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RD Chapter 16- Permutations Ex-16.2 Interview Questions Answers

Question 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?

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 27C1 and similarly the numberof ways to select one girl is 14C1.

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

14C1 × 27C1 =14 × 27 = 378 ways.

Question 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?

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 10C1 and similarlythe number of ways to select one ball pen is 12C1 andnumber of ways to select one pencil from 5 pencils is 5C1

Hence, the number ofways to select one fountain pen, one ball pen and one pencil from a stationeryshop is 10C1 × 12C1 × 5C1= 10 × 12 × 5 = 600 ways.

Question 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?

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 2C1

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

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

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

Question 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?

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.

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

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, 5C1.Similarly, for other parcels also it can be posted in 5C1 ways.

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

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

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

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.

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

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: 2C1 × 2C1 × 2C1 × 2C1 × 2C1 × 2C1 × 2C1 × 2C1 × 2C1 × 2C1 =2×2×2×2×2×2×2×2×2×2 = 1024 ways

Question 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?

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.

Question 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?

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 = 4C1 × 4C1 × 4C1 =4 × 4 × 4 = 64

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

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

Hence, total possibleoutcomes possible are 64 × 8= 512

Question 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?

Answer 10 :

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

In order to buy onemathematics book, number of ways is 5C1 similarlyto buy one physics book number of ways is 6C1

Hence, the number ofways a student buy a Mathematics book and a Physics book is 5C1 × 6C1 =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 11C1 = 11


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RD Chapter 16- Permutations Ex-16.2 Contributors

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