RD Chapter 16- Permutations Ex-16.1 |
RD Chapter 16- Permutations Ex-16.3 |
RD Chapter 16- Permutations Ex-16.4 |
RD Chapter 16- Permutations Ex-16.5 |

**Answer
1** :

Given:

27 boys and 14 girls.

Here the teacher hasto

(i) Select a boy among27 boys, and

(ii) Select a girlamong 14 girls.

Number of ways toselect one boy is ^{27}C_{1} and similarly the numberof ways to select one girl is ^{14}C_{1}.

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

^{14}C_{1} × ^{27}C_{1} =14 × 27 = 378 ways.

**Answer
2** :

Given:

10 fountain pens, 12ball pens, and 5 pencil

Here the person has to

(i) Select a ball penfrom 12 ball pens.

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

(iii) Select a pencilfrom 5 pencils.

The number of ways toselect one fountain pen is ^{10}C_{1} and similarlythe number of ways to select one ball pen is ^{12}C_{1} andnumber of ways to select one pencil from 5 pencils is ^{5}C_{1}

Hence, the number ofways to select one fountain pen, one ball pen and one pencil from a stationeryshop is ^{10}C_{1} × ^{12}C_{1} × ^{5}C_{1=} 10 × 12 × 5 = 600 ways.

**Answer
3** :

Given:

The number of rootsfrom Goa to Bombay is air and sea.

So, the number of waysto go from Goa to Bombay is ^{2}C_{1}

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

So, the number of waysto go from Bombay to Delhi is ^{3}C_{1}

Hence, the number ofways to go from Goa to Delhi via Bombay is ^{2}C_{1} × ^{3}C_{1 }= 2× 3 = 6 ways.

**Answer
4** :

The mint has toperform

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

(ii) Select the firstday of February.

Now,

In 2 ways mint canselect 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 7possibilities.

Hence, the number oftypes calendars should it prepare to serve for all the possibilities in futureyears is 7 × 2 = 14.

**Answer
5** :

Given:

Total number ofparcels = 4

Total number ofpost-offices = 5

One parcel can beposted in 5 ways that is in either of the one post offices. So, ^{5}C_{1}.Similarly, for other parcels also it can be posted in ^{5}C_{1} ways.

Hence the number ofways the parcels be sent by registered post is

^{5}C_{1} × ^{5}C_{1} × ^{5}C_{1} × ^{5}C_{1 }= 5× 5 × 5 × 5 = 625 ways.

**Answer
6** :

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.

**Answer
7** :

Given:

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

The number of ways foran examinee to answer a set of ten true/false type questions are: ^{2}C_{1} × ^{2}C_{1} × ^{2}C_{1} × ^{2}C_{1} × ^{2}C_{1} × ^{2}C_{1} × ^{2}C_{1} × ^{2}C_{1} × ^{2}C_{1} × ^{2}C_{1} =2×2×2×2×2×2×2×2×2×2 = 1024 ways

**Answer
8** :

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

The number ofsuccessful attempts to open the lock = 1

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

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

**Answer
9** :

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

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

Total number of waysto answer the first 3 questions = ^{4}C_{1} × ^{4}C_{1} × ^{4}C_{1} =4 × 4 × 4 = 64

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

Total number of waysto answer the next 3 questions = ^{2}C_{1} × ^{2}C_{1} × ^{2}C_{1} =2 × 2 × 2 = 8

Hence, total possibleoutcomes possible are 64 × 8= 512

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?

**Answer
10** :

(i) Given: there are5 books of mathematics and 6 books of physics.

In order to buy onemathematics book, number of ways is ^{5}C_{1} similarlyto buy one physics book number of ways is ^{6}C_{1}

Hence, the number ofways a student buy a Mathematics book and a Physics book is ^{5}C_{1} × ^{6}C_{1} =5 × 6 = 30

(ii) Given: there isa total of 11 books.

So in order to buyeither a Mathematics book or a Physics book it means that only one book out ofeleven is bought.

Hence, the number ofways in which a student can either buy either a Mathematics book or a Physicsbook is ^{11}C_{1} = 11

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