RD Sharma Solutions Class 11 Maths Chapter 16 Exercise 16.2 (Updated for 2024)

RD Sharma Solutions Class 11 Maths Chapter 16 Exercise 16.2: RD Sharma Solutions are one of the most useful sources for students to score good marks in Class 12 Maths board exams. These solutions are created by our experts with the aim of making the Class 12 Mathematics RD Sharma Solutions Class 11 Maths Chapter 16 Exercise 16.2 easier. You can easily download RD Sharma solutions from the link provided below. 

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1. In a class, there are 27 boys and 14 girls. The teacher wants to select 1 boy and 1 girl to represent the class in a function. In how many ways can the teacher make this selection?

Solution:

Given:

27 boys and 14 girls.

Here the teacher has to

(i) Select a boy among 27 boys, and

(ii) Select a girl among 14 girls.

The number of ways to select one boy is ²⁷C₁ and similarly, the number of ways to select one girl is ¹⁴C₁.

Hence, the number of ways to select 1 boy and 1 girl to represent the class in a function is

¹⁴C₁ × ²⁷C₁ = 14 × 27 = 378 ways.

2. A person wants to buy one fountain pen, one ball pen, and one pencil from a stationery shop. If there are 10 fountain pen varieties, 12 ball pen varieties and 5 pencil varieties, in how many ways can he select these articles?

Solution:

Given:

10 fountain pens, 12 ball pens, and 5 pencil

Here the person has to

(i) Select a ball pen from 12 ball pens.

(ii) Select a fountain pen from 10 fountain pens, and

(iii) Select a pencil from 5 pencils.

The number of ways to select one fountain pen is ¹⁰C₁ and similarly, the number of ways to select one ball pen is ¹²C₁ and the number of ways to select one pencil from 5 pencils is ⁵C₁

Hence, the number of ways to select one fountain pen, one ball pen and one pencil from a stationery shop is ¹⁰C₁ × ¹²C₁ × ⁵C_{1 =} 10 × 12 × 5 = 600 ways.

3. From Goa to Bombay there are two roots; air, and sea. From Bombay to Delhi there are three routes; air, rail, and road. From Goa to Delhi via Bombay, how many kinds of routes are there?

Solution:

Given:

The number of roots from Goa to Bombay is air and sea.

So, the number of ways to go from Goa to Bombay is ²C₁

Given: The number of roots from Bombay to Delhi are air, rail, and road.

So, the number of ways to go from Bombay to Delhi is ³C₁

Hence, the number of ways to go from Goa to Delhi via Bombay is ²C₁ × ³C₁ = 2 × 3 = 6 ways.

4. A mint prepares metallic calendars specifying months, dates and days in the form of monthly sheets (one plate for each month). How many types of calendars should it prepare to serve for all the possibilities in future years?

Solution:

The mint has to perform

(i) Select the number of days in the month of February (there can be 28 or 29 days), and

(ii) Select the first day of February.

Now,

In 2 ways, the mint can select the number of days in February and for selecting first day of February, it can start from any of one of the seven days of the week, so there are 7 possibilities.

Hence, the number of types of calendars it should prepare to serve for all the possibilities in future years is 7 × 2 = 14.

5. There are four parcels and five post offices. In how many different ways can the parcels be sent by registered post?

Solution:

Given:

Total number of parcels = 4

Total number of post-offices = 5

One parcel can be posted in 5 ways, that is in either of the one post offices. So, ⁵C₁. Similarly, for other parcels also, it can be posted in ⁵C₁ ways.

Hence the number of ways the parcels be sent by registered post is

⁵C₁ × ⁵C₁ × ⁵C₁ × ⁵C₁ = 5 × 5 × 5 × 5 = 625 ways.

6. A coin is tossed five times, and outcomes are recorded. How many possible outcomes are there?

Solution:

Given:

A coin is tossed 5 times, so each time, the outcome is either heads or tails, so two possibilities are possible.

The total possible outcomes are:

²C₁ × ²C₁ × ²C₁ × ²C₁ × ²C₁ = 2 × 2 × 2 × 2 × 2 = 32 outcomes.

7. In how many ways can an examinee answer a set of ten true/false type questions?

Solution:

Given:

An examinee can answer a question either true or false, so there are two possibilities.

The number of ways for an examinee to answer a set of ten true/false type questions are: ²C₁ × ²C₁ × ²C₁ × ²C₁ × ²C₁ × ²C₁ × ²C₁ × ²C₁ × ²C₁ × ²C₁ = 2×2×2×2×2×2×2×2×2×2 = 1024 ways

8. A letter lock consists of three rings each marked with 10 different letters. In how many ways it is possible to make an unsuccessful attempt to open the lock?

Solution:

The total number of ways to make an attempt to open the lock is = 10 × 10 × 10 = 1000

The number of successful attempts to open the lock = 1

The number of unsuccessful attempts to open the lock = 1000 – 1 = 999

Hence, the required number of possible ways to make an unsuccessful attempt to open the lock is 999.

9. There are 6 multiple choice questions in an examination. How many sequences of answers are possible, if the first three questions have 4 choices each and the next three have 2 each?

Solution:

Given: Multiple choice question, only one answer is correct of the given options.

For the first three questions, only one answer is correct out of four. So it can be answered in 4 ways.

Total number of ways to answer the first 3 questions = ⁴C₁ × ⁴C₁ × ⁴C₁ = 4 × 4 × 4 = 64

Each of the next 3 questions can be answered in 2 ways.

Total number of ways to answer the next 3 questions = ²C₁ × ²C₁ × ²C₁ = 2 × 2 × 2 = 8

Hence, the total possible outcomes possible are 64 × 8= 512

10. There are 5 books on Mathematics and 6 books on Physics in a book shop. In how many ways can a student buy:
(i) a Mathematics book and a Physics book
(ii) either a Mathematics book or a Physics book?

Solution:

(i) Given: there are 5 books of mathematics and 6 books of physics.

In order to buy one mathematics book, the number of ways is ⁵C₁ similarly, to buy one physics book number of ways is ⁶C₁

Hence, the number of ways a student buy a Mathematics book and a Physics book is ⁵C₁ × ⁶C₁ = 5 × 6 = 30

(ii) Given: there is a total of 11 books.

So in order to buy either a Mathematics book or a Physics book, it means that only one book out of eleven is bought.

Hence, the number of ways in which a student can either buy either a Mathematics book or a Physics book is ¹¹C₁ = 11

11. Given 7 flags of different colours, how many different signals can be generated if a signal requires the use of two flags, one below the other?

Solution:

Given: seven flags are available and out of which two are needed to make a signal.

From this, we can say that we have to select two flags out of seven and arrange these two flags to get one signal.

Seven flags of different colours are available, so the first flag can be selected in 7 ways.

Now, the second flag can be selected from any one of the remaining flags in 6 ways.

Hence, the required number of signals is 7 × 6 = 42.

12. A team consists of 6 boys and 4 girls, and other has 5 boys and 3 girls. How many single matches can be arranged between the two teams when a boy plays against a boy, and a girl plays against a girl?

Solution:

Given:

One team consists of 6 boys and 4 girls, and the other team has 5 boys and 3 girls.

Let team 1 be = 6 boys and 4 girls

Team 2 be = 5 boys and 3 girls

Singles matches are to be played, either a boy plays against a boy, and a girl plays against a girl.

So, the number of ways to select a boy from team 1 is ⁶C₁. Similarly, the number of ways to select a boy from team 2 is ⁵C₁

Hence number of singles matches between boys is ⁶C₁ × ⁵C₁ = 6 × 5 = 30

The number of ways to select a girl from team 1 is ⁴C₁. Similarly, the number of ways to select a girl from team 2 is ³C₁

Hence number of singles matches between girls is ³C₁ × ⁴C₁ = 4 × 3 = 12

∴ The total number of single matches are = 30 + 12 = 42

13. Twelve students compete in a race. In how many ways first three prizes be given?

Solution:

Given:

Twelve students compete in a race.

The number of ways to select the winner of the first prize is ¹²C₁

The number of ways to select the winner of the second prize is ¹¹C₁ (11, since one student is already given a prize)

The number of ways to select the winner of the third prize is ¹⁰C₁ (10, since two students are already given a prize)

Hence, the total number of ways is ¹²C₁ × ¹¹C₁ × ¹⁰C₁ = 12 × 11 × 10 = 1320.

14. How many A.P.’s with 10 terms are there whose first term is in the set {1, 2, 3} and whose common difference is in the set {1, 2, 3, 4, 5}?

Solution:

We know that each AP consists of a unique first term and a common difference.

So, the number of ways to select the first term of a given set is ³C₁ = 3

And the number of ways to select a common difference of a given set is ⁵C₁ = 5

Hence, the total number of APs possible are ³C₁ × ⁵C₁ = 3 × 5 = 15

15. From among the 36 teachers in a college, one principal, one vice-principal and the teacher-in-charge are to be appointed. In how many ways can this be done?

Solution:

The number of ways to appoint one principal, vice-principal and the teacher incharge is equal to the number of ways to select the three teachers from 36 members.

So, a total of three positions are to be appointed.

The number of ways to select principal is ³⁶C₁ = 36

The number of ways to select vice-principal is ³⁵C₁ = 35 (35, since one position is already given)

The number of ways to select teacher in charge is ³⁴C₁ (34, since two positions are already given)

Hence, the number of ways to appoint three teachers is ³⁶C₁ × ³⁵C₁ × ³⁴C₁ = 36 × 35 × 34 = 42840.

16. How many three-digit numbers are there with no digit repeated?

Solution:

Let us assume we have three boxes.

The first box can be filled with any one of the nine digits (0 not allowed in the first place).

So, the possibilities are ⁹C₁

The second box can be filled with any one of the nine digits

So the available possibilities are ⁹C₁

The third box can be filled with any one of the eight digits

So the available possibilities are ⁸C₁

Hence, the total number of possible outcomes are ⁹C₁ × ⁹C₁ × ⁸C₁ = 9 × 9 × 8 = 648.

17. How many three-digit numbers are there?

Solution:

Let us assume we have three boxes.

The first box can be filled with any one of the nine digits (zero not allowed at first position)

So, possibilities are ⁹C₁

The second box can be filled with any one of the ten digits

So the available possibilities are ¹⁰C₁

The third box can be filled with any one of the ten digits

So the available possibilities are ¹⁰C₁

Hence, the total number of possible outcomes are ⁹C₁ × ¹⁰C₁ × ¹⁰C₁ = 9 × 10 × 10 = 900.

18. How many three-digit odd numbers are there?

Solution:

We know that in odd numbers, the last digit consists of (1, 3, 5, 7, 9).

Let us assume we have three boxes.

The first box can be filled with any one of the nine digits (zero not allowed at first position)

So the possibilities are ⁹C₁

The second box can be filled with any one of the ten digits

So the available possibilities are ¹⁰C₁

The third box can be filled with any one of the five digits (1,3,5,7,9)

So the available possibilities are ⁵C₁

Hence, the total number of possible outcomes are ⁹C₁ × ¹⁰C₁ × ⁵C₁ = 9 × 10 × 5 = 450

19. How many different five-digit number license plates can be made if
(i) the first digit cannot be zero, and the repetition of digits is not allowed,
(ii) the first-digit cannot be zero, but the repetition of digits is allowed?

Solution:

(i) We know that zero cannot be the first digit of the license plates. And the repetition of digits is not allowed.

Let us assume five boxes, now the first box can be filled with one of the nine available digits, so the possibility is ⁹C₁

Similarly, the second box can be filled with one of the nine available digits, so the possibility is ⁹C₁

The third box can be filled with one of the eight available digits, so the possibility is ⁸C₁

The fourth box can be filled with one of the seven available digits, so the possibility is ⁷C₁

The fifth box can be filled with one of the six available digits, so the possibility is ⁶C₁

Hence, the total number of possible outcomes is ⁹C₁ × ⁹C₁ × ⁸C₁ × ⁷C₁ × ⁶C₁ = 9 × 9 × 8 × 7 × 6 = 27,216

(ii) We know that zero cannot be the first digit of the license plates. And the repetition of digits is allowed.

Let us assume five boxes, now the first box can be filled with one of the nine available digits, so the possibility is ⁹C₁

Similarly, the second box can be filled with one of the ten available digits, so the possibility is ¹⁰C₁

The third box can be filled with one of the ten available digits, so the possibility is ¹⁰C₁

The fourth box can be filled with one of the ten available digits, so the possibility is ¹⁰C₁

The fifth box can be filled with one of the ten available digits, so the possibility is ¹⁰C₁

Hence, the total number of possible outcomes is  ⁹C₁ × ¹⁰C₁ × ¹⁰C₁ × ¹⁰C₁ × ¹⁰C₁ = 9 × 10 × 10 × 10 × 10 = 90,000

20. How many four-digit numbers can be formed with the digits 3, 5, 7, 8, 9 which are greater than 7000, if repetition of digits is not allowed?

Solution:

The required numbers are greater than 7000.

So, the thousand’s place can be filled with any of the 3 digits: 7, 8, 9.

Let us assume four boxes, now in the first box can either be one of the three numbers 7, 8 or 9, so there are three possibilities which are ³C₁

In the second box, the numbers can be any of the four digits left, so the possibility is ⁴C₁

In the third box, the numbers can be any of the three digits left, so the possibility is ³C₁

In the fourth box, the numbers can be any of the two digits left, so the possibility is ²C₁

Hence, the total number of possible outcomes is ³C₁ × ⁴C₁ × ³C₁ × ²C₁ = 3 × 4 × 3 × 2 = 72.

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FAQ: RD Sharma Solutions Class 11 Maths Chapter 16 Exercise 16.2