MATH 110 Sec 12.2 (S2020): Permutations and Combinations

Slightly modified version of the F2019 video

Today we'll discuss
section 12.2 and continue

our study of combinatorics.

And I'll begin by
reminding you that we've

talked about using lists, and
tables, and tree diagrams.

And more importantly, we
introduced a counting principle

as a counting technique.

I want to add to that.

Our tool box for counting
things will continue to grow.

Suppose we are arranging
a promotional photo

shoot for three celebrities.

And if you like
Star Wars, you'll

recognize these as
Luke, Han, and Yoda.

And I'm asking the question, in
how many ways can we line them

up together for a photo?

If you're just going to
list them by brute force,

one way is to just leave them
in the order they appear now.

You've got Luke, Han and Yoda.

And then you need to figure
out how many ways you

can move things around to
get a different arrangement.

Suppose I decide to keep
Luke in the first position

and switch Han and Yoda.

That would put Yoda
in the second position

and Han in the third position.

And there's no other
arrangement that leaves

Luke in the first position.

So maybe now I'll
slide Luke over

into the second position,
put Luke in the middle.

That means I could put Han
on the left of Luke and Yoda

on the right.

Or I could switch that and
put Yoda on the left and Han

on the right.

If you think about it,
there's no other arrangement

that leaves Luke in the middle.

So let's move Luke
over to the far right.

Now let's put Han
first, Yoda second,

leaving Luke in
the third position.

The only other thing I can
do is switch Han and Yoda.

And that's a sixth
way to do this.

And as you can see, I put Luke
in every possible position.

And then I've switched
the other two around Luke,

so there's no
other way to do it.

And the answer would be six.

You could do this with
a tree diagram as well.

If you were to break
it out that way,

you could start off
with Luke and branch

to either Han, and
Yoda, and so forth.

But the point is,
either way you do it,

there are six possible ways of
arranging three people in line

to take a photo.

Each of those six
orderings is called

a permutation of those objects.

So there are six permutations
of three things taken together.

Fortunately, there's a
quick way to calculate

the number of permutations,
and it involves

something called a factorial.

And the definition involves
an exclamation point.

Here it says, n
exclamation point.

That's read, n factorial.

And n factorial is defined
to be n times n minus 1,

times n minus 2, continuing that
pattern to count down to one.

Now, hold that thought.

We're going to go
off and explore

this idea of a factorial, and
we'll come back to this spot.

So remember this spot,
and we'll come back to it.

Going back and diverting and
talking about factorials,

the n exclamation point
is read, n factorial.

And the calculation
is simply whatever

number is to the left of
the exclamation point.

You start with that,
and then you count down

until you get to one,
and you take the product.

So 4 factorial is 4
times 3 times 2 times 1,

which happens to be 24.

6 factorial would be 6 times 5
times 4 times 3 times 2 times

1, which is 720.

1 factorial-- you start
at 1, and you count down

till you get to 1.

Well, you're already at 1,
so 1 factorial is just 1.

There is one
special case that we

need that doesn't
follow that pattern.

As long as n is 1, or 2, or 3,
or 4, or any counting number,

you can use this formula.

But there is a need
occasionally for us

to define what happens when
you get zero factorial.

And it turns out that zero
factorial is, by definition,

equal to one.

I'm not going to talk about why.

At this point, it
doesn't matter.

But in order for our
factorials to work

the way we want them
to work, we have

to define zero
factorial to be one.

That's the one that you
just have to memorize.

It doesn't follow the
pattern for the others.

Now, you can play with this.

But you probably can see,
just by the definition,

that factorials can
get big really fast.

So it's nice to know that
your Casio FX-260 does

have a Factorial button.

If you look at your calculator,
the Factorial button

is above the number
three on the calculator.

So that means you have to
press Shift to get to it.

So if I ask you to
calculate 8 factorial,

you would press eight.

Then you would do Shift-3.

Shift-3 takes you to factorial.

And you don't even
have to press Equal To,

and you'll get 40,320.

Now that we know how to
calculate factorial, let's

go back to the spot I asked
you to remember earlier

and pick up from there.

Remember, we were
here, just after we'd

done these permutations
of Luke, Han, and Yoda

and found out that there were
six ways we could line them up

for a photo.

The important point
here, though, I

want to emphasize
is that it turns out

that the number of permutations
is actually n factorial where

n is how many distinct
objects you have.

So here, you have three people,
so it will be 3 factorial.

And 3 factorial is 3 times
2 times 1, which is 6.

Notice it matches
what we got earlier.

So rather than having
to actually write out

every permutation, we can
just use the factorial formula

and get there very quickly.

And again, just for emphasis,
that's exactly the same answer

we got when we just
did it by brute force.

Now, you need to remember this.

This is a very
important result. If you

want to count permutations
of indistinct objects,

all you need to do is
calculate n factorial.

Occasionally though, we're
interested in counting

the number of
different arrangements

of a group of objects.

Now let's go back to the
example that we began with.

That's Luke, Han, and Yoda.

But what if we added
a fourth person?

What would that do?

What if we added Princess Leia?

Well, now that we
know that factorials

do the counting for
us, all we have to do

is calculate 4 factorial
instead of 3 factorial.

The 4 factorial is 4 times 3
times 2 times 1, which is 24.

So there are 24 different
arrangements possible

if you got four people.

And knowing that
form is good enough

when the number of
objects you're arranging

is the same as the
number of objects you

have to choose from because--

For example, in the last
exercise, we had four people,

and we lined them all
four up for a photo.

But what if you were doing
something slightly differently?

Suppose we had a
bicycle built for two,

and we wanted to pick
two of those four people

and put them on the
bicycle built for two.

Also keep in mind, if you're
looking for the number of ways

you could do that, a person
sitting on the front seat

would be different than
that same person sitting

on the back seat.

So we will look at the
number of arrangements.

We're choosing from
the four celebrities,

but we can only take them two
at a time to get a photo of them

on a bicycle built for two.

So this time, the count
is slightly different.

I don't want to take
them all four at time.

How do you deal with that?

Well, let's play with this.

I'm going to just
randomly reach up and grab

two of the celebrities.

I'll take Luke and Leia.

Put Luke on the front seat
and Leia on the back seat--

so that would be my
first arrangement.

Now you need a strategy to
go through every possibility

in an organized fashion.

So I decided I'm going to
leave Leia on the back seat

and swap everyone else out.

So if Leia stays on the back
seat, and I take Luke off,

I can put Han in Luke's place.

That's the second possibility.

Again, I want to keep
Leia on that back seat

as long as possible.

So now the only
other way to do that

is to put Yoda on
the front seat.

That's a third possibility.

At this point, Leia has
to come off the back seat,

or I don't get any more
unique arrangements.

So I'll take Leia
off the back seat

and put Luke on the back seat.

And then I'll start
on the left, and I'll

pair Luke on the back seat
with Leia on the front seat.

That's a fourth arrangement.

I'm going to leave
Luke on the back seat

and take everyone else
on the front seat.

I've already used Leia, so I'll
move Leia off and take Han.

So now I'm leaving
Luke on the back seat,

Han goes in the front seat.

Again, if I leave
Luke where he is

and take Han off
and put Yoda, that

gives me yet a
sixth possibility.

There's nothing else
I can do and leave

Luke on the back seat.

So I'll take Luke
off the back seat now

and put Han on the back seat.

And I'll start on the left and
let Leia be on the front seat.

That's a seventh arrangement.

Keeping Han in the
back seat, now I

can put Luke on the front seat.

That's an eighth arrangement.

Leaving Han on the back seat, I
can put Yoda on the front seat.

That's a ninth arrangement.

Now there's nothing else
I can do and still leave

Han on the back seat, so I'll
bump Han off the back seat

and put Yoda on the back seat.

And then I'll start
on the left again.

So Han is off.

Yoda goes in the back seat,
Leia on the front seat.

That's a tenth arrangement.

Leaving Yoda in the
back seat, now I'll

put Luke on the front seat.

That's an 11th arrangement.

And finally, leaving
Yoda in the back seat,

I can put Han on the front seat.

And that's a 12th arrangement.

And there's really
nowhere else to go.

So there are 12 permutations
of four objects--

and this is important--
taken two at a time.

So I'm still trying to
arrange four objects,

but I'm only taking
them in pairs.

This is an important distinction
because, if you're not

taking them all, you've
got to know how many you

are taking at a time.

And that's extremely important.

So in the previous example,
when we took four people,

we took all four at a time.

That turned out just
to be a factorial.

4 factorial is 24.

But when we took
them two at a time,

it turned out to be only 12.

We learn, if we do take them
all, we just use the factorial,

as I mentioned.

The question is, is
there also a formula

if you don't take
all of them at once?

And the answer turns
out to be lucky for us--

yes, there is.

And here's what we need.

A permutation is an ordering
of distinct objects.

You often think of them as
being in a straight line.

If we select r distinct objects
from a set of n objects--

in other words, you're only
taking them r at a time--

and arrange them
in some order, we

call this a permutation of
n objects taken r at a time.

It's denoted quite often with
a capital P and subscripts n

on the left and r on the
right, or just a capital P

and have n comma
r in parentheses.

Both of those notations
are common ways

to list a permutation of n
objects taken r at a time.

And here's the formula.

The formula for that is
n factorial up top and n

minus r factorial down bottom.

We're not going to
prove this, but we're

going to use the
result, nevertheless.

Now let's apply that formula
to the bicycle-built-for-two

problem and see if we get that
same answer, which was 12,

that we got earlier.

So let's plug them in.

We had four people, so n is 4.

We took them two at
a time, so r is 2.

So if you plug in 4
for n and 2 for 4,

you get 4 factorial over the
quantity 4 minus 2 factorial.

Or 4 factorial over
4 minus 2 factorial

is 4 factorial over 2 factorial.

4 factorial's 4 times
3 times 2 times 1.

2 factorial's 2 times 1.

That's 24 over 2, which is 12--

gives us exactly the same
answer we got before.

And just like for the
factorial, there's

a Permutation button on
your calculator as well.

It's above the number one.

That is the permutation button.

It's above the one, so you have
to press Shift to access it.

So if I want to
calculate permutation

of four things taken two at a
time, here are the keystrokes.

Press the four.

Do Shift-1, which takes
you to the permutation.

Then you put in the
two, which is the r.

And in order to get
that to calculate,

you have to press Equal.

For the factorial you
didn't have to press Equal.

You just have one
number, and you

hit the Factorial-Shift
combination.

But when you're
plugging in two numbers,

you have to press Equal
after the second number

to get the calculation
to take place.

And in this case,
you would end up

with 12, which again is the
same answer we got earlier.

So a very important formula--

this is the formula for
the permutation of it

n things taken all r at a time.

Just to practice, if I ask you
to evaluate the permutation

to four things taken two at
a time times the permutations

of eight things
taken four at a time,

you would calculate the
permutation to four combination

two first, which is
four, Shift-1, press two.

Then you want to hit Equal To.

Then you do the
multiplication symbol.

Then you do a permutation
of eight things,

so you press eight.

Then you do Shift-1 again.

Then you press four.

And to make it do the
calculation, you press Equal.

And you'll get 20,160.

So that's just practice
using the calculator

with permutations.

There is one more interesting
fact that I like mentioning.

In that formula above,
if n is equal to r,

you would get the permutations
of n things taken n at a time.

And if you plug that
into the formula above,

you'll get n factorial
over n minus n factorial.

Well, n minus n is 0,
and 0 factorial is 1.

So you get n factorial over
1, which is just n factorial.

So if you don't believe it,
try it on your calculator.

If n and r are the same
number for a permutation,

it's just the same
thing as just doing

the factorial of that number.

You don't have to know that, but
it is a nice, interesting fact.

Now, suppose we've got two
of these four celebrities.

And we're going to choose them
as before on the bicycle built

for two except, this time, we're
not putting them on a bicycle.

We're going to give them a star
on the Hollywood Walk of Fame.

The difference is important.

On the bicycle built for
two, the order mattered.

Sitting on the front seat
was different from sitting

on the back seat.

But if I'm just choosing
two people to get a star,

it doesn't matter what
order I chose them in.

It's an important distinction.

So for instance, if the two
chosen were Han and Yoda,

it doesn't matter if I
do Han/Yoda or Yoda/Han.

In the example on the bicycle
built for two, it did matter.

It decided which one was
on the front of the bike.

But for the stars,
it doesn't matter.

Problems like that, where
the order doesn't matter,

has a special name.

It's called a combination.

Those type problems are
combination problems.

Now, we've already written
out the 12 arrangements

for the bicycle built for two.

And we can see that what
happens is, with the stars,

the order doesn't matter.

So we can actually
look back at the answer

we got for the
bicycle built for two

and just eliminate
the duplicates.

In other words, if
you've got Leia and Han,

and you've got Han and Leia, if
you're just giving them a star,

it doesn't matter what
order they were chosen in.

So really, four is
a duplicate of one,

so you can just get rid of four.

And the same thing is
true going down the list.

If you've got Leia
and Han, that's

the same as Han and Leia,
which is number seven.

So you can get rid of the
seven, and you keep moving down.

Number three is Leia and
Yoda, so you can also

get rid of number 10,
which is Yoda and Leia.

So you don't need
that duplicate.

Giving them stars, the
order doesn't matter.

And as you move down the
list, you've got Luke and Han.

And the number eight
is Han and Luke,

so you can get rid
of number eight.

Number six is Luke and Yoda,
and number 11 is Yoda and Luke,

so you can get rid
of that number 11.

It's a duplicate.

And then, finally,
you've got Han and Yoda,

and number 12 is Yoda and Han.

So you can get rid of number 12.

And that leaves
you simply with six

where you're just
giving them a star.

The order doesn't matter.

Is there a formula for that?

Yes, there is.

It has to do with
the combination.

A combination is a subset
of a set of distinct objects

where order is not important.

If we select r objects
from a set of n objects

without regard to order,
we call this a combination

of n objects taken r at a time.

And it's often written exactly
as for a permutation, except we

use capital C
instead of capital P.

So we could put
the capital C there

with the subscript of n on
the left and r on the right.

Or we can write capital C
with n comma r in parentheses.

So the capital P for
permutation changes

to capital C for combination.

And there is your formula.

Combination of n
things take r at a time

is n factorial in the numerator.

And the denominator is the
product of r factorial and n

minus r factorial.

That's the formula when
order doesn't matter.

If we apply that to that
Hollywood Walk of Fame star

problem, remember, we
got an answer of six

when we did it by brute force.

Let's see if we get the
same answer by plugging in.

In this problem, if we
had four people taken two

at a time where order
doesn't matter--

so it's a combination--

plugging into that
formula for a combination,

we get 4 factorial over the
product of 2 factorial and 4

minus 2 factorial.

Well, that's 4 factorial over
2 factorial times 2 factorial.

4 factorial is 4 times
3 times 2 times 1,

and 2 factorial is 2 times 1.

If you write all that out, 4
times 3 times 2 times 1 is 24.

And 2 times 1 times
2 times 1 is 4.

And 24 divided by 4--

big surprise-- is six.

It's exactly the same answer as
when we did it by brute force.

So these formulas allow
you to get to the count

much faster than having
to do it by brute force.

And again, you won't
be surprised to know

that there is a button on
your calculator for that.

It's above the
number two, and you

have to press the
Shift key to access it.

So if I were to calculate
the combination of four

things taken two at a
time, I would press four,

then I'd do Shift-2.

Then I would press
two, which is the r.

And to calculate it,
you hit Equal To,

and you'll come up
with the number six.

Again, it's the same
answer we got before.

Now, one of the things
you'll come across

when you're doing
these is that, when

you read a problem, sometimes
you have to figure out,

is it a permutation
or is a combination?

For example, suppose there are
10 different books in a box,

and four of them are to
be arranged on a shelf.

How many different
arrangements are possible?

You're taking 10 books.

And you're going to
take four of them

and arrange them on a shelf.

Well, if you're arranging them
on a shelf, by definition,

order matters.

So that makes it a--

right.

It makes it a permutation,
because order matters.

If you get that
wrong, of course,

your answer will be wrong.

You have to know here
that order matters.

You're making arrangements.

By definition, that's order.

So I use the
permutation formula,

plugging 10 in for
n and 4 in for r--

10 books, four at a time.

So that's 10 factorial
over 6 factorial.

And if you do the calculation,
that comes out of a 5,040.

There are 5,040
different arrangements

of 10 books taking four
at a time on a bookshelf.

You could also do this
problem, just FYI,

by using the counting
principal directly.

You could think of it as a
fourth-stage experiment where

you're choosing book number one,
book number two, book number

three, and book number four.

And if you think about
it, you have 10 books.

So for your first stage
of the experiment,

you can choose any of the 10.

But once that first
book has been chosen,

you only have nine left to
choose for the second position,

and eight for the
third position,

and seven for the
fourth position.

And 10 times 9 times 8
times 7, believe it or not,

is also 5,040.

So you could actually
do that problem

by the counting principal.

And you'll find that
to be true quite a lot.

The counting principal is
kind of a bedrock principle.

And sometimes, you
can use it in place

of using one of
these other formulas,

if you really want to.

But it is the same answer,
either way you do it.

If you wanted to do that on your
calculator, you'd type in 10,

hit Shift-1, which is the
permutation combination.

Then you'd press four, and
that would give you 5,040.

Let's do another example.

Suppose we got that 10
different books again.

But this time, Jan wants
to choose four of them

to read while
going on some trip.

Maybe she's taking them
into the car with her.

How many different
arrangements are possible?

Now, how is that different from
putting books on a bookshelf?

If you're putting
books on a bookshelf,

you're arranging the books
in a different order.

Here, Jan doesn't care
which book is chosen first.

She is simply going
to grab four books

and bring them with
her on the trip.

Order doesn't matter at all.

The order she picks those books
is totally irrelevant to Jan.

That means it's a combination.

Again, if you get
that wrong, you're

going to get the answer wrong.

It's a combination because
order does not matter.

Once you figure that out, you
use a combination formula--

again, plugging 10 in
for n and 4 in for r.

And you end up with 10
factorial over the product

of 4 factorial and 6 factorial.

And that comes out to be 210.

And again, if you wanted
to do that calculation

with your calculator--

10 factorial over 6 factorial.

And if you want to do that
calculation in your calculator,

you'd type in 10,
hit Shift-2, which

gives you the combination.

And then you'd do
four, and Equal To,

and that gives you
the same answer.

So even though I'm giving you
the formulas for combinations

of permutations,
really you just need

to know how to do
it on a calculator,

and you can avoid having to
plug into these formulas.

So I'm doing it sort
of just to reinforce

that these formulas really
do give you the right result.

But I'm also pointing out that
the calculator is the fastest

and best way to do it.

Sometimes, a problem requires
the use of more than one

counting tool.

So in order to discuss
problems like that,

it's probably a good time to
go back and remind ourselves

of all the tools we've
developed so far.

Remember, we talked about lists,
tables, and tree diagrams.

We talked about the
counting principle.

We talked about the permutation
formula as a factorial.

Remember, factorial is a special
type of permutation where

you take them all the time.

We've talked about permutations
when you don't take them

all at a time.

You take n objects r at a time.

We've talked about combinations
where order doesn't matter.

So we're building
a tool box of tools

that help us count arrangements
of things and different items

that we want to count.

Let's look at a
particular example.

Suppose a committee
of four people

is chosen from eight
women and eight men.

How many different committees
are possible consisting

of two women and two men?

Well, here, order isn't
important on the committee.

You pick two people to be on
a committee, or four people,

or eight people.

But what order you put
them on the committee

is totally irrelevant,
so we're going

to use combinations
to count this.

But there's really
two different stages.

One stage is to
choose the women,

and the other stage
is to choose the men.

So even though,
for choosing women,

we know order doesn't matter,
we'll use combinations.

And for choosing men,
the order doesn't matter,

so we'll use combinations.

But since it's a
two-stage experiment,

it will be embedded inside
the counting principal.

In other words, the first
stage of the process

is the number of ways
to choose the two women.

The second stage of the
process is the number

of ways to choose the men.

And we find those
two quantities.

We multiply together by
the counting principle.

And that gives us the
total arrangements.

So because the women being
chosen is a combination--

again, because order
doesn't matter--

we use the combination formula.

There are eight women.

Who want to choose two.

So that's eight women,
chosen two at a time.

And it's a combination,
because order doesn't matter.

Then, if you go to the men,
there are also eight men.

It doesn't have to be the same.

But in this case, there
are also eight men.

And you're also
choosing two men.

So that would be the combination
of eight men taken two

at a time.

And the product of
those two calculations

give you the total.

So now we've got to calculate
the combination of eight

things taken two at a time.

Well, that is eight,
Shift-2 for combination.

Press two, because that's the r.

And hit Equal To, and
that gives you 28.

And just by happenstance, notice
that the other combination

is also a combination of eight
things taken two at a time.

So it turns out that
that's also going to be 28.

It doesn't have to be the same.

But in this case, it is.

So there are a total
of 28 times 28, which

is 784 different ways you can
choose two women and two men

to be on the committee if there
are eight women and eight men

to start with.

And notice that the
combination calculation

was used for each stage in
the counting principle, which

had two stages.

Choosing the women
was the first stage.

Choosing the men was
the second stage.

That's an important
point to remember.

This is a process.

This is an example
where you actually

have two of your tools
going on at the same time.

You're using the
combination formula,

but you're also embedding
that inside of the counting

principle.

If five cards are
randomly chosen

from a standard deck of
cards, how many hands

contain four queens?

Now, you've got five cards.

And they told you that you
had to have four queens.

So even though they
didn't come right out

and say it, by
implication, there's

got to be another card
that's not a queen that's

chosen as the fifth card.

You sort of have to read
between the lines there.

You're choosing five cards.

They told you what four
of them have to be.

So there's still a
fifth card involved.

So you're still
choosing five cards.

They just told you what they
wanted four of them to be.

But the basic thing here
is, order isn't important.

So we can use combinations
to count this.

We don't care what order
we draw these cards in.

We're just going to
put them into our hand,

and they're going
to be in our hand.

It doesn't matter what
order we drew them.

So it's a combination.

But like the previous
problem, it really

involves two subtasks.

There are two stages
to this experiment.

One stage is choosing
the four queens,

and the other stage is choosing
the one that wasn't a queen--

the fifth card.

So if you think about
queens versus non-queens,

every standard deck
has four queens.

That means-- 52 minus 4.

That means 48 cards
are not queens.

So I'm thinking the deck is
being partitioned into queens

and not queens.

There are four queens in the
deck, which means the other 48

cards are non-queens.

So embedding this in
the counting principle,

I want to choose
my queens first.

So the first stage
of the process

is the number of ways to
choose the four queens.

Once I've chosen
the four queens,

then I want to choose
one of the other cards.

There are only four
queens in the deck,

so the other card has to be
something other than a queen.

So since it's a combination,
if I'm choosing queens,

there are only four
queens in the deck,

and I want to
choose four queens.

So that's a combination of four
queens taken four at a time.

For the non-queens, again,
order doesn't matter.

But this time, there
are 48 non-queens,

and I only want to
choose one of them.

So that's a combination of 48
things taken one at a time.

And if you do the
calculation for a combination

of four things taken four at
a time, that's four Shift-2.

And then r is also 4,
and you hit Equal To,

and it comes out to be 1.

And if you do combination
of 48 things taken one

at a time, you type
in 48, you Shift-2.

Then you press 1,
which is the r.

And you press Equal To,
and you end up with 48.

And 1 times 48 is 48--

again, another problem where
you're using two of your tools.

You're using the
combination tool inside

of the counting principle tool.

If five cards are
randomly chosen

from a standard deck of
cards, how many hands

contain exactly two
twos and two eights?

It's very similar to
the other problem.

And again, for the
same reason as was

true for the other problem,
order doesn't matter.

So these are going
to be combinations.

And also like the
last problem, it's

actually going to be a counting
principle problem at the base.

But this time, there
are three subtasks,

or there are three stages.

This is a three-stage
experiment.

The first stage, you're
going to choose your twos.

The second stage, you're
going choose your eights.

And if you're going to choose
two twos and two eights,

and there are five
cards, that means

you also got to
choose a fifth card.

So think about this.

You're going to take the number
of ways to choose your twos--

your two twos.

You're going to find the number
of ways to choose your eights.

And then you're going to
find the number of ways

to choose your one more card.

That makes the fifth card.

That can't be a two or
an eight, because we only

want exactly two twos
and exactly two eights.

So if that third card were
either a two or an eight,

we would have more than two
twos or more than two eights.

So going back,
choosing your twos

is a combination, because
order doesn't matter.

There are four twos
in the deck, and we

want to choose two of them.

So that's a combination of four
things taken two at a time.

Then we want to move the eights.

Again, there four
eights in the deck,

and we want to
choose two of them.

So that's also a combination
of four things taken two

at a time.

Now, the last one--
think about it.

You want it to not be
an eight, not be a two.

Well, there are four twos,
and there are four eights.

So for that fifth card,
order still doesn't matter.

But there are only 44 of them,
because 52 minus 4 minus 4--

take away the four twos
and the four eights--

leave only 44 cards.

So that last card has to
come from the 44 cards that

aren't twos or eights.

And when you do the
calculation, the combination

of four things
taken two at a time

is four, Shift-2, two, Equals.

And that comes out to be six.

Well, that actually--
for the eights,

it comes out to be exactly
the same-- four, Shift-2,

two, Equals.

That's also six, because it is
exactly the same combination.

And then, finally, combination
of 44 things taken one

at a time-- that's 44,
Shift-2, one, Equals.

And that comes out to be 44.

And then, by the
counting principal,

you multiply those
numbers together--

6 times 6 times 44--

and you end up with 1,584.

So if you choose five cards
randomly from a standard deck

of cards, there are
1,584 different hands

that contain exactly
two twos and two eights.

Let's look at a couple
of extra problems.

Suppose I ask you to
evaluate this product.

It's the product of
two combinations.

The first combination is eight
things chosen five at a time.

The second combination is nine
things taken four at a time.

So the easiest way is
to use your calculator.

And remember that the Shift-2
combination is the combination

key.

So you type in eight,
Shift-2, five, hit Equal To--

that gives you the first
combination value--

then times.

Now you start the
second combination.

That will be nine,
Shift-2, four.

When you hit Equal, you will
get the final answer of 7,056.

How about this?

The planning committee
for a class reunion

must elect a president,
secretary, and treasurer.

In how many ways can
this be accomplished

if there are 25 members?

I'm actually going to show you a
couple of ways you can do this.

First of all, it
is a permutation,

because order matters.

The president,
secretary, and treasurer

is different than if
you switch positions.

That's a different arrangement.

So order matters.

It's a permutation.

In fact, it's
equivalent to a problem

we've done earlier where we
were taking a group of people,

and pulling some out, and
lining them up for photos.

And we wanted to know how
many ways we could do that--

same thing here with the
president, secretary,

and treasurer.

So this is a permutation of 25
things taken three at a time.

With the calculator and
the Shift-1 keystroke

that gives you the
permutation, we

type in 25 Shift-1 for
permutation, then enter tree.

And when you hit Equal
To, you get 13,800.

So there are 13,800 ways
you could take 25 members

and make them
president, secretary,

and treasurer of the group.

But there is a second way
you could look at this.

You could think of it as
a straightforward counting

principle problem.

If you did it that way, you
could think of the number

of ways to make these selections
as being a three-stage

process--

the number of ways to choose
the president, times the number

of ways to choose the secretary,
times the number of ways

to choose a treasurer.

It's just a standard
counting principal problem.

If you do it that
way, obviously,

when you're choosing
a president,

you can choose any
of the 25 members.

But the president can't also
be secretary or treasurer.

So when you go to
the next position,

you have to eliminate the
person you chose for president.

So you only have 24 members to
choose from for the secretary.

And then, by the time you get
to the treasurer position,

you've already
chosen two people.

So those two people are out of
the running to be treasurer.

That leaves only 23 people.

And if you multiply
25 times 24 times

23, which is what the counting
principle tells you to do,

you'll get the same 13,800 ways.

So either way is
fine-- whichever

way makes most sense to you.

Along the way, I've been
building up our tool box.

I want to go through
this one more time.

We started off with a list,
and tables, and tree diagrams.

Then we moved to the
counting principle.

Then we talked
about n factorial,

which is a special permutation
where you take everything

all at once, like n
things n at a time.

Then you have the permutation,
where you have n objects,

but you're only taking
them r at a time.

Then you move to
the combinations,

where you're taking n
objects r at a time,

but order doesn't matter.

That's where we stand
with our tool box

for doing these
type of problems.