Combination - Probability, Business Mathematics and Statistics Video Lecture - Business Mathematics and Statistics - B Com

welcome to the video my name is fender
Gill and today we are going to learn
permutations and combinations so this is
the first video on permutations and
combinations and this is my email
address here so in case you want to
contact me you can contact me in the
email address so let's go ahead and get
started okay so let's see the simple
question over here a David has three
shirts and two sweaters and how many
ways he can either wear a shirt or a
sweater right so this is a simple
question for us so david has three
shirts let's suppose he has these three
shirts and he has three sweaters let's
suppose I'm sorry two sweaters let's
suppose he has these two sweaters so the
question asked is in how many ways can
he either wear a shirt or a sweater so
he can either wear this shirt or he can
wear this shirt or he can wear this
shirt or he can wear this sweater or he
can wear this sweater so the number of
ways in which he can either wear a shirt
or a sweater it's 3 plus 2 that is 5
ways right so I hope everyone is
understanding this is simple basic stuff
right what if I twist the question a
little bit
I said David has 3 shirts in 2 sweaters
in how many ways he can either wear he
can wear one shirt and one sweater so
instead of an or what I have done is
I've placed a and right so he has to
wear one shirt and one sweater so how
many ways are there so with this shirt
he can wear these two sweaters so two
ways with one shirt and with this shirt
as well again two sweaters and this
shirt as well again two sweaters so that
means the number of ways he can wear
either one shirt and one sweater is 3
into 2 that is 6 ways right so this is
simple basic stuff if you want to wear
either a shirt or a sweater then there
are 5 ways and if he wants to wear a
short and a sweater then there are six
ways right so hope everyone is
understanding that if even a can be done
in MVS and even be
II can be done in n ways then exactly
one of the evens can be done in M plus n
ways right if even a and B together can
be done in M into n ways right so I hope
everyone is understanding till now it's
pretty basic stuff you must be knowing
all this even before watching the video
so let's go ahead and get started so
let's just simply let's see the simple
question let me see there's at least
three set is city a city B and city see
the number of ways of going from City a
to city B are three so there are three
ways by which you can go from City a to
city B and similarly there are two ways
by which you can go from City B to City
FC now if I ask you how many number of
their ways are there to go from City a
to city C Y up B so you have to go
through B so there are three ways here
there are two ways here yeah the total
number of ways would be 3 into 2 because
you have to go through B so 3 into 2
that is 6 ways let me say there is
another city called City D as well and C
D D has 4 ways for for it to go from CDC
so from CDC to City D there are 4 ways
now if I ask you how many number of ways
are there to go from City a to city D Y
a B and C so that would be 3 into 2 into
4 that is 24 ways right so I hope
everyone is understanding my point here
what I'm trying to prove that if even a
can be done in MVS and B can be done in
n ways then both the events together can
be done in M into n ways right so you
understand the logic behind that why
because see if we let's suppose we want
to go from City a to city C so three
ways are here and two ways are here each
way here is gonna give us two ways so
three ways are going to give us 3 into 2
that is 6 ways so that is how we
calculate the answer right so I hope
everyone is understanding let's move
forward to some complex scenarios ok
Oh in how many different ways can five
different chocolates can be distributed
to four children
such that any child can get any number
of chocolates right to do that okay
let's see this question right so a new
flag with vertical stripes is to be
designed using some of or all of the
colors yellow green blue red and the
number of ways this can be done such
that no two adjacent stripes have the
same color so now what kind of question
is this earlier we were just
distributing sweaters and shirts to to
the to the students now over here there
is a condition
so what whenever there is a condition we
have to meet the condition now he is
saying that no two adjacent stripes
should have the same color and there is
a flag with six vertical stripes so the
flag is going to look something like
this right so what in this kind of
question what you have to do is you have
to start from anywhere you know let's
start right from here so how many colors
can you give this stripe you can give
them either yellow you can give it
either green you can give it either blue
or you can give it either red so there
are four ways in which you can give
color to this stripe now we have to meet
the condition and the condition is that
no two adjacent stripes should have the
same color now if one color has been
given to this stripe that means this
stripe cannot have the color which was
given to this strike so that means the
number of ways which we can give color
to this tribe is three ways right now
let's look at the next stripe as well
now in the next stripe we will be able
to give the color that we gave to this
stripe but not the color to this stripe
so let's suppose this was given yellow
and this was given red so now we have a
choice of yellow green and blue from
this try because we can already give him
yellow because no two adjacent stripes
should have the same color so again
three ways similarly next stripe also
three ways next stripe also three ways
next stripe also three ways and you get
your answer right so hope
one is understanding what I'm trying to
establish here right so the the thing
that I'm trying to establish is you have
to go with the question in a structured
manner start with the structural manner
draw the diagram in case it is needed
and start from one point and that is
going to lead you to next points right
so now we are going to make it more
structured let's see what is that so in
how many ways five prizes can be given
to five girls such that one girl gets
one price right so this is what we have
to do there are five five prizes let me
say the prizes are one two three four
and five and then there are five girls a
B C D and E these are the five girls now
we have to distribute the five prizes to
five girls now what do we do is now over
here we are distributing the prizes
rather than girls coming and taking
their prizes so now we are distributing
the prizes now what do we have to do is
draw the number of slots of number of
prizes right so first you have to
distribute the first price so in how
many ways can you distribute the first
prize you can either give it to girl a
or to girl beat girl see your girl D or
girl E
so there are five ways in which you can
distribute the first prize now let's
suppose girl a gets the first prize now
in how many ways can you distribute
these second prize because each girl has
to get one prize girl a cannot get more
than one price so similarly for prizes
here for three three prizes here then
two prizes here and one prizes here so
our answer is nothing but if we
calculate it it is nothing but five
factorial this is equal to five
factorial so I hope you are
understanding right I'm trying to build
something here this is a very simple
question five prices to be distributed
among five students so our first first
price can be given to any five of the
girls second price can be given to any
four of the girls third price can be
given to any three of the girls and so
on right so let's
forward so in how many ways can five
boys stand in a row so this how would
you do this question now this is again a
similar kind of question right so there
are five positions right first position
second position third position fourth
position fifth position and they're
about to stand in a row said how many
ways can an boy come on the first
position five ways because there are
five boys right so first position can be
assigned to any of the five boys second
position can be assigned to any of the
four boys any of the three boys any of
the two boys any of the one voice yes
the answer is five factorial right now
let's see the next question the next
question is and how many ways can five
boys sit on five chairs which is a
similar kind of question guys the
question is same as just the wording
that is going a bit different so again
the answer is five factorial so what I'm
trying to do here why I am a little bit
fast here it's because if you want to
arrange any objects in n places the
answer is always going to be n factorial
you want to give three prizes to three
girls such that each girl gets one price
three factorial
you wanna give you wanna make five boys
stand in a row 5 factorial six girls
stand in the line six factorial so if
you want to arrange n objects in n
places then the answer is always going
to be n factorial right so I hope
everyone is understanding now let's go
ahead and build on what we have learned
here right okay so pretty is asking us a
question okay what if there are some
conditions in the question good question
for thee so if there are conditions in
the question you must satisfy the
conditions first now let's see this
question in how many ways a five prices
can be given to five girls such that
each girl gets one price F first price
cannot be given to girl A or B so I hope
everyone is understanding this question
this question is pretty simple
now as pretty asked us the question that
first pry what if there are conditions
in the question and I answered as you
have to satisfy the conditions first and
then go about with the question so these
are five prices to be given price number
one
two three four and five and these are
the five girls which are there right
okay now what is the condition first
price cannot be given to girl a or B now
let's give the first price the first
price cannot be given to girl a and it
cannot be given to girl B so how many
number of a's are there to give the
first price yes there are three number
of ways in which I can distribute the
first price it can be distributed to
either C either D or either E right now
let's go ahead and solve the next one
now let's suppose any one of them gets
the first price let's suppose he gets
the first price now we have four prizes
left and four girls left so that would
be distributed in four factorial ways if
you are understood it right for girls
for prizes for factorial so whenever
there is a condition what do you need to
do is you need to satisfy the condition
first so if you satisfy the condition
first and then in the end you'll be left
with the simple question for prizes for
girls right so I hope everyone is
understanding this is we are building on
it so that's why it must be looking
simple for you right okay all right
Preeti has another question what if
there are two conditions good question
very good question because you would be
getting questions in which there are you
know more than two conditions so what
should we do so there are certain there
are sometimes certain number of
conditions certain types of conditions
that are there in the question right so
let me go ahead and explain you those
types of conditions first so there are
two types of condition the first are
positive conditions and the second type
of conditions are negative conditions so
what is the difference between them the
positive conditions are the conditions
relating to must be or should be stuff
like that right let's suppose if five
boys wants to sit on five chairs and I
say boy a wants to sit on chair number
one so this is
[Music]
what kind of condition this is a
positive condition right
so if boy a wants to sit on the first
chair okay that is a positive condition
if he if he does not want to sit on the
first chair this is a negative condition
right so if he does not want to sit or
does not want to sit on the first chair
then it is a negative condition so I
hope everyone is understanding what I am
trying to establish here so there are
two types of conditions now this must be
coming into your mind which condition to
satisfy first so whether to satisfy the
positive condition first or whether to
satisfy the negative condition first so
now let's say this is there are five
chairs a wants to sit on either second
or third chair and B does not want to
sit on fourth or I'm sorry it's its
fourth fourth or fifth chair sorry about
that so there are five chairs a wants to
sit on either second or third chair and
B does not want to sit on fourth or
fifth chair so let's see let's see this
way let us satisfy the positive
condition first here and in the nether
scenario let us satisfy the negative
condition first and see which one is
giving us a more efficient answer right
let's see which one is more efficient so
let's let's satisfy the negative
condition first now girl sorry Bob a
wants to sit on either second or third
chair and B does not want to sit on
fourth or fifth chair right so let's see
that now these are our five chairs chair
number one chair number two J number
three jail number four chair number five
now B does not want to sit does not want
to sit on any of these chairs now let's
suppose V sits on this second chair
right now if B sits on the second chair
then a has only one choice left which is
wrong a is supposed to have two choices
because he wants to sit either on second
or I'm sorry B has only one choice left
which is this
so a wants to sit either on second or
third chair which means
that uh which is wrong actually because
we are supposed to give a two choices
and we are satisfying B but in the end
we are forcing a to sit on the third
chair I hope your understanding right so
in this if we satisfy the positive
condition first and then there are five
chairs so if we satisfy the positive
condition first and then why a wants to
sit on either second or third chair so
make him sit on any of these chairs okay
so we are making him set on the second
chair right and now what B does not
wants to sit on fourth and fifth chair
which is fine we won't allow him to sit
on fourth and fifth chair and he has
another option to choose so whenever you
are satisfying a condition never satisfy
the negative condition first always
satisfy the positive condition first
always satisfy the positive condition
first and then satisfy the negative
condition so I hope your understanding
once you do this you can't do a question
wrong you you can't do a question or it
will be very very simple for you so
let's go ahead and move forward right
again how many ways five prices can be
given to five girls such that each girl
gets one price if first price cannot be
given to a or b and second price must be
given to girls C right so now there are
five prices right so I hope you are
understanding the question because we
are just revolving around the same
question once again so these are five
prices and these are five girls right so
now five prices have have to distributed
among five girls right such that each
girl gets one price now what kind of
conditions are there first price can not
be given to girl a or B so this is a
negative condition it cannot be given B
second price must be given to girl C so
according to the question we should be
satisfying the positive conditions for
so these are the five prices to be given
away so in how many ways can you give
the second price second price can only
be given to girls
so there is only one way to give the
second price you cannot give a second
price in any other way right so that
means second prize is solved now let's
satisfy the negative conditions now
first price cannot be given to grill A
or B now first price cannot be given to
a or b no how many ways are there to
give the first price now first price has
already been given already been given to
girl a hey this price also has already
been given to girl has should not be
given to girl B and this price has been
given to girls heat so that means first
price there are only two options left so
I hope everyone is understanding we
cannot give first price to A or B C has
already been given to second second
price has already been given to goal C
so now we are left with only two options
girl D and E right so now what now is
there any condition left no there is no
condition left there are three girls
three prices which can be given in three
factorial ways so this is our answer
right so hope everyone is understanding
what I did here I just solved this I
just went with the event with the
question I i satisfy the positive
conditions first I satisfy the negative
conditions first and then I satisfied
then when there was no condition left
the same number of chairs the same
number of girls that is the factorial
right so I hope everyone is
understanding the question okay so now
let's see this question once again in
how many ways five girls can sit on five
chairs
if girl eight doesn't want to sit on
first or second chair and girl B wants
to sit and third or fourth chair so once
again it's a similar kind of question so
these are the five chairs right and
these are the five girls so these are
the five chairs and these are the five
girls so let us see the condition girl
it doesn't want to sit on first or
second chair girl a doesn't want to sit
so it's a negative condition girl B
wants to sit that means it's a positive
condition so what do we do we satisfy
the positive condition first now girl B
wants to sit on third
or forth cheer now let's see let's
satisfy girl Bea's condition first now
how many ways can you make girl be said
she can either sit on the third chair or
she can either sit on the fourth chair
so there are two ways in which goby can
sit let's suppose she sits on the third
chair so third chair is already gone
right now girl aid does not want to sit
on first or second chair now girl a does
not want to sit on the first chair she
does not want to sit on the second chair
third chair has already been taken by
girls see that girl B that means there
are only two places left for girl a to
set right and now once again there are
three chairs left and there are three
girls left which can be solved in three
factorial ways so this is our answer
right so hope everyone is understanding
when I'm trying to be do here let me
just take a quick recap there are five
chairs five girls right so first we
satisfy the positive condition girl B
wants to sit on third or fourth chair so
let us say that make her happy
let let girl B sit on third or fourth
chair so we give her a choice third or
fourth two ways to do that she sits on
the third alright fine move on now girl
a does not want to sit on first or
second chair girl it does not want to
sit on first or on second third has
already been taken that means there are
two choices left four girl a now there
are three girls left three chairs left
so that means the answer would be three
factorial into four right so hope
everyone is understanding this a going
in a structured manner that's why it
must be looking really simple for you
right
okay now let's see this question okay
now this is again a difficult question a
cat question in fact right so there are
six dots and six persons task one cannot
be assigned to either to person 1 or
person to task 2 must be assigned to
either to person 3 or 4 every person is
assigned one task how many weight this
assignment can be done right so there is
task 1 task to task three task 4
task 5 and task 6 right and then there
are six people 1 2 3 I'm sorry
4 5 6 so these six tasks have to be
assigned right so since we are assigning
the tasks so we would be giving away the
tasks rather than people coming and
choosing their tasks right so task 1
cannot be so it's a negative condition
right task 2 must be so it's a positive
condition now let's satisfy the positive
condition first task 2 must be assigned
either to person 3 or 2 percent for so
task 2 must be assigned either to person
three or person 4 so how many ways are
there to assign tasks to there are two
ways it can be either assigned to person
3 or two person 4 so that means let's
suppose task 3 is assigned to person 3
now let's move on now he's saying task 1
cannot be assigned to person 1 or person
- but this task 1 cannot be assigned to
person 1 it cannot be assigned to person
to task 3 has already been assigned to
person 3 so that means there are three
ways left to assign the task 1 right so
I hope everyone is understanding right
so now how many tasks at left there are
four tasks left and how many people are
left there are four people left no
condition for factorial right so I hope
everyone is understanding so even if you
calculate that that is six times a 1
into 2 into 3 into 4 right so this is
the definition of 4 factorial so that is
12 and 72 and 144 right so I hope
everyone is understanding what I'm
trying to establish here now this is a
cat question so I hope you're seeing it
the toughest test in the nation has such
simple questions now the questions are
not that simple but what happens is the
the if you go with a structured way if
you if you plan the question accordingly
if you develop some habits some good
habits just the way I did here just the
way I wrote the number of tasks
I wrote the
of people then I draw the number of
slots and I drew the number of slots and
then I did did the question in a
structured manner you'll be able to do
that pretty easily right so let's go
ahead and move forward oh god someone is
really bored he's saying oh my god how
many more conditions okay hold on there
is one more to go fine now let's see
that
now what if conditions collide I'm
amazed free didn't ask me a question
what if conditions collide now if
conditions cry what do you mean by that
now let me show you there are five
chairs a wants to sit on either second
or third chair and B does not want to
sit on first or second chair so this
these are the five chairs one two three
four five so if we go by the question we
satisfy this wants to girl a wants to
sit on for a second or third chair let's
suppose girl a in one scenario sit on
the second chair right now if girl a
sits on the second chair
how many choices this B has now B does
not wants to sit on first or second a is
already sat on second chair so that
means B has three choices left for him
now what if a sits on the first year he
wants to study sits on the third chair
he wants to sit on either second or
third once him once we make him sit on
second general we make him sit on the
third chair now if he sits on the third
chair now B is only left now he can't
sit on one he can't sit on two theory
has already been taken by a that means
there are only two choices left for B
now you see the conditions are colliding
the choice of a is changing the choice
for B so that means in this kind of
questions we will make cases so what we
will do in this kind of questions we
have to make different cases and add
those cases right which was pretty much
the same thing in the previous time
because since all the cases were the
same so we were multiplying by the
number of the number of ways in which
the cases would be made but over here
since the cases are
friend so we have to add the number of
cases right so I hope everyone is
understanding right now this is not
simple stuff guys right this most
probably you won't be able to if you
don't understand the previous video
right so let's go ahead and move forward
okay now let's see this question same
kind of question five girls wants to sit
on five chairs girl a does not want to
sit on first or second girl B wants to
sit on second and third chair now if you
see over here the conditions are pretty
much colliding right
so these are the five girls girl number
a b c d e and this is one two these are
those five chairs three four and five
right so these are the five chairs and
these are the five girls and let's see
now what do you do in such kind of
questions you make cases you calculate
the different cases for the positive one
okay now make B sit on the second chair
how many ways are there to make B sit on
the second chair only one way right so
you make be set on the second chair okay
all right now let's suppose B has sat on
the second chair now a does not want to
sit on first or second chair now a over
here cannot sit here he cannot sit here
so he has these three choices left so he
has three choices left for him right now
girl left for her I'm sorry now there
are three girls left and three chairs
left with no conditions so three
factorial
now let's add another case now let's
make girls sit girl be sick on the third
chair how many ways to do that just one
way to do that ya want to make go B sit
on the on one chair so one way to do
that now let's see further now let's see
how many choices this girl a has now B
has already sat on the third chair right
I'm sorry Abby has already sat on the
third chair now girl it does not want to
sit on the first she does not want to
sit on the second the third chairs
already been taken she's she
left with two choices and now there are
three girls left and three chairs left
which is nothing but can be solved with
three factorial right so I hope everyone
is understanding so this is way the way
you solve the question so this is
nothing but three factorial into 2 plus
3 into 5 that is our answer
so you're everyone is understanding when
the cases collide you were supposed to
make different cases and you start with
the positive case because the positive
case gives you and how many gives you
the different number of cases and you go
about and you solve the question right
now let's make it more clear to you
let's let's solve some questions on it
right so that you understand it ok now
this is a typical kind of question find
in how now whenever these kind of
questions come the cases are already
there you know that if I'm trying to
make a three digit number zero can't
come in the beginning of him trying to
make an even number two that's supposed
to be some two four six or zero at the
end so cases are already there so in
these kind of questions
cases are predefined right so you're
supposed to know that fine so in how
many integers greater than are you
saying how many integers greater than
nine nine nine but less than 4-thousand
right so that means from thousand to
three nine nine nine so how many such
integers can be formed using these
numbers if the repetition of digits is
allowed right so the repetition of
digits is allowed so this has to be a
four-digit number right because you're
forming a four-digit number had been
there there been a three-digit number as
well and a four-digit number as well
then there were two cases right so it's
a four digit number and let's start with
this lot here so in how many ways can
you put numbers here so zero can't come
here because if zero comes here then
this will become a three-digit number
which we do not want so zero can't come
here either one can come either two can
come
either three can come can four come no
four can't come because we're forming
numbers from one thousand to three nine
nine so only one two and three come here
so there are three ways to fill this
slot right so hope everyone is
understanding now over here you can fill
any of the five numbers fine over here
you can fill any of the five numbers
similarly here as well you can fill any
of the five numbers here as well you can
fill any of the five numbers so because
the numbers can be anything it can be 3
1 2 3 1 2 4 or 3 4 4 4 since the
repetition of digits is allowed so your
answer is nothing but 375 right so I
hope everyone is understanding and this
is a pretty simple scenario you will be
able to solve this question pretty
easily now let's see let's now this is a
cat 2008 question guys I hope you are
understanding how the toughest test in
the nation the test even tougher than
the GMAT gives you some such simple
questions fine ok let's move forward
let's see some more questions 8 boys
have to be seated in eight chairs number
one to eight in a row in how many ways
the seating can be done such that if a
particular boy does not want to sit on
the first 4 chairs and other boy wants
to sit on either set seventh or eighth
chair now there are again two cases
right he does not want to sit it's a
negative case he wants to sit that means
it's a positive case so what we'll do
will satisfy the positive case first so
let's say there are eight boys let me
not name them like this a b c d e f g h
right so one two three four five six
seven eight eight boys and let's say
these are the eight chairs so eight boys
and eight chairs
now let's satisfy the condition for the
positive case now one boy let's called a
let's call that guy a he wants to sit on
either seventh or eighth chair and there
is another boy who does not want to sit
on the first four chair that means the
cases are not colliding here right so he
wants to sit so let me make eight slots
1 2 3 4 5 6 7 8 right so this boy a
wants to sit on either sex 7th or 8th
year so he has to number of ways in
which he can sit let's suppose he sits
on the 8th chair now a particular boy
let's say B does not want to sit on
first 4th chair so if he does not want
to sit on the first four chairs he can
either sit on the fifth he can either
sit on the sixth he can either sit on
the seventh chair so three ways he has
got to set now is there any condition
left no there is no condition left no
number of chairs on how many there are
six number of chairs left and six
numbers of people left so nothing but
six factorial so my answer is 6 into 6
factorial so it looks an easy question
once you go with a structured way you
satisfy the positive conditions first
let me satisfy negative conditions first
and then you go about with a question
fine okay let's see another question
right now in eight boys have to be
seeded in eight chairs numbered one to
eight in a row in how many ways the
seating can be done such that if a
particular boy does not want to sit on
even number chair and another boy wants
to sit at extremes now if you see
carefully first of all you have to
decide either whether the cases are
colliding or not now let's see that
let's see these are the five chairs 1 2
3 4 5 6 7 8 so these are the five chairs
1 2 3 4 5 6 7 8 now for the sake of
saving time let me say there are two
boys lives who have conditions a and B a
over here does not want to sit at
even-numbered chair so he does not want
to sit at even number chair and another
boy wants to sit at extremes so he wants
to sit at extremes so how do we go about
that so let's suppose let's satisfy the
positive condition first
now if let's suppose this guy over here
there are two extremes
and eight right so if boy be sets on the
first chair then a has how many choices
left he can either sit on two we can
either sit on four he can either sit on
six he can either sit on eight so he has
four choices with him
but if boy B sits on the eighth chair
how many choices does a have left with
it he has just three choices two four
and six so that means the cases are
colliding if B sets on the first chair
is left with four choices if B sits on
the eighth here is left with three
choices now using this we can form out
an equation now let let us make set be
on the first chair how many number of
ways to do that one way because we are
making set be on the first year and then
how many choices does a have now four
choices right fine plus six boys left
six chairs left six factorial would be
the number of phase right sorry this is
a multiply I'm sorry let me do it that
way okay plus this is a multiply because
the same case is going on now we are
coming to a different case let let us
make be set on the eights chair one way
to do that he has a has only three
choices left three choices into six boys
left six chairs left six factorial so
this is nothing but your answer is six
factorial into seven right so I hope
everyone understood you know this
question is way too difficult if you
don't know the exact structure to solve
this question right so this can come in
your GMAT examination this can either
come in your cat examination yeah for
GRE yeah can come but you know the
chances are pretty slim right so let's
go ahead and move forward with that fine
so our answer is nothing but 7 into 6
factorial find so let's see this
question so and how many four-digit
numbers can be formed using the digits 2
4 5 8 if the repetition of digits is
allowed right so over here repetition of
digit is allowed and 4 digit number so
this is a four digit number which can be
formed right so four digit number has to
be formed using this two four five and
eight and the repetition of digit is
allowed means you
use these digit any number of times you
want so how many numbers can come here
any four numbers 2 4 5 8 because em
four-digit number can contain anything
over here any phone numbers over here
any phone numbers over here any for
answer my answer is 256 easy question
right why I put an easy question in
between of these tough questions just to
take a mind away from you know the DOS
tough scenario so for you to understand
it pretty more clearly right so let's
move forward okay so how many four digit
numbers can be formed using the digits
exactly once right now this is a
different question exactly once means
you cannot use them more than one time
so you're forming a four digit number
right so that means four slots so over
here you can use any four of them
because it's a four digit number now if
one of them is used that means you can't
use that one once more one more time and
that means your answer is nothing but
four factorial right easy question right
a pretty simple stuff let's move forward
okay now how many four digit numbers can
be formed using this desert if the
repetition of digits is allowed right
now this is again a sim similar kind of
question but there is just one twist in
that the twist is that over here the
condition is you can't use zero if you
use zero the number will be a
three-digit number
so over here you can only use this four
numbers right one two five six over here
you can only use these for numbers no
repetition of digits is allowed that
means you can use any five numbers here
you can use any five numbers here you
can use any five numbers here so that
means your answer is nothing but 125
times four which is equal to 500 right
so I hope everyone is understanding the
level of difficulty that we have done in
the previous is just in the video that
this is nothing at compared to that
right okay let's move forward so how
many four digit numbers can be formed
using the digits if the repetition of
the digits is not allowed so I've just
taken the previous question just one
step forward now over here repetition of
digits is not allowed so again for slots
alright
now over here you can't use zero fine
so let's satisfy this first you can use
any four numbers out of here okay now
over here you can use zero but you can't
use the number that you used there so
again for number of days right because
let's suppose you give one to this
number you can use one here because
repetition is not allowed fine
now how many numbers are left so there
were five numbers two numbers have gone
three numbers left and over here two
numbers left so your answer is nothing
but it's a 3 times 2 times 4 times 4
which is 16 into 6 that is 96 right so I
hope everyone is understanding this
question it's a it's a easy question but
you have to be careful fine okay how
many four digit even numbers can be
formed using these digits if the
repetition of digits is allowed right
now over here repetition of digits is
allowed whenever this is given our
answer becomes the solution becomes a
little bit easy right so repetition of
digits is allowed now there is a
condition on two cases over here zero
can't come but over here only zero two
and four can come right so over here you
can fill only three numbers because
since you're forming even numbers right
so over here only zero to four and come
over here a zero can come so you can
fill in any of the four numbers over
here you can fill any of the five
numbers any of the five number since the
repetition of digits is allowed right so
your answer is nothing but 300 so I hope
everyone is understanding what I'm
trying to establish here it's a pretty
easy question let's go ahead and move
forward fine okay so now let's take the
previous question to a next level right
how many four digit even numbers can be
formed using this digit such that we
should not be using numbers more than
once so that means the repetition of
digits is not allowed right so let's try
solving this
now since the repetition of digits is
not allowed
we have to see whether the cases are
colliding or not so over here you can
use 0 2 & 4 over here you can't use zero
right now let's suppose let me let me
assume if I use 0 here right let me call
this a slot number 1 slot number 2 slot
number 3 and slot number 4 if in slot
number 4
I give 0 then how many ways I can fill
the slot number 1 if I give it zero to
it I can fill 1 2 4 & 5 that means there
are 4 ways
fine if give I for this is slot number
right and these are number of ways
so if I give any of them any of 2 or 4
to slot number 4 then how many ways will
slot number 1/2 now over here slot
number 1 can have only 3 numbers why
because one of the numbers from 2 & 4
have been given to slot number 4 that
means slot number one will be left with
only three choices rather than four
choices so you're seeing a difference of
scenarios that means the cases are
colliding now in the previous questions
the the colliding of cases was pretty
clear to you right in front of you 1 & 2
2 & 3 but over here the question is
different but we're using the same
concept here the cases are colliding now
let's so use this to solve our question
now let's give 0 to slot number 4 how
many number of ways to do that one way
to do that right now if we give one zero
to slot number four how many ways will
slot number one has four ways right so
now how many digits are left we have
already distributed two digits that
means three digits are left and two
slots are left so any three to this one
any to do this one right plus let's give
two or four to slot number four how many
ways to do that two ways to do that
similarly how many choices this slot
number one has three choices right again
two choices left three digits left so 3
into 2
right so this becomes pretty simple if
you solve it this way so if you solve it
it's 4 times 3 12 is 24 24 plus plus 3
into 3 9 9 4 36 so 24 plus 36 that means
it's equal to 60 right so we got ourself
an answer so if you see if you don't
follow the structural manner in this
question there is no way going forward
right so you have to go in a structured
way solve the posit see whether the
cases are colliding and the cases are
colliding start with the positive case
once you start with the positive case
see what all cases are becoming and draw
this draw the search flow chart because
this is very important once you draw
this flow chart from this you're able to
come through this and from this you get
your answer right ok thank you ok so if
you have understood all the concept so
what should it be a next step your next
step should be practice so this is a
we've just done regarding the cases the
number of cases that it should take care
of so we have only only only covered the
number of cases so you should practice
on that once you practice you will be
able to understand it with more clarity
and with more depth right so you can
watch this video as many number of as
many times you want you can even
download it in case you need right so
this was about the video thank you very
much for the video this is my email
address where you can send me your
valuable feedback so in the coming
videos we'll be doing some combination
problems permutation problems wherein
there are different number of cases
different scenarios but we'll be
building on our first video so I would
recommend you to watch this first video
and then go towards the other videos
right there will be about more videos to
come and permutation combinations so
thank you very much and see you next
video