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Welcome back. So, we have been studying how
to count or we have been dealing with the
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subject of combinatorics.
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So, combinatorics is the subject of branch
of mathematics that involve in counting. And
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a typical question in this regard is given
a set S what is the size of the set or what
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is the cardinality of S? Or how many elements
are there in S?
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Now, the question to be asked is how is the
set given? In most of the cases, the set is
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given implicitly by describing the set and
not explicitly and that is what makes the
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problem hard. So, the set is usually described
in words and we have to count the number of
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elements in this set.
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So, for example one can ask, how many elements
of a universe set satisfy a certain set of
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conditions? Or equivalently, you can ask things
like how many ways can you draw an element
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from universe such that the element satisfies
certain set of conditions. So, these are both
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counting problems and these are equivalent
questions. And these are the kind of questions
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that we would like to answer in the subject
of combinatorics.
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So, here are some examples that we discussed
last class or let me repeat it all over again.
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So, how many n digit numbers are there where
there are no consecutive digits are same?
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Similarly, the next question is how many functions
are there from the set 1 to n, to 1 to k that
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are non-decreasing? That means if x is less
then y then f of x is less than or equal to
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f of y. Or how many ways can you distribute
n identical toffees among k kids?
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And the last one is how many strings of length
n, how many zero one strings of length n are
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there which does not have any consecutive
zero. So, these are some of the problems that
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we should keep in our mind and we have been
looking at some of the general theories of
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attacking counting problems.
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Now, while we can do a lot of theories about
counting problems one thing to remember always
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is that each problem is unique and each has
to be solved applying a technique that fits
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it. So, different problems require different
techniques and that you have to solve that
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you have to find out yourself. Counting is
in fact one of the most challenging subjects
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in mathematics in fact big names like Srinivasa
Ramanujan spent a lot of his time solving
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problems in counting.
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There are some handy tricks and tools to attack
this problems and that is what we have been
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looking at in this set of lectures.
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In the last video we looked at the problem
of how many ways can we select k objects from
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n objects and there are four cases that we
have to look at question is or alright two
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cases that we have to look at. Case one, are
we allowed to pick objects multiple times
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meaning, is repetition allowed? And the second
one is inside the selected set of objects
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does ordering matter or does it not matter?
Is it just a group or is it a kind of an ordered
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set and depending on that we have got this
four different counts.
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So, in this video we will start looking at
a kind of similar problem but slightly different
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it is called the distribution problem or in
other words, how many ways can you distribute
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n balls in k bins, in k baskets? So you have
say n balls and you have to distribute it
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in k baskets. Now, just like last time here
there are few questions to be asked. First
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question, are the bins distinguishable or
are they indistinguishable? Or in other word
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is it like said okay I have the basket 1,
basket 2, basket 3, basket 4 or are we just
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asking them to club some of balls into groups
and I do not care which is group 1, which
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group 2.
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Similarly, are the balls distinguishable or
indistinguishable? Meaning are all the balls
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same identical, so it does not matter whether
I put the first ball in the first basket or
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the second ball in the first basket as long
as one of the ball goes in the first basket
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set. And in case if the balls are distinguishable
the inside a particular bin does the order
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matter or not? So, inside a bin do we care
about how the balls have been placed or which
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ball gone there first and which ball has gone
there second and so on or it does not matter
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it is just another group of balls.
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Next question is that are we allowed to keep
some of the bins empty? Are we allowed to
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keep some of the baskets empty? Just ignore
some of the baskets. And are there any other
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restrictions basically. So, we will be looking
at this problem and this is in fact a very
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general class of problems. How to distribute
n balls into k bins. We call it as a balls
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and bins problem and we will be through various
cases and see how to attack each of the cases.
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Some of them is easy. Some of them are notoriously
hard.
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So let us start with the first problem that
we have. So, this is the same kind of problem
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that we were doing in the last video. So,
here we have n constables in your police station
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and you have to divide the n constable into
4 projects, Project 1, Project 2, Project
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3, Project 4. No, constable can be in part
of more than one project and every constable
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should be part of some project. How many ways
can you form the groups? Now, this is basically
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a balls and bins problem.
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So, you have say Project 1, Project 2, Project
3, Project 4. Now, And you have the constables
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C 1, to C n. Now this constable you can put,
you have to divide them into this baskets
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basically, right. I do not care at this point
about what is that ordering of the things
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in the Project 1 of the constables that goes
in Project 1 and so on. So, I just have to
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divide into groups. How many ways can I do
it? So, the idea is that the constable number
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1 can be either we put into Project 1 or Project
2, or Project 3 or Project 4.
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So, there are four possibilities for constable
1. Still constable 2 can be put into Project
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1 or Project 2 or Project 3 or Project 4.
So, there are four possibilities of constable
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2 also. And Similarly, for every of the constable
there are four possibilities and that means
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the total number of possibility that are there
is 4 times 4 n times which is 4 power n. So,
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the number of ways of placing this or dividing
this n constable to four projects is 4 power
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n. Speaking of a general problem
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If you have n distinguishable balls now distinguishable
balls what do I mean distinguishable balls?
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Means the balls are different like in the
constables, constable 1 is different from
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constable 2, right and you have k labeled
bins meaning Project 1, Project 2, Project
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3, Project 4. Constable 1 going to Project
1 is different from constable 1 going to Project
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2 and I do not care about the ordering inside
the bins. So, in that case the total number
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of ways of taking n balls into k bins by the
same argument the first ball can go to any
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of the k bins, second ball can go any of the
k bins.
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Third ball can go to any of the k bins. So,
all total it is k power n. Now, couple of
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things to note here it is very much possible
that if say I have this four buckets and I
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have the n constable and I might also put
all of the constables into one basket and
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none in the other baskets. This is a valid
possibility. So, that means it is possible
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that some of the bins are left empty. The
problem as such does not stop us, does not
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restrict us from doing so. So, we are fine
but it is something to note that in this particular
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way of dividing it the bins can be made empty.
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Now, one more problem and this time we have
talked about giving gifts to constables. So,
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we have say n constable in your police station
and there are 100 identical gifts think of
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shopping coupons. You have 100 shopping coupons
and you want to distribute the coupons among
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your constables. A constable can get any number
of coupons and a constable may or may not
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get a coupon also. How many ways can you give
a coupon?
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Now, this is very similar to the technique
here that we will apply is very similar to
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the one that we apply when we talked about
choosing k persons from n, from a group of
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size n where repetitions are allowed and ordering
matters. So, let us quickly lets see this
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technique again. So, here we have 100 coupons.
So, let me say that they are 100 gifts and
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let me draw them as square as these are the
coupons. Now, all of these coupons are identical
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so it does not matter what type it is?
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How I place them? So, we lay the coupons in
a straight line and now we have to decide
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how to divide the coupons among the n constables.
So one thing that we can do or let us think
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what we will do naturally? So, as soon as
the first constable comes up we will tick,
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okay let us see. This three of them are yours.
The next constable comes, we said okay. This
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two of them now are yours. So, in some case
think of them that a constable comes arise.
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I just take the first three and gives it to,
the constable, first constable then I am left
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with things from here to here.
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So, next constable comes I again said okay
everything here bit aare yours. The next constable
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comes said everything –this three of them
are yours. The fourth constable comes, okay
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I do not like you. I do not give you any coupon.
so I will say okay here is your bar, right.
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So, what am I doing? So, in other words, I
am basically placing some kind of bars between
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these coupons. So, there are 100 coupons,
there are 100 coupons and I am placing some
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kind of bars between these coupons and how
many bars should I put.
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Now, if I put three bars or four bars I say
that everything left of it goes to the constable
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1, everything between these two bars goes
to constable 2, everything between this one
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goes to constable 3, everything between this
two goes to constable 4 and everything right
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of this goes to constable 5. It is like walls
we have created right? So, four walls basically
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is a way of dividing the coupons into 5 constables.
Similarly, if I have n constables I can use
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n minus 1 walls like this and I can place
this wall anywhere I want.
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So, any placement, any placement of the squares
and bars give me a partition of this n identical
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gifts into they can sorry, partition of this
100 identical gifts into n constables. For
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example, if say I suddenly so I look at a
orderalready. So, say this is some ordering
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of the square and bars and what does this
say? This says that constable 1 gets 2 coupons,
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constable 2 gets of course there is nothing
here zero coupons. Constable 3 gets zero coupons,
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constable 4 gets 2 coupons, constable 5 gets
4 coupons, constable 6 gets 3 coupons and
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constable 7 gets zero coupons.
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So, this is a way of distributing this 11
coupons into 7 constables. So, in other words
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what we are talking about is that every distribution
of this 100 identical objects into n constables
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can be represented as an ordering of 100 squares
and n minus 1 walls, right n minus 1 walls.
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At any ordering of this 100 squares and n
minus 1 walls gives me a distribution of 100
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coupons to n constables. So, the answer to
this question of how many ways can we distribute
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100 coupons to n constable is basically the
number of ways we can order 100 squares and
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n minus 1 bars.
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In how many ways can that we done? So how
many objects are there, 100 squares and n
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minus 1 bars. It is n plus 100 minus 1 right.
So, this many objects are there so I ordered
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this many objects, how many ways can I do
it? I can do it in this many number of ways,
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factorial number of ways. But again since
all the coupons are all identical so we have
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to factor them out we should not consider
the fact the permutation between them.
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So, you should factor out that by dividing
by n factorial, sorry by dividing by 100 factorial
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and we should also factor out that fact that
all the walls are identical, they are just
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the walls have nothing written on them especially.
So, we should factor that out also and we
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get n minus 1. So, that is the answer to this
question. It is very similar to the technique
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that we did last time in the last video for
selecting k people from n, from a people group
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of size n where repetitions are allowed and
ordering matters.
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So, in general what we are talking about here?
If you have n indistinguishable balls and
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k labeled bins, now they are labeled bins
because once we are doing this kind of parts
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I have labeled this goes to bin number 1.
This goes to bin number 2. So, here I am already
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talking about some kind of a natural labeling
that is coming up because of the placement
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of the bars. So, if we have n indistinguishable
balls like the coupons and k labeled bins
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like the k constables.
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In how many ways can you place the n balls
in k bins by ordering inside the bins does
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not matter? It is what we just wrote which
is n + 1, n + k minus 1 factorial divided
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by n factorial to take care of the fact that
the balls are all indistinguishable and k
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minus 1 factorial to take care of the bars
or the walls are indistinguishable. Which
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actually in terms of notation is same as n
+ k minus 1 choose k minus 1. Ok, Note here
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that since two walls can be next to each other
so there can be a possibility that a bin remains
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completely empty.
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Now, what happens if we add the constraint
that every constable must get at least one
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coupon. Now, there are two ways of doing this
whole thing two ways of calculating it. Number
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one thing is that you can first give all the
n constables one coupon each, right. So, in
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that case I will be left with 100 minus n
coupons and now divide this n coupons in whatever
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way you want in among the n minus 1 constables
which is by the earlier rules.
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This factorial by n minus 1 factorial times
100 minus n factorial. Or which just comes
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down to 100 minus 1 factorial by n minus 1
factorial in 100 minus n factorial which is
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equals to 100 minus n choose n minus 1. So,
idea is that just keep one coupon to all the
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constables to start with so everybody now
has at least one coupon now whatever is left
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divide among the constables in whatever way
you want and which is this many number of
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ways.
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Now, there is one other way of solving the
same or reaching the same answer it is like
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this bars and coupons problem. Again we lay
out the coupons in a straight line and again
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we have to basically put up n minus 1 horizontal
bars sorry vertical bars. But where do we
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put them up since we cannot put two of the
vertical bars next to each other. So, that
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means between any gap between this two consecutive
coupons there can be at most one horizontal
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bar, right.
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So, this n minus 1 horizontal bars should
be placed among this set of all the possible
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positions. How many positions are there? How
many gaps are between two consecutive coupons
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since there are 100 coupons the number of
position between the coupons is 100 minus
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1. I can choose any one, any n minus 1 of
them and place the bars there and I get affirmed.
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I get a distribution of the coupons into n
constables so that every constable gets at
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least one coupon.
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So, the number we get again is same 100 minus
1 choose n minus 1, right. So, this of course
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we can generalize again to
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If we have n indistinguishable balls and k
labeled bins how many ways can it place the
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n balls in k bins when ordering in the bin
does not matter but no bins can be made empty
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it is arrange the balls in a straight line
there are n minus 1 gaps between them. How
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many ways can you place the k minus 1 balls
in between among this n minus 1 spaces which
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is n minus 1 choose k minus 1. So, ok, till
now we were not looking at the ordering in
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the bins at all. What happens if we look at
the ordering inside the bin?
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For example, say, in the constable problem
we want to divide this n constable into 4
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projects, Project 1, Project 2, Project 3,
Project 4. No, constable can be part of more
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than one project every constable should be
part of some project but you also have to
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decide the ranking of the constable in each
of the project. So, note that I can do the
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same trick that I did which is –now I can
place the constables in a particular ordering
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so I have Constable 1, Constable 2, till Constable
n.
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Take any ordering of the constables and if
I can partition, if I can draw this since
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there are four projects if I can draw three
bars along them and I say that okay these
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00:27:36,960 --> 00:27:42,059
set of constable is in project 1, these of
them are in project 2, these are in project
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00:27:42,059 --> 00:27:50,419
3 and these are in project 4 then we get a
one way of partitioning it and any permutation
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00:27:50,419 --> 00:28:00,679
of C 1 to C n will give me a different way
of partitioning them into this 4 Projects
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00:28:00,679 --> 00:28:10,500
and clearly for example C 1 C 2 as this thing
is same as sorry different as C 2 C 1 and
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00:28:10,500 --> 00:28:11,500
this thing.
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00:28:11,500 --> 00:28:21,820
So, a different permutation gives me a different
–all though the group might be same but
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a different ordering of or different ranking
of constable in each of the projects, right.
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00:28:28,299 --> 00:28:40,269
So, we can do the same idea about like the
balls and bars issue and I get –what do
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00:28:40,269 --> 00:29:00,399
I get? Okay, number of ways I can place this
–so I have the n constables and I have 3
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00:29:00,399 --> 00:29:05,909
bars. So, I have n + 3 objects. I factorize
them I do the –these are the numbers of
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00:29:05,909 --> 00:29:15,659
ways I can just order them and I just have
to just take care of the fact that the 3 bars
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00:29:15,659 --> 00:29:19,020
are identical the three bars are the three
objects.
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00:29:19,020 --> 00:29:23,720
I shouldn’t different between them but I
do not need to divide it by n factorial because
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the every different ordering of the constable
give me a different way of putting the constables
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00:29:31,999 --> 00:29:40,360
in different projects and different ranking.
So generalizing this, we have
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00:29:40,360 --> 00:29:51,350
If we have n distinguishable balls and k labeled
bins then the numbers of ways we can place
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00:29:51,350 --> 00:30:01,210
n balls in k bins when ordering inside the
bins matter is just n + k minus 1 factorial
215
00:30:01,210 --> 00:30:10,150
divided by k mins 1 factorial. So, so we have
seen quite different versions of them different
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00:30:10,150 --> 00:30:14,500
version of distributing k n balls into k bins
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00:30:14,500 --> 00:30:22,110
But till now what we have seen is that if
the items are indistinguishable and bins are
218
00:30:22,110 --> 00:30:31,619
labeled we know how to solve it, n + k minus
1 choose k minus 1 when they can be empty
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00:30:31,619 --> 00:30:41,630
and bins are labeled and they cannot be empty
then I have n minus 1 choose k minus 1. When
220
00:30:41,630 --> 00:30:49,899
the items are distinguishable and ordering
inside the bins does not matter in that case
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00:30:49,899 --> 00:30:59,239
when the bins are empty, can be empty then
total number is k power n where as otherwise
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00:30:59,239 --> 00:31:03,840
we just now saw it is n + k minus 1 factorial
by k minus 1 factorial.
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00:31:03,840 --> 00:31:16,609
As you can see in this diagram at least two
of the boxes are left empty namely what happens
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00:31:16,609 --> 00:31:27,070
when the items are distinguishable and ordering
inside the bins does not matter but the bins
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00:31:27,070 --> 00:31:34,519
cannot be empty and similarly the order inside
the bins matters but the bins cannot be empty.
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00:31:34,519 --> 00:31:42,960
There are three more things that are also
which are the fact what happens if the bins
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00:31:42,960 --> 00:31:48,539
are unlabeled? What happens when the –I
am not at all interested in the labeling of
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00:31:48,539 --> 00:31:54,730
the bins but I just want to know how many
groups in which you can form it.
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00:31:54,730 --> 00:32:00,259
So, we are left with five options, five places
to solve and we will continue with our understanding
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00:32:00,259 --> 00:32:09,349
of trying to solve this particular matrix
in the
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00:32:09,349 --> 00:32:10,610
next class. Thank you.