Introduction to Combinations Video Lecture | Mathematics (Maths) for JEE Main & Advanced

In the last video we figured
out how many different ways
could 5 people sit in 3 chairs.
So for example, if this is
chair one, this is chair
two, this is chair three.
We said, well, in chair one
we could put 5 people.
No one's sitting down.
Then there's only going to be 4
people left, so we could put 4
different people in chair 2.
And then there would be 3
people left that we could
put in chair three.
So the total permutations, the
different ways that people
could sit in the different
yours if we cared about the
order-- if we cared about which
chair they were sitting down
in-- it would be 5
times 4 times 3.
And another way of thinking
about that-- 5 times 4 times
3-- that's the same thing.
That equals 5 times 4 times 3
times 2 times 1 over what?
Over 2 times 1.
And that's the same thing as 5
factorial over 2 factorial.
Now where does
this 2 come from?
How is 2 related
to the 5 and 3?
What's the difference
between the two?
So this is the same thing,
this is equal to 5 factorial
over 5 minus 3 factorial.
And that in general is how
we figure out, how many
permutations can 5 things
place themselves or be
placed into 3 positions?
And the general formula
is-- and we learned
this in the last video.
And I will switch colors.
If we want to put n things into
k positions, and k has to
be less than or equal to n.
Well actually, it
doesn't have to.
But for our purposes right now
we'll assume it is because our
formula might break
down if it weren't.
And that equals n factorial
over n minus k factorial.
I always find this harder
to memorize than just
thinking about the spot.
Then just saying, oh well,
you know, 5 people, 5 of
the things could be here.
And then once one thing is
here, there's 4 possibilities
left and then there's
3 possibilities left.
The way I think of it I
take the first k terms
of the n factorial.
Or in this case, I take
the first three terms
of the 5 factorial.
5 times 4 times 3.
That's how I think
of permutations.
This is great if we cared--
let's say these are
people A, B, C, D, E.
So these are the 5 people that
are going to sit in the chairs.
This is great if we wanted to
count the permutation ABC as
different from the permutation
ACB, as different from the
permutation-- I don't know--
BAC as different from
the permutation BCA.
Because remember, when we did
this we actually cared about
where they're sitting.
And in the previous video we
double counted everything
because it matters if person
A is in seat one and
person B is in seat two.
And then if they switch
we recounted, right?
This is where they switch.
But what if we didn't
care about that?
What if we didn't care
who's in what seat?
We just wanted to know how
many different ways can
the 5 people sit down?
So we want to count all the
situations where person A, B,
and C are sitting down as
essentially one situation.
We don't care who's
sitting in which chair.
We just care that those are
the 3 people sitting down.
That's the set, the subset
of the people sitting down.
And so the question then
becomes not how many different
permutations or how many
different ways can the people
sit down, the question becomes,
how many subsets of 3 can
we take out of a set of 5?
And I know I'm kind of jumping
up around a little bit, but
that's essentially what
a combination is.
A combination is a permutation
where you don't care
about the order.
So how do we figure it out?
Well, when we figured out the
permutations using this formula
we counted-- for example, we
counted ABC, ACB, BAC,
BCA, and let's see.
There should be two
more permutations.
CAB and CBA.
We counted all 6 of these
as different permutations.
But in our combinations we're
going to want to-- this is all
essentially the same
combination because we don't
care about the order.
So for any 3 different people
that are in these seats,
there's actually going to be 6
permutations that we're
counting when we do
the permutations.
So if we want the combinations
we'll just divide by the number
of ways we can rearrange
3 people into 3 seats.
That's essentially
what we did here.
So how many different
ways can you arrange 3
people into 3 seats?
Well, this is kind of another
permutation problem.
The first seat you could put 3
different people, the second
seat you put 2 different
people, and the last seat--
well, there's only
1 person left.
So it equals 3 factorial,
which is equal to 6.
This is equal to 3 factorial,
which is equal to 6.
Hope I'm not confusing.
What I'm just trying to say is
when you did a permutation we
counted all of the different
orders of how people could
arrange themselves.
And what I'm saying now is
well, how many different ways
can people arrange themselves?
Well, it's going to be the
number of places factorial
because if we have 3 people
in 3 spots or let's say,
4 people in 4 spots.
The fourth spot can have 4
people, the second spot can
have 3, and so forth, the third
spot could have 2, and the
last spot will only have 1.
So it's the number of spots
factorial is how many
permutations we're counting.
When we have just the same
people, they're just playing
musical chairs in the
exact same seats.
So on order to figure out the
combination, so if we wanted to
say how many people-- let's
say if we had 5 people.
How many different groups
of 3 can be seated?
And we don't want to double.
We don't want to more
than double count.
I don't know what the
word is for counting
something six times.
Well, it's just going to be the
same thing as the permutation
divided by all of the
extra counting we did.
We'll just divide by the number
of ways that 3 people can
arrange themselves in 3 seats.
And that's 3 factorial.
And I hope I'm making sense.
Maybe I'll do a couple more
examples in other videos.
And definitely request
it if you think this
is extra confusing.
So in general, if we say, what
are the different ways that
n things can be chosen?
Or the number of combinations
that n things can be chosen
into sets of r, where r is
less than or equal to n.
It's equal to the number of
permutations you could create
of putting n things into r
spots, divided by r factorial.
We're going to divide by the
number of ways the r spots
themselves could be rearranged
because we don't want to
count those as extra.
And so if we go back to this
formula up here, well, this
was a k, but now we're
saying it's an r.
This is the same thing as--
so the permutations was
n factorial over n
minus r factorial.
And now we're dividing
everything by r factorial.
So that equals-- let
me just write this.
This is often written
as n choose r.
Another way it's
written is n choose r.
This is called the binomial
coefficient and we'll do a
whole series of modules on that
as well because this actually
shows up in polynomial
expansion when you take
polynomials to powers.
But this is equal to n
factorial over r factorial
divided by n minus r factorial.
You could memorize this.
You know, it's useful if
you want to do things
quickly on tests.
But it's very important to
think about where it came from.
The n factorial over the n
minus r factorial-- this
is just the permutation.
And what is that?
Well, that was just the first
r-- I guess you could call
it the first r factors.
The r largest factors
of n factorial.
That's all that is.
And then when we do the
combinations we divided by r
factorial because we want to
divide it by all the different
arrangements that the people
could seat themselves
in r seats.
Or, the balls could
be placed in r cups.
So in this situation if we want
to know how many different
groups of 3 can be selected
from 5 people or from 5
letters, it's going to be 5
factorial over 3 factorial
times 5 minus 3 factorial.
And that's 5 times 3 times
2 times 1, over-- 3
factorial is just 6.
We'll put that aside
for a second.
Divided by-- this
is 3 factorial.
2 times 1.
So notice, this is a
permutation part right here.
This term just gets rid of
the two lowest factors.
You get 5 times 3.
Oh, sorry there's a 4.
5 times 4 times 3, which is
the number of permutations.
And then we divide by 6 because
we get 6 permutations for
really every combination.
Maybe that confused you.
But anyway, so we get 5 times
4 times 3 divided by 6.
And that's what?
5 times 12 divided by 6,
which is equal to 5 times 2.
There's 10 possible ways
that we can take sets of 3
from a group of 5 things.
See you in the next video.