This exercise deals with two fundamental principles (i) principle of addition and (ii) principle of multiplication. These two principles will enable us to understand the concept of permutations and combinations. This exercise contains various solved examples before the exercise wise problems start, in order to help students practice numerous problems to obtain a better academic score. The solutions are formulated by teachers after conducting vast research on each concept so that students improve their logical approach and their abilities to analyse the type of questions that would appear in the board exam. For a better conceptual idea, students can access RD Sharma Class 11 Maths Solutions free pdf, from the links, which are 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.

Number of ways to select one boy isÂ ^{27}C_{1}Â and similarly the number of ways to select one girl isÂ ^{14}C_{1}.

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

^{14}C_{1} Ã— ^{27}C_{1} = 14 Ã— 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 isÂ ^{10}C_{1}Â and similarly the number of ways to select one ball pen isÂ ^{12}C_{1}Â and number of ways to select one pencil from 5 pencils isÂ ^{5}C_{1}

Hence, the number of ways to select one fountain pen, one ball pen and one pencil from a stationery shop isÂ ^{10}C_{1}Â Ã—Â ^{12}C_{1}Â Ã—Â ^{5}C_{1 =}Â 10 Ã— 12 Ã— 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 isÂ ^{2}C_{1}

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 isÂ ^{3}C_{1}

Hence, the number of ways to go from Goa to Delhi via Bombay isÂ ^{2}C_{1}Â Ã—Â ^{3}C_{1 }=Â 2 Ã— 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 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 calendars should it prepare to serve for all the possibilities in future years is 7 Ã— 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,Â ^{5}C_{1}. Similarly, for other parcels also it can be posted inÂ ^{5}C_{1}Â ways.

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

^{5}C_{1}Â Ã—Â ^{5}C_{1}Â Ã—Â ^{5}C_{1}Â Ã—Â ^{5}C_{1 }=Â 5 Ã— 5 Ã— 5 Ã— 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:

^{2}C_{1}Â Ã—Â ^{2}C_{1}Â Ã—Â ^{2}C_{1}Â Ã—Â ^{2}C_{1}Â Ã—Â ^{2}C_{1}Â = 2 Ã— 2 Ã— 2 Ã— 2 Ã— 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: ^{2}C_{1}Â Ã—Â ^{2}C_{1}Â Ã—Â ^{2}C_{1}Â Ã—Â ^{2}C_{1}Â Ã—Â ^{2}C_{1}Â Ã—Â ^{2}C_{1}Â Ã—Â ^{2}C_{1}Â Ã—Â ^{2}C_{1}Â Ã—Â ^{2}C_{1}Â Ã—Â ^{2}C_{1}Â = 2Ã—2Ã—2Ã—2Ã—2Ã—2Ã—2Ã—2Ã—2Ã—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 Ã— 10 Ã— 10 = 1000

The number of successful attempts to open the lock = 1

The number of unsuccessful attempts to open the lock = 1000 â€“ 1 = 999

Hence, 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 = ^{4}C_{1}Â Ã—Â ^{4}C_{1}Â Ã—Â ^{4}C_{1}Â = 4 Ã— 4 Ã— 4 = 64

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

Total number of ways to answer the next 3 questions = ^{2}C_{1} Ã—Â ^{2}C_{1}Â Ã—Â ^{2}C_{1}Â = 2 Ã— 2 Ã— 2 = 8

Hence, total possible outcomes possible areÂ 64Â Ã—Â 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, number of ways isÂ ^{5}C_{1}Â similarly to buy one physics book number of ways isÂ ^{6}C_{1}

Hence, the number of ways a student buy a Mathematics book and a Physics book isÂ ^{5}C_{1}Â Ã—Â ^{6}C_{1}Â = 5 Ã— 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 isÂ ^{11}C_{1}Â = 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 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 areÂ 7Â Ã— 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:

A team consists of 6 boys and 4 girls, and 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, numberÂ of ways to select a boy from team 1 isÂ ^{6}C_{1}. Similarly, Number of ways to select a boy from team 2 isÂ ^{5}C_{1}

Hence number of singles matches between boys isÂ ^{6}C_{1}Â Ã—Â ^{5}C_{1}Â = 6 Ã— 5 = 30

A numberÂ of ways to select a girl from team 1 isÂ ^{4}C_{1}. Similarly, Number of ways to select a girl from team 2 isÂ ^{3}C_{1}

Hence number of singles matches between girls isÂ ^{3}C_{1}Â Ã—Â ^{4}C_{1}Â = 4 Ã— 3 = 12

âˆ´ 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.

Number of ways to select the winner of the first prize isÂ ^{12}C_{1}

Number of ways to select the winner of the second prize isÂ ^{11}C_{1}Â (11, since one student is already given a prize)

Number of ways to select the winner of the third prize isÂ ^{10}C_{1}Â (10, since two students are already given a prize)

Hence, total number of ways isÂ ^{12}C_{1}Â Ã—Â ^{11}C_{1}Â Ã—Â ^{10}C_{1}Â = 12 Ã— 11 Ã— 10 = 1320.

**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:**

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

So, number of ways to select the first term of a given set isÂ ^{3}C_{1}Â = 3

And, number of ways to select a common difference of given set isÂ ^{5}C_{1}Â = 5

Hence, total number of APâ€™s possible areÂ ^{3}C_{1}Â Ã—Â ^{5}C_{1}Â = 3 Ã— 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:**

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.

Number of ways to select principal isÂ ^{36}C_{1} = 36

Number of ways to select vice-principal isÂ ^{35}C_{1}Â = 35 (35, since one position is already given)

Number of ways to select teacher in charge isÂ ^{34}C_{1}Â (34, since two positions are already given)

Hence, number of ways to appoint three teachers isÂ ^{36}C_{1}Â Ã—Â ^{35}C_{1}Â Ã—Â ^{34}C_{1}Â = 36 Ã— 35 Ã— 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 at first place).

So, the possibilities areÂ ^{9}C_{1}

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

So the available possibilities areÂ ^{9}C_{1}

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

So the available possibilities areÂ ^{8}C_{1}

Hence, the total number of possible outcomes areÂ ^{9}C_{1}Â Ã—Â ^{9}C_{1}Â Ã—Â ^{8}C_{1}Â = 9 Ã— 9 Ã— 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 areÂ ^{9}C_{1}

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

So the available possibilities areÂ ^{10}C_{1}

Third box can be filled with any one of the ten digits

So the available possibilities areÂ ^{10}C_{1}

Hence, the total number of possible outcomes areÂ ^{9}C_{1}Â Ã—Â ^{10}C_{1}Â Ã—Â ^{10}C_{1}Â = 9 Ã— 10 Ã— 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 areÂ ^{9}C_{1}

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

So the available possibilities areÂ ^{10}C_{1}

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

So the available possibilities areÂ ^{5}C_{1}

Hence, the total number of possible outcomes areÂ ^{9}C_{1}Â Ã—Â ^{10}C_{1}Â Ã—Â ^{5}C_{1}Â = 9 Ã— 10 Ã— 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 isÂ ^{9}C_{1}

Similarly, the second box can be filled with one of the nine available digits, so the possibility isÂ ^{9}C_{1}

The thirdÂ box can be filled with one of the eight available digits, so the possibility isÂ ^{8}C_{1}

The fourthÂ box can be filled with one of the seven available digits, so the possibility isÂ ^{7}C_{1}

The fifthÂ box can be filled with one of the six available digits, so the possibility isÂ ^{6}C_{1}

Hence, the total number of possible outcomes isÂ ^{9}C_{1}Â Ã—Â ^{9}C_{1}Â Ã—Â ^{8}C_{1}Â Ã—Â ^{7}C_{1}Â Ã—Â ^{6}C_{1}Â = 9 Ã— 9 Ã— 8 Ã— 7 Ã— 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 isÂ ^{9}C_{1}

Similarly, the second box can be filled with one of the ten available digits, so the possibility isÂ ^{10}C_{1}

The thirdÂ box can be filled with one of the ten available digits, so the possibility isÂ ^{10}C_{1}

The fourthÂ box can be filled with one of the ten available digits, so the possibility isÂ ^{10}C_{1}

The fifthÂ box can be filled with one of the ten available digits, so the possibility isÂ ^{10}C_{1}

Hence, the total number of possible outcomes isÂ ^{ 9}C_{1}Â Ã—Â ^{10}C_{1}Â Ã—Â ^{10}C_{1}Â Ã—Â ^{10}C_{1}Â Ã—Â ^{10}C_{1}Â = 9 Ã— 10 Ã— 10 Ã— 10 Ã— 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 thousandâ€™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 areÂ ^{3}C_{1}

In the second box, the numbers can be any of the four digits left, so the possibility isÂ ^{4}C_{1}

In the third box, the numbers can be any of the three digits left, so the possibility isÂ ^{3}C_{1}

In the fourth box, the numbers can be any of the two digits left, so the possibility isÂ ^{2}C_{1}

Hence, the total number of possible outcomes isÂ ^{3}C_{1}Â Ã—Â ^{4}C_{1}Â Ã—Â ^{3}C_{1}Â Ã—Â ^{2}C_{1}Â = 3 Ã— 4 Ã— 3 Ã— 2 = 72.

**21. How many four-digit numbers can be formed with the digits 3, 5, 7, 8, 9 which are greater than 8000, if repetition of digits is not allowed?**

**Solution:**

The required numbers are greater than 8000.

So, the thousandâ€™s place can be filled with 2 digits: 8 or 9.

Let us assume four boxes, now in the first box can either be one of the two numbers 8 or 9, so there are two possibilities which isÂ ^{2}C_{1}

In the second box, the numbers can be any of the four digits left, so the possibility isÂ ^{4}C_{1}

In the third box, the numbers can be any of the three digits left, so the possibility isÂ ^{3}C_{1}

In the fourth box, the numbers can be any of the two digits left, so the possibility isÂ ^{2}C_{1}

Hence total number of possible outcomes isÂ ^{2}C_{1}Â Ã—Â ^{4}C_{1}Â Ã—Â ^{3}C_{1}Â Ã—Â ^{2}C_{1}Â = 2 Ã— 4 Ã— 3 Ã— 2 = 48.

**22. In how many ways can six persons be seated in a row?**

**Solution:**

Let us assume there are six seats.

In the first seat, any one of six members can be seated, so the total number of possibilities isÂ ^{6}C_{1}

In the second seat, any one of five members can be seated, so the total number of possibilities isÂ ^{5}C_{1} ways.

In the third seat, any one of four members can be seated, so the total number of possibilities isÂ ^{4}C_{1} ways.

In the fourth seat, any one of three members can be seated, so the total number of possibilities isÂ ^{3}C_{1} ways.

In the fifth seat, any one of two members can be seated, so the total number of possibilities isÂ ^{2}C_{1} ways.

In the sixth seat, only one remaining person can be seated, so the total number of possibilities isÂ ^{1}C_{1} ways.

Hence, the total number of possible outcomes =Â ^{6}C_{1}Â Ã—Â ^{5}C_{1}Â Ã—Â ^{4}C_{1}Â Ã—Â ^{3}C_{1}Â Ã—Â ^{2}C_{1}Â Ã—Â ^{1}C_{1}Â = 6 Ã— 5 Ã— 4 Ã— 3 Ã— 2 Ã— 1 = 720.

**23. How many 9-digit numbers of different digits can be formed?**

**Solution:**

In a nine-digit number, 0 cannot appear in the first digit and the repetition of digits is not allowed. So, the number of ways of filling up the first digit is ^{9}C_{1}= 9

Now, 9 digits are left including 0. So, the second digit can be filled with any of the remaining 9 digits in 9 ways.

Similarly, the thirdÂ box can be filled with one of the eight available digits, so the possibility isÂ ^{8}C_{1}

The fourthÂ digit can be filled with one of the seven available digits, so the possibility isÂ ^{7}C_{1}

The fifthÂ digit can be filled with one of the six available digits, so the possibility isÂ ^{6}C_{1}

The sixthÂ digit can be filled with one of the six available digits, so the possibility isÂ ^{5}C_{1}

The seventhÂ digit can be filled with one of the six available digits, so the possibility isÂ ^{4}C_{1}

The eighthÂ digit can be filled with one of the six available digits, so the possibility isÂ ^{3}C_{1}

The ninthÂ digit can be filled with one of the six available digits, so the possibility isÂ ^{2}Â C_{1}

Hence the number of total possible outcomes isÂ ^{9}C_{1}Â Ã—Â ^{9}C_{1}Â Ã—Â ^{8}C_{1}Â Ã—Â ^{7}C_{1}Â Ã—Â ^{6}C_{1}Â = 9 Ã— 9 Ã— 8 Ã— 7 Ã— 6 Ã— 5 Ã— 4 Ã— 3 Ã— 2 = 9 (9!)