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:

Number of boys in the class = 27

Therefore number of ways to select a boy = 27

Number of girls in the class = 14

Therefore number of ways to select a girl = 14

Therefore the number of ways to select 1 boy and 1 girl to represent the class in a function = 27 \times 14 = 378 .

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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:

Number of ways fountain pen varieties = 10

Therefore number of ways to select a fountain pen = 10

Number of ways ball pen varieties = 12

Therefore number of ways to select a ball pen = 12

Number of ways pencil varieties = 5

Therefore number of ways to select a pencil = 5

Hence the total number of ways to select a fountain pen, a ball pen and a pencil = 10 \times 12 \times 5 = 600

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Question 3: From Goa to Bombay there are 2 roots; air, and sea. From Bombay to Delhi there are 3 routes; air, rail and road. From Goa to Delhi via Bombay, how many kinds of routes are there?

Answer:

Number of routes from Goa to Bombay = 2

Number of routes from Bombay to Delhi = 3

Therefore the number of routes from Goa to Delhi via  Bombay  = 2 \times 3 = 6

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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:

There are two types of years that the mint needs to take care. Non-leap year and leap year.

Each year can start with any of the 7 days.

Hence, the mint needs to prepare 2 \times 7 = 14 types of calendars to serve for all the possibilities in future years

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

Answer:

Total number of parcels = 4

Total number of post offices = 5

Since a parcel can be sent by registered post by any of the post office, the number of different ways can the parcels be sent by registered post = 5 \times 5 \times 5 \times 5 = 5^4 = 625 

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

Answer:

There are 2 possible outcomes when a coin is tossed.

If the coin is tossed 5 times, then the possible outcomes = 2 \times 2 \times 2 \times 2 \times 2 = 2^5 = 32

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

Answer:

There are two possible ways to answer any question i.e. \text{TRUE} or \text{FALSE}

Therefore the total possible ways to answer all 10 \text{ TRUE / FALSE} questions = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 2^{10} = 1024

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Question 8: A letter lock consists of 3 rings each marked with 10 different letters. In how many it is possible to make an unsuccessful attempt to open the lock?

Answer:

The number of possible setting on each of the ring = 10

Number of rings = 3

Therefore the total number of combinations = 10 \times 10 \times 10 = 1000

However, there is 1 combination that is correct. Hence the number of possible unsuccessful attempts = 1000 - 1 = 999

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Question 9: There are 6 multiple choice questions in an examination. How many sequences of answers are possible, if the first 3 questions have 4 choices each and the next 3 have 2 each?

Answer:

Number of ways you could answer the first 3 questions = 4

Number of ways you could answer the next 3 questions = 2

Therefore the total possible ways of answering the 6 questions = 4 \times 4 \times 4 \times 2 \times 2 \times 2 = 512

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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 ii) either a Mathematics book or a Physics book?

Answer:

Number of mathematics books = 5

Therefore the number of ways of buying a Mathematics book = 5

Number of Physics books = 6

Therefore the number of ways of buying a Physics book = 5

i) The number of ways of buying a Mathematics book and a Physics = 6 \times 5 = 30

ii) The number of ways of buying a Mathematics book or a Physics book = 6+5 = 11

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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?

Answer:

Number of flags = 7

Number of ways of selecting the first flag = 7

Number of ways of selecting the second flag = 6

Therefore the number of different signals can be generated if a signal requires the use of two flags, one below the other = 7 \times 6 = 42

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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?

Answer:

Number of ways to select a boy from first team = 6

Number of ways to select a boy from second team = 5

Therefore the number of ways of arranging singles matches between boys of the first team and the second team = 6 \times 5 = 30

Number of ways to select a girl from first team = 4

Number of ways to select a girl from second team = 3

Therefore the number of ways of arranging singles matches between girls of the first team and the second team = 4 \times 3 = 12

Therefore the total number of singles matches that can be arranged = 30 + 12 = 42

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Question 13: 12 students compete in a race. In how many ways can the first three prizes can be given.

Answer:

The first prize can be given in 12 ways.

The second prize can be given in 11 ways.

The third prize can be given in 10 ways.

Hence the first three prizes = 12 \times 11 \times 10 = 1320 ways

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Question 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 \} ?

Answer:

Number of ways of selecting the first term = 3

For each of the first terms, there can be 5 common differences that can be chosen.

Hence the number of A.P’s would be 3 \times 5 = 15

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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?

Answer:

Number of teachers in the college = 36

Number of ways of selecting a Principal = 35

Number of ways of selecting a Vice-Principal = 34

Number of ways of selecting a teacher in-charge = 33

Hence the number of ways this can be done = 36 \times 35 \times 34 = 42840

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

Answer:

We have to form all possible 3 digit numbers where no digit is repeated.

A three digit number has a 100^{th} space, 10^{th} space and a units place.

In all there are 10 digits \{ 0, 1, ,2 ,3 ,4 ,5,6 ,7,8 ,9 \}

Hence the 100^{th} space can be filled in 9 possible ways. We cannot have 0 in the 100^{th} space for the number to be a three digit number.

The 10^{th} space can be filled in 9 different ways.

The units place can be filled in 8 different ways.

Therefore the number of  three-digit numbers are there with no digit repeated = 9 \times 9 \times 8 = 648

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Question 17: How many three-digit numbers are there?

Answer:

We have to form all possible 3 digit numbers.

A three digit number has a 100^{th} space, 10^{th} space and a units place.

In all there are 10 digits \{ 0, 1, ,2 ,3 ,4 ,5,6 ,7,8 ,9 \}

Hence the 100^{th} space can be filled in 9 possible ways. We cannot have 0 in the 100^{th} space for the number to be a three digit number.

The 10^{th} space can be filled in 10 different ways. The digits are allowed to repeat.

The units place can be filled in 10 different ways. The digits are allowed to repeat.

Therefore the number of  three-digit numbers are there with no digit repeated = 9 \times 10 \times 10 = 900

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Question 18: How many three-digit odd numbers are there?

Answer:

We have to form all possible 3 digit odd numbers.

A three digit number has a 100^{th} space, 10^{th} space and a units place.

In all there are 10 digits \{ 0, 1, ,2 ,3 ,4 ,5,6 ,7,8 ,9 \}

Hence the 100^{th} space can be filled in 9 possible ways. We cannot have 0 in the 100^{th} space for the number to be a three digit number.

The 10^{th} space can be filled in 10 different ways. The digits are allowed to repeat.

The units place can be filled in 5 different ways. The digits can be from \{ 1, 3, 5, 7, 9 \} for the number to be odd.

Therefore the number of  three-digit numbers are there with no digit repeated = 9 \times 10 \times 5 = 450

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Question 19: How many different five-digit number license plates can be made if

i) 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?

Answer:

i) The first digit can be filled in 9 different ways. 0 cannot be considered.

The second digit can be filled in 9 different ways.

The third digit can be filled in 8 different ways.

The fourth digit can be filled in 7 different ways.

The fifth digit can be filled in 6 different ways.

Five-digit number license plates can be made if the first digit cannot be zero and the repetition of digits is not allowed = 9 \times 9 \times 8 \times 7 \times 6  = 27216

ii) The first digit can be filled in 9 different ways. 0 cannot be considered.

The second digit can be filled in 10 different ways.

The third digit can be filled in 10 different ways.

The fourth digit can be filled in 10 different ways.

The fifth digit can be filled in 10 different ways.

Five-digit number license plates can be made if the first-digit cannot be zero, but the repetition of digits is allowed = 9 \times 10 \times 10 \times 10 \times 10  = 90000

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Question 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?

Answer:

The first digit can be filled in 3 ways. Only 7, 8 or 9 can go there as the number should be greater than 7000 .

The second digit can be filled in 4 different ways. Remember,  digits cannot be repeated.

The third digit can be filled in 3 different ways. Remember,  digits cannot be repeated.

The fourth digit can be filled in 2 way. Remember,  digits cannot be repeated.

Therefore the four-digit numbers that can be formed with the digits 3,5,7,8,9 which are greater than 7000 with no repetition of digits = 3 \times 4  \times 3 \times 2 = 72

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

Answer:

The first digit can be filled in 2 ways. Only 8 or 9 can go there as the number should be greater than 8000 .

The second digit can be filled in 4 different ways. Remember,  digits cannot be repeated.

The third digit can be filled in 3 different ways. Remember,  digits cannot be repeated.

The fourth digit can be filled in 2 way. Remember,  digits cannot be repeated.

Therefore the four-digit numbers that can be formed with the digits 3,5,7,8,9 which are greater than 8000 with no repetition of digits = 2 \times 4  \times 3 \times 2 = 48

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Question 22: ln how many ways can 6 persons be seated in a row?

Answer:

The first seat can be filled in 6 ways.

The second seat can be filled in 5 ways.

The third seat can be filled in 4 ways.

The fourth seat can be filled in 3 ways.

The fifth seat can be filled in 2 ways.

The sixth seat can be filled in 1 ways.

Therefore the number of ways 6 persons be seated in a row = 6 \times 5 \time 4 \times 3 \times 2 \times 1 = 720

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Question 23: How many 9-digit numbers of different digits can be formed?

Answer:

Note: Repetition is not allowed as ‘9-digit numbers of different digits’

The first digit can be filled in 9 different ways.

The second digit can be filled in 9 different ways.

The third digit can be filled in 8 different ways.

The fourth digit can be filled in 7 different ways.

The fifth digit can be filled in 6 different ways.

The sixth digit can be filled in 5 different ways.

The seventh digit can be filled in 4 different ways.

The eight digit can be filled in 3 different ways.

The ninth digit can be filled in 2 different ways.

Therefore the number of  9-digit numbers of different digits can be formed = 9 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 = 3265920 or 9(9!)

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Question 24: How many odd numbers less than 1000 can be formed by using the digits 0,3, 5,7 when repetition of digits is not allowed?

Answer:

Note: Repetition is not allowed. Also for number to be less than 1000 , it can be a three digit number, two digit number and a single digit number.

Three digit number

The first digit can be filled in 3 different ways. We can only select from 3, 5, 7

The second digit can be filled in 2 different ways.

The third digit can be filled in 2 different ways. We cannot have 0 in the last digit as the number is odd.

Therefore the number of three digits  odd numbers less than 1000 can be formed by using the digits 0,3, 5,7 when repetition of digits is not allowed = 3 \times 2 \times 2  = 12

Two digit number

The first digit can be filled in 3 different ways. We can only select from 3, 5, 7 .

The second digit can be filled in 2 different ways. No repetition is allowed and 0 cannot be considered as the number needs to be odd.

Therefore the number of two digits  odd numbers less than 1000 can be formed by using the digits 0,3, 5,7 when repetition of digits is not allowed = 3 \times 2 =6

Single digit number

The unit digit can be filled in 3 different ways.

Therefore the total number of numbers which are odd and less than 1000 that can be formed from 0, 3, 5, 7 = 12 + 6 + 3 = 21

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Question 25: How many 3-digit numbers are there, with distinct digits, with each digit odd?

Answer:

The odd digits are 1, 3, 5, 7, 9 . There is no repetition allowed ‘distinct digits’.

The first digit can be filled in 5 different ways.

The second digit can be filled in 4 different ways.

The third digit can be filled in 3 different ways.

Therefore the number of 3-digit numbers are there, with distinct digits, with each digit odd = 5 \times 4 \times 3 = 60

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Question 26: How many different numbers  of six digits each can be formed from the digits 4, 5, 6, 7, 8, 9 when repletion of digits is not allowed.

Answer:

Note: Repetition is not allowed.

The first digit can be filled in 6 different ways.

The second digit can be filled in 5 different ways.

The third digit can be filled in 4 different ways.

The fourth digit can be filled in 3 different ways.

The fifth digit can be filled in 2 different ways.

The sixth digit can be filled in 1 different ways.

Therefore the numbers  of six digits each can be formed from the digits 4, 5, 6, 7, 8, 9 when repletion of digits is not allowed = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 6! = 720

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Question 27:  How many different numbers  of six digits each can be formed from the digits 3, 1, 7, 0,  9,5 when repletion of digits is not allowed.

Answer:

Note: Repetition is not allowed.

The first digit can be filled in 5 different ways.

The second digit can be filled in 5 different ways. 0 is not allowed in the sixth digit.

The third digit can be filled in 4 different ways.

The fourth digit can be filled in 3 different ways.

The fifth digit can be filled in 2 different ways.

The sixth digit can be filled in 1 different ways.

Therefore the numbers  of six digits each can be formed from the digits 3, 1, 7, 0,  9,5 when repletion of digits is not allowed = 5 \times 5 \times 4 \times 3 \times 2 \times 1 = 600

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Question 28: How many four digit different numbers, greater than 5000 can be formed with the digits 1, 2, 5 , 9, 0 when repetition of digits is not allowed?

Answer:

The first digit can be filled in 2 different ways. Only 5 and 9 can be considered.

The second digit can be filled in 4 different ways.

The third digit can be filled in 3 different ways.

The fourth digit can be filled in 2 different ways.

Therefore the four digit different numbers, greater than 5000 can be formed with the digits 1, 2, 5 , 9, 0 when repetition of digits is not allowed = 2 \times 4 \times 3 \times 2 = 48

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Question 29: Serial numbers for an item produced in a factory are to be made using two letters followed by four digits (0 \text{to} 9) . If the letters are to be taken from six letters of English alphabet without repetition and the digits are also not repeated in numbers a serial number, how many serial are possible?

Answer:

The first digit ( Alphabet) can be filled in 6 different ways.

The second digit ( Alphabet) can be filled in 5 different ways.

The third digit (Number) can be filled in 10 different ways.

The fourth digit (Number) can be filled in 9 different ways.

The fifth digit (Number) can be filled in 8 different ways.

The sixth digit (Number) can be filled in 7 different ways.

Therefore the number of serial numbers that can be created = 6 \times 5 \times 10 \times 9 \times 8 \times 7 = 151200 

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Question 30: A number lock on a suitcase has 3 wheels each labelled with ten digits 0 to 9. If opening of the lock is a particular sequence of three digits with no repeats now many such sequences will be possible? Also, find the number of unsuccessful attempts to open the lock.

Answer:

The number of possible setting on each of the ring = 10

Number of rings = 3

Therefore the total number of combinations = 10 \times 9 \times 8 = 720

However, there is 1 combination that is correct. Hence the number of possible unsuccessful attempts = 720 - 1 = 719

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Question 31: A customer forgets a four digit code for an Automatic Teller Machine (ATM) in a bank. However, he remembers that this code consists of digits 3, 5, 6 \text{and} 9 . Find the largest possible number of trials necessary to obtain the correct code.

Answer:

Note: There is no repetition of the digits.

The first digit can be filled in 4 different ways.

The second digit can be filled in 3 different ways.

The third digit can be filled in 2 different ways.

The fourth digit can be filled in 1 different ways.

Maximum number of code combinations = 4 \times 3 \times 2 \times 1 = 24

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Question 32: In how many ways can three jobs \text{ I, II and III} be assigned to three persons \text{A, B and C} if one person is assigned only one job and all are capable if doing each job?

Answer:

Number of ways \text{A} can be assigned Jobs = 3

Number of ways \text{B} can be assigned Jobs = 2

Number of ways \text{C} can be assigned Jobs = 1

The three jobs \text{ I, II and III} be assigned to three persons \text{A, B and C} = 3 \times 2 \times 1 = 6 possible ways.

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Question 33: How many four digit natural numbers not exceeding 4321 can be formed with the digits 1, 2, 3 and 4 , if the digits can repeat?

Answer:

Four digit natural numbers not exceeding 4321 can be 4 digits, 3 digits, 2 digits and single digits.

4 digit natural numbers not exceeding 4321

The first digit can be filled in 4 different ways.

The second digit can be filled in 4 different ways.

The third digit can be filled in 4 different ways.

The fourth digit can be filled in 4 different ways.

Therefore the total number four digit different numbers = 3 \times 4 \times 4 \times 4 = 256

Now lets calculate the numbers formed greater than $latex 4321.

When the first digit is 4 , then the numbers formed = 1 \times 2 \times 4 \times 4 = 32

Also there are 4311, 4312, 4313, 4314, 4321 are less then $latex 4321.

Hence the total numbers formed less than 4321  =  256 - 32 +5 = 229

3 digit natural numbers not exceeding 4321

The first digit can be filled in 4 different ways.

The second digit can be filled in 4 different ways.

The third digit can be filled in 4 different ways.

Therefore the three digit different numbers = 4 \times 4 \times 4  = 64

2 digit natural numbers not exceeding 4321

The first digit can be filled in 4 different ways.

The second digit can be filled in 4 different ways.

Therefore the three digit different numbers = 4 \times 4   = 16

Single digit natural numbers not exceeding 4321

The single digit can be filled in 4 different ways.

Therefore the three digit different numbers = 4

Therefore the total number of number = 229+64+16+4 = 313

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Question 34: How many numbers of six digits can be formed from the digits 0, 1, 3, 5,7 and 9, when no digit is repeated? How many of them are divisible by 10?

Answer:

The first digit ( Alphabet) can be filled in 5 different ways.

The second digit ( Alphabet) can be filled in 5 different ways.

The third digit (Number) can be filled in 4 different ways.

The fourth digit (Number) can be filled in 3 different ways.

The fifth digit (Number) can be filled in 2 different ways.

The sixth digit (Number) can be filled in 1 different ways.

Therefore the total number of six digit numbers = 5 \times 5 \times 4 \times 3 \times 2 \times 1 = 600

For the numbers to be divisible by 10, the last digit should be 0. Hence,

The first digit ( Alphabet) can be filled in 5 different ways.

The second digit ( Alphabet) can be filled in 4 different ways.

The third digit (Number) can be filled in 3 different ways.

The fourth digit (Number) can be filled in 2 different ways.

The fifth digit (Number) can be filled in 1 different ways.

The sixth digit (Number) can be filled in 1 different ways.

Therefore the total number of six digit numbers = 5 \times 4 \times 3 \times 2 \times 1 \times 1 = 120

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Question 35: Three six face die each marked with numbers 1 to 6 on six faces, are thrown find , the total number of possible outcomes.

Answer:

The number of possible outcomes on one dice = 6

Therefore the total number of possible outcomes when three dices are thrown = 6 \times 6 \times 6 = 6^3 = 216

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Question 36: A coin is tossed three times and the outcomes are recorded. How many possible outcomes are there? How many possible outcomes if the coin is tossed four times ? Five times? N times.

Answer:

Total number of possible outcomes when a coin is tossed = 2

If a coin is tossed 3 times, the possible outcomes = 2 \times 2 \times 2 = 2^3 = 8

If a coin is tossed 4 times, the possible outcomes = 2 \times 2 \times 2 \times 2= 2^4 = 16

If a coin is tossed 5 times, the possible outcomes = 2 \times 2 \times 2 \times 2 \times 2 = 2^5 = 32

If a coin is tossed n times, the possible outcomes = 2 \times 2 \times 2 \times 2 \ldots \text{n times} = 2^n = 2^n

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Question 37: How many numbers of four digits can be formed with the digits 1, 2,3,4,5 if the digits can be repeated in the same number?

Answer:

Note: Repetition of the digits is allowed.

The first digit can be filled in 5 different ways.

The second digit can be filled in 5 different ways.

The third digit can be filled in 5 different ways.

The fourth digit can be filled in 5 different ways.

Maximum number of code combinations = 5 \times 5 \times 5 \times 5 = 625

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Question 38: How many three digit numbers can be formed by using the digits 0, 1,3,5, 7 while each digit may be repeated any number of times?

Answer:

Note: Repetition of the digits is allowed.

The first digit can be filled in 4 different ways.

The second digit can be filled in 5 different ways.

The third digit can be filled in 5 different ways.

Maximum number of code combinations = 4 \times 5 \times 5 =100

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Question 39: How many natural numbers less than 1000 can be formed from the digits 0,1,2,3,4, 5 when a digit may be repeated any number of times?

Answer:

Four digit natural numbers not exceeding 1000 can be 3 digits, 2 digits and single digits.

3 digit natural numbers not exceeding 1000

The first digit can be filled in 5 different ways.

The second digit can be filled in 6 different ways.

The third digit can be filled in 6 different ways.

Therefore the three digit different numbers = 5 \times 6 \times 6  = 180

2 digit natural numbers not exceeding 1000

The first digit can be filled in 5 different ways.

The second digit can be filled in 6 different ways.

Therefore the three digit different numbers = 5 \times 6   = 30

Single digit natural numbers not exceeding 1000

The single digit can be filled in 5 different ways. ( 0 is not a natural number)

Therefore the three digit different numbers = 5

Therefore the total number of number = 180+30+5= 215

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Question 40: How many five digit telephone numbers can be constructed using the digits 0 to 9. If each number starts with 67 and no digit appears more than once?

Answer:

Note: Repetition of the digits is not allowed.

The first digit can be filled in 1 different ways.  Only 6 can be chosen.

The second digit can be filled in 1 different ways.  Only 7 can be chosen.

The third digit can be filled in 8 different ways.

The fourth digit can be filled in 7 different ways.

The fifth digit can be filled in 6 different ways.

Maximum number of code combinations = 1 \times 1 \times 8 \times 7 \times 6   = 336

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Question 41: Find the number of ways in which 8 distinct toys can be distributed among 5 children.

Answer:

Total number of toys = 8

Total number of children = 5

Therefore number of ways in which 8 distinct toys can be distributed among 5 children = 5 \times 5\times 5 \times 5 \times 5 \times 5 \times 5 \times 5 = 5^8

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Question 42: Find the number of ways in which one can post 5 letters in 7 letter boxes.

Answer:

Total number of letters = 5

Total number of letter boxes = 7

Therefore number of ways in which one can post 5 letters in 7 letter boxes = 7 \times 7 \times 7 \times 7 \times 7  = 7^5

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Question 43: Three dice are rolled. Find the number of possible outcomes in which at least one dice shows 5 .

Answer:

Total number of dice = 3

Therefore the number of possible outcomes = 6 \times 6 \times 6 = 216

Total number of outcomes when 5 does not show up in any of the dice = 5 \times 5 \times 5 = 125

Therefore the required number of possible outcomes = 216 - 125 = 91

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Question 44: Find the total number of ways in which 20 balls can be put into 5 boxes so that first box contains just one ball.

Answer:

Total number of balls =20

Total number of boxes = 5

The first ball can be put in the first box in 20 ways.

Now the remaining 19 balls are to be put in the four boxes.  This can be done in 4^{19} ways because there are 4 choices for each ball.

Therefore the required number of ways = 20 \times 4^{19}

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Question 45: In how many ways can 5 different balls be distributed among three boxes?

Answer:

Each ball can be distributed in 3 ways.

Therefore the required number of ways 5 different balls be distributed among three boxes = 3 \times 3 \time 3 \times 3 \times 3 = 3^5

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Question 46: In how many ways can 7 letters be posted in 4 letter boxes?

Answer:

Each letter can be posted in 4 different ways.

Therefore the required number of ways to post 7 letters in 4 letter boxes = 4 \times 4 \time 4 \times 4 \times 4 \times 4 \time 4 = 4^7

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Question 47: In how many ways can 4 prizes be distributed among 5 students, when

i) no student gets more than one prize ?

ii) a student may get any number of prizes ?

iii) no student gets all the prizes?

Answer:

i)       no student gets more than one prize

Number of ways the prizes can be distributed = 5 \times 4 \times 3 \times 2 =120

ii)      a student may get any number of prizes

Number of ways the prizes can be distributed = 5 \times 5 \times 5 \times 5 =5^4

iii)     no student gets all the prizes

Number of ways where a student receives all the prices = 5

Number of ways 4 prizes be distributed among 5 students, when no student gets all the prizes = 625 - 5 = 620

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Question 48: There are 10 lamps in a hall. Each one of them can be switched on independently. Find the number of ways in which the hall can be illuminated.

Answer:

The hall would be lighted even when one bulb is lighted.

Number of ways the hall can be lighted = 2^{10}-1 = 1023