Class 11 RD Sharma Solutions- Chapter 16 Permutations – Exercise 16.2 | Set 1

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?

Solution:

Given: Total number of boys = 27

Total number of girls = 14

So, ways to select a boy = 27 P1 = 27

Ways to select a girl = 14P1 = 14

Ways for selecting a pair of 1 boy, 1 girl = 27 x 14 = 378

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?

Solution:

Given: Total number of fountain pen = 10

Total number of ball pen = 12

Total number of fountain pencil = 5

Person wants to buy only one fountain pen, one ball pen, and one pencil

So, ways to select a pen = 10P1 = 10

Ways to select a ball pen = 12P1 = 12

Ways to select a pencil = 5P1 = 5

Ways for selecting the desired triplet = 10 x 12 x 5 = 600

Question 3. From Goa to Bombay there are two routes; 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: From Goa to Bombay two routes = air, and sea

From Bombay to Delhi there are three routes = air, rail, and road

So, the routes from Goa to Bombay = 2P1 = 2

Routes from Bombay to Delhi = 3P1 = 3

Total different routes from Goa to Delhi = 2 x 3 = 6

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?

Solution:

We need to find the different total number of calendars so that all the years can be represented by any one of these.

Case 1: Leap year

A leap year may start with any of 7 possible days (Monday to Sunday) = 7 options

Case 2: Ordinary year

An ordinary year may start with any of 7 possible days (Monday to Sunday) = 7 options

Total calendars = 7 + 7 = 14

Question 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

Each of the four parcels have 5 options of post-offices.

So, each parcel can be sent in 5P1 ways.

Hence, the total ways = 5P1 x 5P1 x 5P1 x 5P1 = 5 x 5 x 5 x 5 = 625

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

Solution:

Each toss can result in 2P1 = 2 ways.

Five tosses = 2 x 2 x 2 x 2 x 2 = 32 ways of outcomes

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

Solution:

For answering one question: 2 P1 = 2 ways

For answering 10 questions: 2 x 2 x 2 x……2 (10 times) = 210 = 1024 possibilities

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?

Solution:

Number of possibilities for a single ring = 10 ways

For 3 rings: 10 x 10 x 10 = 1000 ways

Out of these, 1 way will be the correct password

So, the number of unsuccessful attempts = 1000 – 1 = 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?

Solution:

Possible sequences for first 3 questions = 4P1 x 4P1 x 4P1 = 4 x 4 x 4 = 64

Possible sequences for next 3 questions = 2P1 x 2P1 x 2P1 = 2 x 2 x 2 = 8

Total possibilities = 64 x 8 = 512

(ii) either a Mathematics book or a Physics book?

Solution:

Given: Total number of Mathematics book = 5

Total number of Physics book = 6

(i) Number of ways of buying Mathematics book = 5P1   = 5

Number of ways of buying Physics book = 6P1  = 6

Total possibilities = 5 x 6 = 30

(ii) Number of ways of buying a book (can be any Maths or Physics) = 11 P1 = 11

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

Solution:

Ways to select 2 flags out of 7 = 7P2 = 7x(7 – 1)/2 = 21

Ways of generating different signals from these 2 selected flags = 2

(say, A and B are the colors selected then A can be above and B below; and vice versa so 2 ways)

Total distinct signals possible = 21 x 2 = 42

Question 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:

Case 1: A boy plays against a boy

Select a boy from team 1 and a boy from team 2

Team 1: 6P1 = 6

Team 2: 5P1 = 5

Total ways of a boy playing against a boy = 6 x 5 = 30

Case 2: A girl plays against a girl

Select a girl from team 1 and a boy from team 2

Team 1: 4P1 = 4

Team 2: 3P1 = 3

Total ways of a girl playing against a girl = 4 x 3 = 12

Total ways of signal matches = Ways of boy playing against boy + Ways of girl playing against girl

= 30 + 12

= 42

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

Solution:

Number of ways of selecting 3 winners = 12P3 = 12 x 11 x 10 / (3 x 2 x 1) = 220

For 3 winners selected, different ways of assigning the position

For first position we have 3 possibilities of people,

then for second we have 2 possibilities (other than the one already given first position)

and for third we have 1 possibility (other than the ones declared second and first positions)

So, 3 x 2 x 1 = 6 possibilities of assigning these 3 positions to the three selected people

Total ways of giving 3 prizes = No. of ways of selecting 3 people x Assigning 3 positions to the 3 people

= 220 x 6

= 1320

Question 14. How many A.P.â€™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:

Number of ways of selecting first term = 3P1 = 3

Number of ways of selecting common difference = 5P1 = 5

Total different A.P. series = 3 x 5 = 15

Question 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:

Number of ways of selecting 3 people = 36P3

= 36 x 35 x 34 / (3 x 2 x 1)

= 7140

For 3 people selected, different ways of assigning the posts

For the principal post we have 3 possibilities of people,

then for vice-principal post we have 2 possibilities (other than the one already given principal post)

and for teacher-in-charge we have 1 possibility (other than people declared principal and vice-principal)

So, 3 x 2 x 1 = 6 possibilities of assigning these 3 posts to the three selected people

Total ways of assigning 3 posts = No. of ways of selecting 3 people x Assigning 3 posts to 3 people

= 7140 x 6

= 42840

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

Solution:

Ways of selecting 100th place = 9P1 = 9 (Selecting from all digits except 0)

Ways of selecting 10th place = 9P1 = 9 (Selecting from all digits except the digit placed at 100th position)

Ways of selecting unit pace = 8P1 = 8 (Selecting from all digits except those at 100th and 10th places)

Total 3-digit numbers possible with no digit repeated = 9 x 9 x 8 = 648

Previous
Next
Share your thoughts in the comments
Similar Reads