\text{a student can choose a programme of 5 courses in 35 ways.}\end{array}$

Q.32 How many words, with or without meaning, each of 2 vowels and 3 consonants can be formed from the letters of the word DAUGHTER?

Ans

$\begin{array}{l}\mathrm{Number}\text{of letters in the word ‘DAUGHTER’}=8\\ \mathrm{Number}\text{of vowels in the word ‘DAUGHTER’}=3\\ \mathrm{Number}\text{of consonants in the word ‘DAUGHTER’}=5\\ \mathrm{Ways}\text{to select 2 vowels and 3 consonants}={\text{\hspace{0.17em}}}^{3}{\mathrm{C}}_{2}×{\text{\hspace{0.17em}}}^{5}{\mathrm{C}}_{3}\\ =\frac{3!}{2!\left(3-2\right)!}×\frac{5!}{3!\left(5-3\right)!}\\ =\frac{1}{2!\text{\hspace{0.17em}}×\text{\hspace{0.17em}}1!}×\frac{5!}{2!}\\ =\frac{120}{4}\\ =30\\ \mathrm{Number}\text{of ways to arrange 2 vowels and 3 consonants}\\ =30×5!\end{array}$ $\begin{array}{l}=30×120\\ =3600\\ \mathrm{Thus},\text{required number of different words is 3600.}\end{array}$

Q.33 How many words, with or without meaning, can be formed using all the letters of the word EQUATION at a time so that the vowels and consonants occur together?

Ans

$\begin{array}{l}\mathrm{Number}\text{of letter in the word ‘EQUATION’}=8\\ \mathrm{Number}\text{of vowel in the word ‘EQUATION’}=5\\ \mathrm{Number}\text{of ways to arrange vowels}={\text{ }}^{5}{\text{P}}_{5}\\ =5!\\ =120\\ \mathrm{Number}\text{ of ways to arrange 3 consonants}={\text{ }}^{3}{\text{P}}_{3}\\ =3!\\ =6\\ \mathrm{Since},\text{vowels and consonants will occur together, both vowels}\left(\mathrm{EUAIO}\right)\text{ and consonants}\\ \left(\mathrm{QTN}\right)\text{can be considered two objects. So, number of ways to arrange two objects}\\ ={\text{ }}^{2}{\text{P}}_{2}=2\\ \mathrm{Thus},\text{number of words formed by using}=120×6×2\\ =1440\end{array}$

Q.34 A committee of 7 has to be formed from 9 boys and 4 girls. In how many ways can this be done when the committee consists of:

(i) exactly 3 girls ?
(ii) atleast 3 girls ?
(iii) atmost 3 girls ?

Ans

$\begin{array}{l}\mathrm{Number}\text{of members in a commitee}=7\\ \text{\hspace{0.17em}}\mathrm{Number}\text{of total boys for commitee}=9\\ \text{\hspace{0.17em}\hspace{0.17em}}\mathrm{Number}\text{of total girls for commitee}=4\\ \left(\mathrm{i}\right)\mathrm{If}\text{​ there are exactly 3 girls in commitee.}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} Number of ways to select 3 girls}={\text{\hspace{0.17em}}}^{4}{\mathrm{C}}_{3}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{4!}{3!\left(4-3\right)!}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=4\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} Number of ways to select 4 boys}={\text{\hspace{0.17em}}}^{9}{\mathrm{C}}_{4}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{9!}{4!\left(9-4\right)!}\end{array}$ $\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{9×8×7×6×5!}{4×3×2×1×5!}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=126\\ \mathrm{Number}\text{of ways to form a commitee}=4×126\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=504\\ \left(\mathrm{ii}\right)\mathrm{If}\text{​ there are atleast 3 girls in commitee.}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Number of ways to form commitee}={\text{\hspace{0.17em}}}^{4}{\mathrm{C}}_{3}{×}^{9}{\mathrm{C}}_{4}{+}^{4}{\mathrm{C}}_{4}{×}^{9}{\mathrm{C}}_{3}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{4!}{3!\left(4-3\right)!}×\frac{9!}{4!\left(9-4\right)!}\\ +\frac{4!}{4!\left(4-4\right)!}×\frac{9!}{3!\left(9-3\right)!}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{4!}{3!\text{\hspace{0.17em}\hspace{0.17em}}1!}×\frac{9!}{4!\text{\hspace{0.17em}\hspace{0.17em}}5!}+\frac{4!}{4!\text{\hspace{0.17em}\hspace{0.17em}}0!}×\frac{9!}{3!\text{\hspace{0.17em}\hspace{0.17em}}6!}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=4×126+1×84\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=504+84\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=588\\ \left(\mathrm{iii}\right)\mathrm{If}\text{​ there are atmost 3 girls in commitee.}\\ \text{There are different ways to complete 7 persons in commitee.}\\ \text{When 3 girls and 4 boys are selected}={\text{\hspace{0.17em}}}^{4}{\mathrm{C}}_{3}{×}^{9}{\mathrm{C}}_{4}\\ \text{\hspace{0.17em}\hspace{0.17em}}=\frac{4!}{3!\left(4-3\right)!}×\frac{9!}{4!\left(9-4\right)!}\\ \text{\hspace{0.17em}\hspace{0.17em}}=4×126\text{\hspace{0.17em}\hspace{0.17em}}=504\\ \text{When 2 girls and 5 boys are selected}={\text{\hspace{0.17em}}}^{4}{\mathrm{C}}_{2}×{\text{\hspace{0.17em}}}^{9}{\mathrm{C}}_{5}\\ \text{\hspace{0.17em}\hspace{0.17em}}=\frac{4!}{2!\left(4-2\right)!}×\frac{9!}{5!\left(9-5\right)!}\\ \text{\hspace{0.17em}\hspace{0.17em}}=6×126\end{array}$ $\begin{array}{l}\end{array}$ $\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}}=756\\ \text{When 1 girl and 6 boys are selected}={\text{\hspace{0.17em}}}^{4}{\mathrm{C}}_{1}×{\text{\hspace{0.17em}}}^{9}{\mathrm{C}}_{6}\\ \text{\hspace{0.17em}\hspace{0.17em}}=\frac{4!}{1!\left(4-1\right)!}×\frac{9!}{6!\left(9-6\right)!}\\ \text{\hspace{0.17em}\hspace{0.17em}}=4×84\\ \text{\hspace{0.17em}\hspace{0.17em}}=336\\ \text{When no girl and 6 boys are selected}={\text{\hspace{0.17em}}}^{4}{\mathrm{C}}_{0}×{\text{\hspace{0.17em}}}^{9}{\mathrm{C}}_{7}\\ \text{\hspace{0.17em}\hspace{0.17em}}=\frac{4!}{0!\left(4-0\right)!}×\frac{9!}{7!\left(9-7\right)!}\\ \text{\hspace{0.17em}\hspace{0.17em}}=1×36\\ \text{\hspace{0.17em}\hspace{0.17em}}=36\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Number of ways to form commitee}={\text{\hspace{0.17em}}}^{4}{\mathrm{C}}_{3}{×}^{9}{\mathrm{C}}_{4}+{\text{\hspace{0.17em}}}^{4}{\mathrm{C}}_{2}{×}^{9}{\mathrm{C}}_{5}\\ +{\text{\hspace{0.17em}}}^{4}{\mathrm{C}}_{1}×{\text{\hspace{0.17em}}}^{9}{\mathrm{C}}_{6}+{\text{\hspace{0.17em}}}^{4}{\mathrm{C}}_{0}×{\text{\hspace{0.17em}}}^{9}{\mathrm{C}}_{7}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=504+756+336+36\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=1632\end{array}$

Q.35 If the different permutations of all the letter of the word EXAMINATION are listed as in a dictionary, how many words are there in this list before the first word starting with E?

Ans

$\begin{array}{l}\text{\hspace{0.17em}}\mathrm{Number}\text{of letters in word ‘EXAMINATION’}=11\\ \mathrm{Number}\text{of letter A in word ‘EXAMINATION’}=2\\ \text{\hspace{0.17em}}\mathrm{Number}\text{of letter I in word ‘EXAMINATION’}=2\\ \mathrm{Number}\text{of letter N in word ‘EXAMINATION’}=2\\ \mathrm{The}\text{words starting with A will come first in dictionary from}\\ \text{the words starting with E.}\end{array}$ $\begin{array}{l}\text{So, the number of words starting with letter A}\\ =\frac{10!}{2!\text{\hspace{0.17em}\hspace{0.17em}}2!}\\ =\frac{10×9×8×7×6×5×4×3×2!}{2!\text{\hspace{0.17em}\hspace{0.17em}}2!}\\ =5×9×8×7×6×5×4×3\\ =907200\\ \mathrm{Thus},\text{there are 907200 words listed in dictionary formed by}\\ \text{the letters of ‘EXAMINATION’ before the first word starting with}\\ \text{letter E.}\end{array}$

Q.36 How many 6-digit numbers can be formed from the digits 0, 1, 3, 5, 7 and 9 which are divisible by 10 and no digit is repeated?

Ans

$\begin{array}{l}\mathrm{Since},\text{a number will be divisible by 10 if its last digit is 0.}\\ \text{Number of given digits}=6\\ \mathrm{Last}\text{digit of number formed is fixed i.e., 0.}\\ \text{Since, no digit is repeated.}\end{array}$

 0

$\begin{array}{l}\text{Number of 6-digit numbers formed}=\text{\hspace{0.17em}}\mathrm{ways}\text{to fill remaining places}\\ \text{\hspace{0.17em}of 6-digit numbers}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=5×4×3×2×1\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=120\\ \mathrm{Thus},\text{the number of 6-digit numbers formed by given digits}\\ \text{and divisible by 10 is 120.}\end{array}$

Q.37 The English alphabet has 5 vowels and 21 consonants. How many words with two different vowels and 2 different consonants can be formed from the alphabet?

Ans

$\begin{array}{l}\mathrm{Number}\text{of vowels in the English alphabets}=5\\ \mathrm{Number}\text{of consonants in the English alphabets}=21\\ \mathrm{Number}\text{of words formed by 2 vowels and 2 consonants}\\ ={\text{\hspace{0.17em}}}^{5}{\mathrm{C}}_{2}×{\text{\hspace{0.17em}}}^{21}{\mathrm{C}}_{2}\\ =\text{\hspace{0.17em}}\frac{5!}{2!\left(5-2\right)!}\\ ×\frac{21!}{2!\left(21-2\right)!}\\ =\frac{5!}{2!3!}×\frac{21!}{2!\text{\hspace{0.17em}}19!}\\ =10×210\\ =2100\\ \mathrm{These}\text{2100 words has 4 letters only.}\\ \text{Number of ways to arrange 4 letters}={\text{\hspace{0.17em}}}^{4}{\mathrm{P}}_{4}\\ =\frac{4!}{\left(4-4\right)!}\\ =4!=24\\ \mathrm{So},\text{​ the number of total words formed by 2 vowels and}\\ \text{2 consonants}=2100×24\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=50400\end{array}$

Q.38 In an examination, a question paper consists of 12 questions divided into two parts i.e., Part I and Part II, containing 5 and 7 questions, respectively. A student is required to attempt 8 questions in all, selecting at least 3 from each part. In how many ways can a student select the questions?

Ans

$\begin{array}{l}\mathrm{Number}\text{of questions in a paper}=12\\ \mathrm{Number}\text{of questions in I part}=5\\ \mathrm{Number}\text{of questions in II part}=7\\ \mathrm{Number}\text{of ways to select questions from I part and II part each}\\ \begin{array}{ccc}\begin{array}{l}\mathrm{Questions}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Part}\text{\hspace{0.17em}}\mathrm{I}\end{array}& \begin{array}{l}\mathrm{Questions}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Part}\text{\hspace{0.17em}}\mathrm{I}\end{array}& \mathrm{Ways}\text{to select the questions}\\ 3& 5& {}^{5}{\mathrm{C}}_{3}×{\text{\hspace{0.17em}}}^{7}{\mathrm{C}}_{5}\\ 4& 4& {}^{5}{\mathrm{C}}_{4}×{\text{\hspace{0.17em}}}^{7}{\mathrm{C}}_{4}\\ 5& 3& {}^{5}{\mathrm{C}}_{5}×{\text{\hspace{0.17em}}}^{7}{\mathrm{C}}_{3}\end{array}\\ \mathrm{Number}\text{of ways to select at least 3 questions from each part}\\ ={\text{\hspace{0.17em}}}^{5}{\mathrm{C}}_{3}×{\text{\hspace{0.17em}}}^{7}{\mathrm{C}}_{5}+{\text{\hspace{0.17em}}}^{5}{\mathrm{C}}_{4}×{\text{\hspace{0.17em}}}^{7}{\mathrm{C}}_{4}+{\text{\hspace{0.17em}}}^{5}{\mathrm{C}}_{5}×{\text{\hspace{0.17em}}}^{7}{\mathrm{C}}_{3}\\ =\text{\hspace{0.17em}}\frac{5!}{3!\text{\hspace{0.17em}}\left(5-3\right)!}×\frac{7!}{5!\text{\hspace{0.17em}}\left(7-5\right)!}\\ +\frac{5!}{4!\text{\hspace{0.17em}}\left(5-4\right)!}×\frac{7!}{4!\text{\hspace{0.17em}}\left(7-4\right)!}\\ +\frac{5!}{4!\text{\hspace{0.17em}}\left(5-4\right)!}×\frac{7!}{5!\text{\hspace{0.17em}}\left(7-5\right)!}\\ =\text{\hspace{0.17em}}\frac{5!}{3!\text{\hspace{0.17em}\hspace{0.17em}}2!}×\frac{7!}{5!\text{\hspace{0.17em}\hspace{0.17em}}2!}+\frac{5!}{4!\text{\hspace{0.17em}\hspace{0.17em}}1!}×\frac{7!}{4!\text{\hspace{0.17em}\hspace{0.17em}}3!}\\ +\frac{5!}{4!\text{\hspace{0.17em}\hspace{0.17em}}1!}×\frac{7!}{5!\text{\hspace{0.17em}\hspace{0.17em}}1!}\\ =210+175+35\end{array}$ $\begin{array}{l}=420\\ \mathrm{Thus},\text{a student can select 8 questions from question paper by}\\ \text{420 ways.}\end{array}$

Q.39 Determine the number of 5-card combinations out of a deck of 52 cards if each selection of 5 cards has exactly one king.

Ans

$\begin{array}{l}\mathrm{Total}\text{number of cards}=52\\ \mathrm{Number}\text{of kings}=4\\ \mathrm{A}\text{set of 5 cards have one king and 4 other cards.}\\ \text{Number of ways to make 5-card combinations}\\ {=}^{4}{\mathrm{C}}_{1}×{\text{\hspace{0.17em}}}^{48}{\mathrm{C}}_{4}\\ =\frac{4!}{1!\text{\hspace{0.17em}\hspace{0.17em}}3!}×\frac{48!}{4!\text{\hspace{0.17em}\hspace{0.17em}}\left(48-4\right)!}\\ =\frac{4!}{1!\text{\hspace{0.17em}\hspace{0.17em}}3!}×\frac{48!}{4!\text{\hspace{0.17em}\hspace{0.17em}}44!}\\ =4×194580\\ =778320\\ \mathrm{Thus},\text{required number of 5-card combinations is 778320.}\end{array}$

Q.40 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?

Ans

$\begin{array}{l}\mathrm{Number}\text{of men to seat}=5\\ \mathrm{Number}\text{of women to seat}=4\\ \mathrm{Women}\text{occupy even place in seating arrangement.}\end{array}$

 M W M W M W M W M

$\begin{array}{l}\mathrm{Seats}\text{for women on even places}=\text{4}\\ \mathrm{Seats}\text{for women on odd places}=\text{5}\\ \text{Number of arrangements of seating}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=4!\text{\hspace{0.17em}}×5!\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=24×120\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=2880\\ \mathrm{Thus},\text{5 men and 4 women can sit in 2880 ways.}\end{array}$

Q.41 From a class of 25 students, 10 are to be chosen for an excursion party. There are 3 students who decide that either all of them will join or none of them will join. In how many ways can the excursion party be chosen?

Ans

$\begin{array}{l}\mathrm{Total}\text{number of students in a class}=25\\ \mathrm{Number}\text{of students will join party}=10\\ \mathrm{Either}\text{\hspace{0.17em}}3\text{students will join party or not.}\\ \text{Case\hspace{0.17em}I: When 3 students join party.}\\ \text{Number of students to be select}=10-3\\ =7\\ \mathrm{Number}\text{of ways to select 7 students\hspace{0.17em}}={\text{\hspace{0.17em}}}^{22}{\mathrm{C}}_{7}\\ \text{Case\hspace{0.17em}II: When 3 students are not joining party.}\\ \text{Number of remaining students in class}=\text{22}\\ \text{Number of students to select from class}=10\\ \mathrm{Number}\text{of ways to select 10 students\hspace{0.17em}}={\text{\hspace{0.17em}}}^{22}{\mathrm{C}}_{10}\\ \mathrm{Then},\text{total ways to select 10 students}={\text{\hspace{0.17em}}}^{22}{\mathrm{C}}_{7}{+}^{22}{\mathrm{C}}_{10}\end{array}$

Q.42 In how many ways can the letters of the word ASSASSINATION be arranged so that all the S’s are together?

Ans

$\begin{array}{l}\mathrm{Number}\text{of letters in the word ‘ASSASSINATION}‘=13\\ \mathrm{Number}\text{of letter\hspace{0.17em}A in the word ‘ASSASSINATION}‘=3\\ \mathrm{Number}\text{of letter\hspace{0.17em}S in the word ‘ASSASSINATION}‘=4\\ \text{\hspace{0.17em}}\mathrm{Number}\text{of letter\hspace{0.17em}I in the word ‘ASSASSINATION}‘=2\\ \mathrm{Number}\text{of letter\hspace{0.17em}N in the word ‘ASSASSINATION}‘=2\\ \mathrm{If}\text{all S’s are taken together and all S’s are considered}\\ \text{a single letter.}\\ \text{Number letters to arrange in the ‘ASSASSINATION}‘=13-4+1\\ =10\\ \mathrm{Number}\text{of words when all S’s are taken together}=\frac{10!}{3!\text{\hspace{0.17em}\hspace{0.17em}}2!\text{\hspace{0.17em}\hspace{0.17em}}2!}\\ =\frac{10×9×8×7×6×5×4×3!}{3!\text{\hspace{0.17em}}×2!\text{\hspace{0.17em}}×2!}\\ =151200\\ \mathrm{Thus},\text{\hspace{0.17em}the letters of the word\hspace{0.17em}ASSASSINATION can be arranged}\\ \text{in 151200 ways so that all the S}’\text{s}\mathrm{}\text{\hspace{0.17em}are together.}\end{array}$

## 1. Is it important to use the NCERT Solutions Class 11 Mathematics Chapter 7?

The NCERT Solutions are very important from an exam point of view. Chapter 7 is one of the most important chapters, and to gain high marks in it, students must understand all concepts thoroughly. Practising the problems included in the NCERT Solutions Class 11 Mathematics Chapter 7, students will be able to solve questions of all difficulty levels without making any silly mistakes.

## 2. Do Extramarks provide notes for all chapters of Class 11 Mathematics?

Extramarks is a leading website which provides a wholesome learning experience. NCERT Solutions are well-structured and detailed. The following chapter notes are available on the Extramarks:

Ch 1: Sets

Ch 2: Relations and Functions

Ch 3: Trigonometric Functions

Ch 4: Principle of Mathematical Induction

Ch 5: Complex Numbers and Quadratic Equations

Ch 6: Linear Inequalities

Ch 7: Permutations and Combinations

Ch 8: Binomial Theorem

Ch 9: Sequences and Series

Ch 10: Straight Lines

Ch 11: Conic Sections

Ch 12: Introduction to Three Dimensional Geometry

Ch 13: Limits and Derivatives

Ch 14: Mathematical Reasoning

Ch 15: Statistics

Ch 16: Probability