\displaystyle \textbf{Question 1: } \text{In a class there are }27\text{ boys and }14\text{ girls. The teacher wants to select}
\displaystyle \text{one boy and one girl to represent the class in a function. In how many ways}
\displaystyle \text{can the teacher make this selection?}
\displaystyle \text{Answer:}
\displaystyle \text{Number of ways to select one boy}=27
\displaystyle \text{Number of ways to select one girl}=14
\displaystyle \therefore \text{Required number of ways}=27\times14=378
\displaystyle \\

\displaystyle \textbf{Question 2: } \text{A person wants to buy one fountain pen, one ball pen and one pencil.}
\displaystyle \text{If there are }10\text{ fountain pen varieties, }12\text{ ball pen varieties and }5\text{ pencil}
\displaystyle \text{varieties, in how many ways can he select these articles?}
\displaystyle \text{Answer:}
\displaystyle \text{Number of ways to select a fountain pen}=10
\displaystyle \text{Number of ways to select a ball pen}=12
\displaystyle \text{Number of ways to select a pencil}=5
\displaystyle \therefore \text{Required number of ways}=10\times12\times5=600
\displaystyle \\

\displaystyle \textbf{Question 3: } \text{From Goa to Bombay there are }2\text{ routes: air and sea. From Bombay}
\displaystyle \text{to Delhi there are }3\text{ routes: air, rail and road. From Goa to Delhi via Bombay,}
\displaystyle \text{how many kinds of routes are there?}
\displaystyle \text{Answer:}
\displaystyle \text{Number of routes from Goa to Bombay}=2
\displaystyle \text{Number of routes from Bombay to Delhi}=3
\displaystyle \therefore \text{Number of routes from Goa to Delhi via Bombay}=2\times3=6
\displaystyle \\

\displaystyle \textbf{Question 4: } \text{A mint prepares metallic calendars specifying months, dates and days in}
\displaystyle \text{the form of monthly sheets. How many types of calendars should it prepare}
\displaystyle \text{to serve all the possibilities in future years?}
\displaystyle \text{Answer:}
\displaystyle \text{There are two types of years: leap years and non-leap years.}
\displaystyle \text{Each type of year can begin on any one of the }7\text{ days of the week.}
\displaystyle \therefore \text{Number of different calendars required}=2\times7=14
\displaystyle \\

\displaystyle \textbf{Question 5: } \text{There are four parcels and five post offices. In how many different}
\displaystyle \text{ways can the parcels be sent by registered post?}
\displaystyle \text{Answer:}
\displaystyle \text{Number of parcels}=4
\displaystyle \text{Number of post offices}=5
\displaystyle \text{Each parcel can be sent through any one of the }5\text{ post offices.}
\displaystyle \therefore \text{Required number of ways}=5\times5\times5\times5=5^4=625
\displaystyle \\

\displaystyle \textbf{Question 6: } \text{A coin is tossed five times and the outcomes are recorded. How many}
\displaystyle \text{possible outcomes are there?}
\displaystyle \text{Answer:}
\displaystyle \text{Each toss has }2\text{ possible outcomes.}
\displaystyle \text{Therefore the total number of possible outcomes}=2^5=32
\displaystyle \\

\displaystyle \textbf{Question 7: } \text{In how many ways can an examinee answer a set of ten true/false type}
\displaystyle \text{questions?}
\displaystyle \text{Answer:}
\displaystyle \text{Each question can be answered in }2\text{ ways: TRUE or FALSE.}
\displaystyle \therefore \text{Total number of ways to answer }10\text{ questions}=2^{10}=1024
\displaystyle \\

\displaystyle \textbf{Question 8: } \text{A letter lock consists of }3\text{ rings each marked with }10\text{ different}
\displaystyle \text{letters. In how many ways is it possible to make an unsuccessful attempt to}
\displaystyle \text{open the lock?}
\displaystyle \text{Answer:}
\displaystyle \text{Each ring has }10\text{ possible settings.}
\displaystyle \text{Number of rings}=3
\displaystyle \therefore \text{Total number of possible settings}=10^3=1000
\displaystyle \text{Only one setting opens the lock.}
\displaystyle \therefore \text{Number of unsuccessful attempts}=1000-1=999
\displaystyle \\

\displaystyle \textbf{Question 9: } \text{There are }6\text{ multiple choice questions in an examination. If the first}
\displaystyle \text{three questions have }4\text{ choices each and the next three have }2\text{ choices each,}
\displaystyle \text{how many sequences of answers are possible?}
\displaystyle \text{Answer:}
\displaystyle \text{Each of the first }3\text{ questions can be answered in }4\text{ ways.}
\displaystyle \text{Each of the next }3\text{ questions can be answered in }2\text{ ways.}
\displaystyle \therefore \text{Total number of answer sequences}=4^3\times2^3=512
\displaystyle \\

\displaystyle \textbf{Question 10: } \text{There are }5\text{ Mathematics books and }6\text{ Physics books in a book}
\displaystyle \text{shop. In how many ways can a student buy (i) a Mathematics book and a}
\displaystyle \text{Physics book (ii) either a Mathematics book or a Physics book?}
\displaystyle \text{Answer:}
\displaystyle \text{Number of Mathematics books}=5
\displaystyle \text{Number of Physics books}=6
\displaystyle \text{i) Number of ways to buy a Mathematics book and a Physics book}=5\times6=30
\displaystyle \\

\displaystyle \text{ii) Number of ways to buy either a Mathematics book or a Physics book}
\displaystyle =5+6=11
\displaystyle \\

\displaystyle \textbf{Question 11: } \text{Given }7\text{ flags of different colours, how many different signals can be}
\displaystyle \text{generated if a signal requires the use of two flags, one below the other?}
\displaystyle \text{Answer:}
\displaystyle \text{Number of ways to choose the upper flag}=7
\displaystyle \text{Number of ways to choose the lower flag}=6
\displaystyle \therefore \text{Number of different signals}=7\times6=42
\displaystyle \\

\displaystyle \textbf{Question 12: } \text{One team consists of }6\text{ boys and }4\text{ girls, and the other consists}
\displaystyle \text{of }5\text{ boys and }3\text{ girls. How many singles matches can be arranged between}
\displaystyle \text{the two teams if a boy plays against a boy and a girl plays against a girl?}
\displaystyle \text{Answer:}
\displaystyle \text{Number of ways to select a boy from the first team}=6
\displaystyle \text{Number of ways to select a boy from the second team}=5
\displaystyle \therefore \text{Number of matches between boys}=6\times5=30

\displaystyle \text{Number of ways to select a girl from the first team}=4
\displaystyle \text{Number of ways to select a girl from the second team}=3
\displaystyle \therefore \text{Number of matches between girls}=4\times3=12

\displaystyle \therefore \text{Total number of singles matches}=30+12=42

\displaystyle \textbf{Question 13: } \text{Twelve students compete in a race. In how many ways can the first three}
\displaystyle \text{prizes be given?}
\displaystyle \text{Answer:}
\displaystyle \text{The first prize can be given in }12\text{ ways.}
\displaystyle \text{The second prize can be given in }11\text{ ways.}
\displaystyle \text{The third prize can be given in }10\text{ ways.}
\displaystyle \therefore \text{Number of ways to give the first three prizes}=12\times11\times10=1320
\displaystyle \\

\displaystyle \textbf{Question 14: } \text{How many A.P.s with }10\text{ terms are there whose first term belongs to}
\displaystyle \{1,2,3\}\text{ and whose common difference belongs to }\{1,2,3,4,5\}\text{?}
\displaystyle \text{Answer:}
\displaystyle \text{Number of ways to select the first term}=3
\displaystyle \text{Number of ways to select the common difference}=5
\displaystyle \therefore \text{Number of such A.P.s}=3\times5=15
\displaystyle \\

\displaystyle \textbf{Question 15: } \text{From among the }36\text{ teachers in a college, one principal, one vice-principal}
\displaystyle \text{and one teacher-in-charge are to be appointed. In how many ways can this be done?}
\displaystyle \text{Answer:}
\displaystyle \text{Number of teachers in the college}=36
\displaystyle \text{Number of ways to select the principal}=36
\displaystyle \text{Number of ways to select the vice-principal}=35
\displaystyle \text{Number of ways to select the teacher-in-charge}=34
\displaystyle \therefore \text{Required number of ways}=36\times35\times34=42840
\displaystyle \\

\displaystyle \textbf{Question 16: } \text{How many three-digit numbers are there with no digit repeated?}
\displaystyle \text{Answer:}
\displaystyle \text{We have to form all possible three-digit numbers with no digit repeated.}
\displaystyle \text{A three-digit number has a hundreds place, a tens place and a units place.}
\displaystyle \text{The available digits are }\{0,1,2,3,4,5,6,7,8,9\}.
\displaystyle \text{The hundreds place can be filled in }9\text{ ways since }0\text{ cannot be used.}
\displaystyle \text{The tens place can be filled in }9\text{ ways.}
\displaystyle \text{The units place can be filled in }8\text{ ways.}
\displaystyle \therefore \text{Number of three-digit numbers}=9\times9\times8=648
\displaystyle \\

\displaystyle \textbf{Question 17: } \text{How many three-digit numbers are there?}
\displaystyle \text{Answer:}
\displaystyle \text{We have to form all possible three-digit numbers.}
\displaystyle \text{A three-digit number has a hundreds place, a tens place and a units place.}
\displaystyle \text{The available digits are }\{0,1,2,3,4,5,6,7,8,9\}.
\displaystyle \text{The hundreds place can be filled in }9\text{ ways since }0\text{ cannot be used.}
\displaystyle \text{The tens place can be filled in }10\text{ ways. Digits may be repeated.}
\displaystyle \text{The units place can be filled in }10\text{ ways. Digits may be repeated.}
\displaystyle \therefore \text{Number of three-digit numbers}=9\times10\times10=900
\displaystyle \\

\displaystyle \textbf{Question 18: } \text{How many three-digit odd numbers are there?}
\displaystyle \text{Answer:}
\displaystyle \text{We have to form all possible three-digit odd numbers.}
\displaystyle \text{A three-digit number has a hundreds place, a tens place and a units place.}
\displaystyle \text{The available digits are }\{0,1,2,3,4,5,6,7,8,9\}.
\displaystyle \text{The hundreds place can be filled in }9\text{ ways since }0\text{ cannot be used.}
\displaystyle \text{The tens place can be filled in }10\text{ ways. Digits may be repeated.}
\displaystyle \text{The units place can be filled in }5\text{ ways using }\{1,3,5,7,9\}.
\displaystyle \therefore \text{Number of three-digit odd numbers}=9\times10\times5=450
\displaystyle \\

\displaystyle \textbf{Question 19: } \text{How many different five-digit number license plates can be made if}
\displaystyle \text{i) the first digit cannot be zero and repetition of digits is not allowed,}
\displaystyle \text{ii) the first digit cannot be zero, but repetition of digits is allowed?}
\displaystyle \text{Answer:}
\displaystyle \text{i) The first digit can be filled in }9\text{ ways since }0\text{ cannot be used.}
\displaystyle \text{The second digit can be filled in }9\text{ ways.}
\displaystyle \text{The third digit can be filled in }8\text{ ways.}
\displaystyle \text{The fourth digit can be filled in }7\text{ ways.}
\displaystyle \text{The fifth digit can be filled in }6\text{ ways.}
\displaystyle \therefore \text{Number of five-digit license plates}=9\times9\times8\times7\times6=27216
\displaystyle \\

\displaystyle \text{ii) The first digit can be filled in }9\text{ ways since }0\text{ cannot be used.}
\displaystyle \text{Each of the remaining four digits can be filled in }10\text{ ways.}
\displaystyle \therefore \text{Number of five-digit license plates}=9\times10\times10\times10\times10=90000
\displaystyle \\

\displaystyle \textbf{Question 20: } \text{How many four-digit numbers greater than }7000\text{ can be formed using the}
\displaystyle \text{digits }3,5,7,8,9\text{ if repetition of digits is not allowed?}
\displaystyle \text{Answer:}
\displaystyle \text{The first digit can be filled in }3\text{ ways using }7,8\text{ or }9.
\displaystyle \text{The second digit can be filled in }4\text{ ways.}
\displaystyle \text{The third digit can be filled in }3\text{ ways.}
\displaystyle \text{The fourth digit can be filled in }2\text{ ways.}
\displaystyle \therefore \text{Number of required four-digit numbers}=3\times4\times3\times2=72
\displaystyle \\

\displaystyle \textbf{Question 21: } \text{How many four-digit numbers greater than }8000\text{ can be formed using the}
\displaystyle \text{digits }3,5,7,8,9\text{ if repetition of digits is not allowed?}
\displaystyle \text{Answer:}
\displaystyle \text{The first digit can be filled in }2\text{ ways using }8\text{ or }9.
\displaystyle \text{The second digit can be filled in }4\text{ ways.}
\displaystyle \text{The third digit can be filled in }3\text{ ways.}
\displaystyle \text{The fourth digit can be filled in }2\text{ ways.}
\displaystyle \therefore \text{Number of required four-digit numbers}=2\times4\times3\times2=48
\displaystyle \\

\displaystyle \textbf{Question 22: } \text{In how many ways can }6\text{ persons be seated in a row?}
\displaystyle \text{Answer:}
\displaystyle \text{The first seat can be filled in }6\text{ ways.}
\displaystyle \text{The second seat can be filled in }5\text{ ways.}
\displaystyle \text{The third seat can be filled in }4\text{ ways.}
\displaystyle \text{The fourth seat can be filled in }3\text{ ways.}
\displaystyle \text{The fifth seat can be filled in }2\text{ ways.}
\displaystyle \text{The sixth seat can be filled in }1\text{ way.}
\displaystyle \therefore \text{Number of arrangements}=6\times5\times4\times3\times2\times1=6!=720
\displaystyle \\

\displaystyle \textbf{Question 23: } \text{How many nine-digit numbers with different digits can be formed?}
\displaystyle \text{Answer:}
\displaystyle \text{Since all the digits are different, repetition of digits is not allowed.}
\displaystyle \text{The first digit can be filled in }9\text{ ways since }0\text{ cannot be used.}
\displaystyle \text{The second digit can be filled in }9\text{ ways.}
\displaystyle \text{The third digit can be filled in }8\text{ ways.}
\displaystyle \text{The fourth digit can be filled in }7\text{ ways.}
\displaystyle \text{The fifth digit can be filled in }6\text{ ways.}
\displaystyle \text{The sixth digit can be filled in }5\text{ ways.}
\displaystyle \text{The seventh digit can be filled in }4\text{ ways.}
\displaystyle \text{The eighth digit can be filled in }3\text{ ways.}
\displaystyle \text{The ninth digit can be filled in }2\text{ ways.}
\displaystyle \therefore \text{Number of nine-digit numbers}=9\times9\times8\times7\times6\times5\times4\times3\times2
\displaystyle =3265920=9\times9!
\displaystyle \\

\displaystyle \textbf{Question 24: } \text{How many odd numbers less than }1000\text{ can be formed using the digits}
\displaystyle 0,3,5,7\text{ if repetition of digits is not allowed?}
\displaystyle \text{Answer:}
\displaystyle \text{The required numbers may be three-digit, two-digit or one-digit numbers.}
\displaystyle \text{Three-digit numbers:}
\displaystyle \text{The units digit can be filled in }3\text{ ways using }3,5\text{ or }7.
\displaystyle \text{The hundreds digit can be filled in }2\text{ ways using the remaining non-zero digits.}
\displaystyle \text{The tens digit can be filled in }2\text{ ways using either remaining digit.}
\displaystyle \therefore \text{Number of three-digit odd numbers}=3\times2\times2=12

\displaystyle \text{Two-digit numbers:}
\displaystyle \text{The units digit can be filled in }3\text{ ways using }3,5\text{ or }7.
\displaystyle \text{The tens digit can be filled in }2\text{ ways using the remaining non-zero digits.}
\displaystyle \therefore \text{Number of two-digit odd numbers}=3\times2=6

\displaystyle \text{One-digit numbers:}
\displaystyle \text{The one-digit odd numbers are }3,5\text{ and }7.
\displaystyle \therefore \text{Number of one-digit odd numbers}=3

\displaystyle \therefore \text{Total number of odd numbers}=12+6+3=21

\displaystyle \textbf{Question 25: } \text{How many three-digit numbers are there with distinct digits,}
\displaystyle \text{if each digit is odd?}
\displaystyle \text{Answer:}
\displaystyle \text{The available odd digits are }1,3,5,7,9.
\displaystyle \text{Since the digits are distinct, repetition is not allowed.}
\displaystyle \text{The first digit can be filled in }5\text{ ways.}
\displaystyle \text{The second digit can be filled in }4\text{ ways.}
\displaystyle \text{The third digit can be filled in }3\text{ ways.}
\displaystyle \therefore \text{Number of three-digit numbers}=5\times4\times3=60
\displaystyle \\

\displaystyle \textbf{Question 26: } \text{How many different six-digit numbers can be formed using the digits}
\displaystyle 4,5,6,7,8,9\text{ if repetition of digits is not allowed?}
\displaystyle \text{Answer:}
\displaystyle \text{The first digit can be filled in }6\text{ ways.}
\displaystyle \text{The second digit can be filled in }5\text{ ways.}
\displaystyle \text{The third digit can be filled in }4\text{ ways.}
\displaystyle \text{The fourth digit can be filled in }3\text{ ways.}
\displaystyle \text{The fifth digit can be filled in }2\text{ ways.}
\displaystyle \text{The sixth digit can be filled in }1\text{ way.}
\displaystyle \therefore \text{Number of six-digit numbers}=6\times5\times4\times3\times2\times1=6!=720
\displaystyle \\

\displaystyle \textbf{Question 27: } \text{How many different six-digit numbers can be formed using the digits}
\displaystyle 3,1,7,0,9,5\text{ if repetition of digits is not allowed?}
\displaystyle \text{Answer:}
\displaystyle \text{Since repetition is not allowed, the six digits can be arranged in }6!\text{ ways.}
\displaystyle \text{However, arrangements with }0\text{ in the first position are not six-digit numbers.}
\displaystyle \text{Number of such arrangements}=5!
\displaystyle \therefore \text{Required number of six-digit numbers}=6!-5!=720-120=600
\displaystyle \\

\displaystyle \textbf{Question 28: } \text{How many different four-digit numbers greater than }5000\text{ can be formed}
\displaystyle \text{using the digits }1,2,5,9,0\text{ if repetition of digits is not allowed?}
\displaystyle \text{Answer:}
\displaystyle \text{The first digit can be filled in }2\text{ ways using }5\text{ or }9.
\displaystyle \text{The second digit can be filled in }4\text{ ways.}
\displaystyle \text{The third digit can be filled in }3\text{ ways.}
\displaystyle \text{The fourth digit can be filled in }2\text{ ways.}
\displaystyle \therefore \text{Number of four-digit numbers}=2\times4\times3\times2=48
\displaystyle \\

\displaystyle \textbf{Question 29: } \text{Serial numbers for an item produced in a factory consist of two letters}
\displaystyle \text{followed by four digits from }0\text{ to }9.\text{ The letters are chosen from six letters of the}
\displaystyle \text{English alphabet without repetition, and the digits are also not repeated. How many}
\displaystyle \text{such serial numbers are possible?}
\displaystyle \text{Answer:}
\displaystyle \text{The first letter position can be filled in }6\text{ ways.}
\displaystyle \text{The second letter position can be filled in }5\text{ ways.}
\displaystyle \text{The first digit position can be filled in }10\text{ ways.}
\displaystyle \text{The second digit position can be filled in }9\text{ ways.}
\displaystyle \text{The third digit position can be filled in }8\text{ ways.}
\displaystyle \text{The fourth digit position can be filled in }7\text{ ways.}
\displaystyle \therefore \text{Number of serial numbers}=6\times5\times10\times9\times8\times7=151200
\displaystyle \\

\displaystyle \textbf{Question 30: } \text{A number lock on a suitcase has three wheels, each labelled with the}
\displaystyle \text{digits }0\text{ to }9.\text{ If the lock opens with a particular sequence of three digits with no}
\displaystyle \text{repetition, how many such sequences are possible? Also, find the number of}
\displaystyle \text{unsuccessful attempts to open the lock.}
\displaystyle \text{Answer:}
\displaystyle \text{The first wheel can be set in }10\text{ ways.}
\displaystyle \text{The second wheel can be set in }9\text{ ways.}
\displaystyle \text{The third wheel can be set in }8\text{ ways.}
\displaystyle \therefore \text{Total number of possible sequences}=10\times9\times8=720
\displaystyle \text{Only one sequence opens the lock.}
\displaystyle \therefore \text{Number of unsuccessful attempts}=720-1=719
\displaystyle \\

\displaystyle \textbf{Question 31: } \text{A customer forgets a four-digit code for an Automatic Teller Machine}
\displaystyle \text{(ATM). However, he remembers that the code consists of the digits }3,5,6\text{ and }9.
\displaystyle \text{Find the largest possible number of trials required to obtain the correct code.}
\displaystyle \text{Answer:}
\displaystyle \text{The four digits }3,5,6\text{ and }9\text{ are used without repetition.}
\displaystyle \text{The first digit can be filled in }4\text{ ways.}
\displaystyle \text{The second digit can be filled in }3\text{ ways.}
\displaystyle \text{The third digit can be filled in }2\text{ ways.}
\displaystyle \text{The fourth digit can be filled in }1\text{ way.}
\displaystyle \therefore \text{Number of possible codes}=4\times3\times2\times1=4!=24
\displaystyle \therefore \text{The largest possible number of trials required is }24.
\displaystyle \\

\displaystyle \textbf{Question 32: } \text{In how many ways can the three jobs I, II and III be assigned to three}
\displaystyle \text{persons A, B and C if each person is assigned only one job and all are capable}
\displaystyle \text{of doing each job?}
\displaystyle \text{Answer:}
\displaystyle \text{Person A can be assigned a job in }3\text{ ways.}
\displaystyle \text{Person B can then be assigned a job in }2\text{ ways.}
\displaystyle \text{Person C can then be assigned the remaining job in }1\text{ way.}
\displaystyle \therefore \text{Number of possible assignments}=3\times2\times1=3!=6
\displaystyle \\

\displaystyle \textbf{Question 33: } \text{How many natural numbers not exceeding }4321\text{ can be formed using}
\displaystyle \text{the digits }1,2,3,4\text{ if the digits may be repeated?}
\displaystyle \text{Answer:}
\displaystyle \text{The required numbers may have }4,3,2\text{ or }1\text{ digits.}
\displaystyle \text{Four-digit numbers not exceeding }4321:
\displaystyle \text{If the first digit is }1,2\text{ or }3,\text{ the remaining three digits can each be chosen in }4\text{ ways.}
\displaystyle \text{Number of such numbers}=3\times4^3=192
\displaystyle \text{If the first digit is }4,\text{ then:}
\displaystyle \text{Second digit }=1\text{ or }2:\;2\times4\times4=32
\displaystyle \text{Second digit }=3:\text{ the valid numbers are }4311,4312,4313,4314,4321,
\displaystyle \text{giving }5\text{ numbers.}
\displaystyle \therefore \text{Four-digit numbers}=192+32+5=229

\displaystyle \text{Three-digit numbers}=4^3=64

\displaystyle \text{Two-digit numbers}=4^2=16

\displaystyle \text{One-digit numbers}=4

\displaystyle \therefore \text{Total number of natural numbers}=229+64+16+4=313

\displaystyle \textbf{Question 34: } \text{How many six-digit numbers can be formed using the digits }0,1,3,5,7,9
\displaystyle \text{if no digit is repeated? How many of them are divisible by }10\text{?}
\displaystyle \text{Answer:}
\displaystyle \text{The first digit can be filled in }5\text{ ways since }0\text{ cannot be used.}
\displaystyle \text{The second digit can be filled in }5\text{ ways.}
\displaystyle \text{The third digit can be filled in }4\text{ ways.}
\displaystyle \text{The fourth digit can be filled in }3\text{ ways.}
\displaystyle \text{The fifth digit can be filled in }2\text{ ways.}
\displaystyle \text{The sixth digit can be filled in }1\text{ way.}
\displaystyle \therefore \text{Number of six-digit numbers}=5\times5\times4\times3\times2\times1=600

\displaystyle \text{For a number to be divisible by }10,\text{ the last digit must be }0.
\displaystyle \text{The remaining five digits can be arranged in }5\times4\times3\times2\times1\text{ ways.}
\displaystyle \therefore \text{Number of six-digit numbers divisible by }10=5\times4\times3\times2\times1=120
\displaystyle \\

\displaystyle \textbf{Question 35: } \text{Three six-faced dice, each marked with the numbers }1\text{ to }6,\text{ are}
\displaystyle \text{thrown. Find the total number of possible outcomes.}
\displaystyle \text{Answer:}
\displaystyle \text{Each die has }6\text{ possible outcomes.}
\displaystyle \therefore \text{Total number of possible outcomes}=6\times6\times6=6^3=216
\displaystyle \\

\displaystyle \textbf{Question 36: } \text{A coin is tossed three times and the outcomes are recorded. How many}
\displaystyle \text{possible outcomes are there? How many possible outcomes if the coin is tossed}
\displaystyle 4\text{ times, }5\text{ times and }n\text{ times?}
\displaystyle \text{Answer:}
\displaystyle \text{Each toss has }2\text{ possible outcomes.}
\displaystyle \text{If a coin is tossed }3\text{ times, the number of possible outcomes}=2^3=8
\displaystyle \text{If a coin is tossed }4\text{ times, the number of possible outcomes}=2^4=16
\displaystyle \text{If a coin is tossed }5\text{ times, the number of possible outcomes}=2^5=32
\displaystyle \text{If a coin is tossed }n\text{ times, the number of possible outcomes}=2^n
\displaystyle \\

\displaystyle \textbf{Question 37: } \text{How many four-digit numbers can be formed using the digits }1,2,3,4,5
\displaystyle \text{if repetition of digits is allowed?}
\displaystyle \text{Answer:}
\displaystyle \text{The first digit can be filled in }5\text{ ways.}
\displaystyle \text{The second digit can be filled in }5\text{ ways.}
\displaystyle \text{The third digit can be filled in }5\text{ ways.}
\displaystyle \text{The fourth digit can be filled in }5\text{ ways.}
\displaystyle \therefore \text{Number of four-digit numbers}=5\times5\times5\times5=5^4=625
\displaystyle \\

\displaystyle \textbf{Question 38: } \text{How many three-digit numbers can be formed using the digits }0,1,3,5,7
\displaystyle \text{if each digit may be repeated any number of times?}
\displaystyle \text{Answer:}
\displaystyle \text{Repetition of digits is allowed.}
\displaystyle \text{The first digit can be filled in }4\text{ ways since }0\text{ cannot be used.}
\displaystyle \text{The second digit can be filled in }5\text{ ways.}
\displaystyle \text{The third digit can be filled in }5\text{ ways.}
\displaystyle \therefore \text{Number of three-digit numbers}=4\times5\times5=100
\displaystyle \\

\displaystyle \textbf{Question 39: } \text{How many natural numbers less than }1000\text{ can be formed using the digits}
\displaystyle 0,1,2,3,4,5\text{ if a digit may be repeated any number of times?}
\displaystyle \text{Answer:}
\displaystyle \text{The required numbers may have }3,2\text{ or }1\text{ digits.}
\displaystyle \text{Three-digit numbers:}
\displaystyle \text{The first digit can be filled in }5\text{ ways since }0\text{ cannot be used.}
\displaystyle \text{Each of the remaining two digits can be filled in }6\text{ ways.}
\displaystyle \therefore \text{Number of three-digit numbers}=5\times6\times6=180

\displaystyle \text{Two-digit numbers:}
\displaystyle \text{The first digit can be filled in }5\text{ ways since }0\text{ cannot be used.}
\displaystyle \text{The second digit can be filled in }6\text{ ways.}
\displaystyle \therefore \text{Number of two-digit numbers}=5\times6=30

\displaystyle \text{One-digit natural numbers:}
\displaystyle \text{The possible one-digit natural numbers are }1,2,3,4,5.
\displaystyle \therefore \text{Number of one-digit natural numbers}=5

\displaystyle \therefore \text{Total number of natural numbers}=180+30+5=215
\displaystyle \\

\displaystyle \textbf{Question 40: } \text{How many five-digit telephone numbers can be constructed using the digits}
\displaystyle 0\text{ to }9\text{ if each number starts with }67\text{ and no digit is repeated?}
\displaystyle \text{Answer:}
\displaystyle \text{The first digit is fixed as }6.
\displaystyle \text{The second digit is fixed as }7.
\displaystyle \text{The third digit can be filled in }8\text{ ways.}
\displaystyle \text{The fourth digit can be filled in }7\text{ ways.}
\displaystyle \text{The fifth digit can be filled in }6\text{ ways.}
\displaystyle \therefore \text{Number of telephone numbers}=1\times1\times8\times7\times6=336
\displaystyle \\

\displaystyle \textbf{Question 41: } \text{Find the number of ways in which }8\text{ distinct toys can be distributed}
\displaystyle \text{among }5\text{ children.}
\displaystyle \text{Answer:}
\displaystyle \text{Each of the }8\text{ distinct toys can be given independently to any one of the }5\text{ children.}
\displaystyle \therefore \text{Number of ways}=5\times5\times5\times5\times5\times5\times5\times5=5^8
\displaystyle \\

\displaystyle \textbf{Question 42: } \text{Find the number of ways in which }5\text{ letters can be posted in }7\text{ letter}
\displaystyle \text{boxes.}
\displaystyle \text{Answer:}
\displaystyle \text{Each of the }5\text{ letters can be posted independently in any one of the }7\text{ letter boxes.}
\displaystyle \therefore \text{Number of ways}=7\times7\times7\times7\times7=7^5
\displaystyle \\

\displaystyle \textbf{Question 43: } \text{Three dice are rolled. Find the number of possible outcomes in which at least}
\displaystyle \text{one die shows }5.
\displaystyle \text{Answer:}
\displaystyle \text{Total number of outcomes}=6^3=216
\displaystyle \text{If no die shows }5,\text{ each die has }5\text{ possible outcomes.}
\displaystyle \text{Number of outcomes in which no die shows }5=5^3=125
\displaystyle \therefore \text{Required number of outcomes}=216-125=91
\displaystyle \\

\displaystyle \textbf{Question 44: } \text{Find the number of ways in which }20\text{ distinct balls can be placed in}
\displaystyle 5\text{ distinct boxes so that the first box contains exactly one ball.}
\displaystyle \text{Answer:}
\displaystyle \text{The ball to be placed in the first box can be selected in }20\text{ ways.}
\displaystyle \text{Each of the remaining }19\text{ balls can be placed in any one of the other }4\text{ boxes.}
\displaystyle \text{Number of ways to place the remaining balls}=4^{19}
\displaystyle \therefore \text{Required number of ways}=20\times4^{19}
\displaystyle \\

\displaystyle \textbf{Question 45: } \text{In how many ways can }5\text{ different balls be distributed among}
\displaystyle 3\text{ distinct boxes?}
\displaystyle \text{Answer:}
\displaystyle \text{Each ball can be placed independently in any one of the }3\text{ boxes.}
\displaystyle \therefore \text{Required number of ways}=3\times3\times3\times3\times3=3^5=243
\displaystyle \\

\displaystyle \textbf{Question 46: } \text{In how many ways can }7\text{ letters be posted in }4\text{ letter boxes?}
\displaystyle \text{Answer:}
\displaystyle \text{Each letter can be posted independently in any one of the }4\text{ letter boxes.}
\displaystyle \therefore \text{Required number of ways}=4\times4\times4\times4\times4\times4\times4=4^7
\displaystyle \\

\displaystyle \textbf{Question 47: } \text{In how many ways can }4\text{ distinct prizes be distributed among }5\text{ students if}
\displaystyle \text{i) no student gets more than one prize,}
\displaystyle \text{ii) a student may get any number of prizes,}
\displaystyle \text{iii) no student gets all the prizes?}
\displaystyle \text{Answer:}
\displaystyle \text{i) The first prize can be awarded in }5\text{ ways.}
\displaystyle \text{The second prize can be awarded in }4\text{ ways.}
\displaystyle \text{The third prize can be awarded in }3\text{ ways.}
\displaystyle \text{The fourth prize can be awarded in }2\text{ ways.}
\displaystyle \therefore \text{Required number of ways}=5\times4\times3\times2=120

\displaystyle \text{ii) Each prize can be awarded independently to any one of the }5\text{ students.}
\displaystyle \therefore \text{Required number of ways}=5\times5\times5\times5=5^4=625

\displaystyle \text{iii) Total number of ways to distribute the prizes}=5^4=625
\displaystyle \text{Number of ways in which one student receives all four prizes}=5
\displaystyle \therefore \text{Required number of ways}=625-5=620
\displaystyle \\

\displaystyle \textbf{Question 48: } \text{There are }10\text{ lamps in a hall. Each lamp can be switched on independently.}
\displaystyle \text{Find the number of ways in which the hall can be illuminated.}
\displaystyle \text{Answer:}
\displaystyle \text{Each lamp has }2\text{ possible states: switched on or switched off.}
\displaystyle \therefore \text{Total number of on-off arrangements}=2^{10}
\displaystyle \text{The arrangement in which all lamps are switched off does not illuminate the hall.}
\displaystyle \therefore \text{Number of ways in which the hall can be illuminated}=2^{10}-1=1023

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