Binomial Theorem (part 2)

Binomial Theorem and Pascal's Triangle

In the last video we saw that
if you wanted to take a plus b

to the nth power, and if n is
larger than, really, 2-- but

really, especially larger than
3-- it is very tedious to

multiply it out, essentially
using the distributive

property, or doing polynomial
multiplication, or FOIL, or

however you learned it.

It is extremely,
extremely tedious.

And then we learned that the
binomial theorem, which said

this, that that is equal to
the sum from k is equal to 0,

to n of n choose k-- right?

Where that was what we learned
in combinatorics as the

binomial coefficient.

And that's why it's called a
binomial coefficient, because

it's actually the coefficient
of the binomial theorem.-- Of

x to the n minus k-- oh,
sorry, I keep writing x.

Let me undo that.

Edit, undo.

Edit, undo.

Oh, it's taking too long.

Oh, let me just-- oh no, that's
not what I wanted to do.

Let me erase it.

OK.

I keep writing x.

It could be an x, but then
this would have to be

an x here, as well.

Maybe I should do that.

--Of a to the n minus
k, times b to the k.

So each term-- you know, the n
stays constant-- but each term,

you start at k equals 0 and
you keep incrementing up.

And we did an example to solve
a plus b to the fourth power

in the previous video.

And as you saw, that was
tedious, but less tedious than

actually multiplying it out.

And if you get really fast
at computing n choose k

for different ns and k, it
could be reasonably fast.

So what I want to do is, I'm
going to show you a slightly

faster method than
what we just did.

Kind of a faster way to compute
the binomial coefficients.

And then after that, I'm going
to show you a super fast way

that, short of memorizing the
coefficients-- which I actually

know some people who've done
that-- is a pretty amazing way

to essentially multiply
out any binomial.

So what's my pseudo
fast way of doing it?

Well, I hinted in the last
presentation that those

coefficients were actually
terms of a Pascal triangle.

So what's a Pascal triangle?

So if we start off with a 1 and
then you just go-- actually,

let me do it-- well, yeah,
let me do it right here.

And then, actually let
me start with two 1s.

And what you do is, you take
the sum of both of these.

So that's a 2.

And then you bring down
a 1, to the left and

the right hand side.

And notice, these are
the coefficients of

a plus b squared.

And these are the
coefficients of a plus b.

You could say a
plus b to the one.

1a plus 1b.

This is a squared-- so you
could rewrite, a plus b

squared. is 1a squared
plus 2ab plus 1b squared.

So these are the coefficients
of a plus b squared.

Let me arbitrarily
switch colors.

And so, 1 plus 2 is 3.

2 plus 1 is 3.

Bring down the 1.

Bring down the 1.

And now we have the
coefficients for a

plus b to the third.

Which we computed in
the-- that was the very

first thing we did.

We actually multiplied it out.

And we just know the pattern.

The first coefficient is 1.

So it's 1a to the third, b to
the zero-- so we don't have

to write the b-- plus 3.

We just decrement
this exponent 1.

3a squared b plus 3ab squared,
and then plus 1a to the zero--

which is just 1-- b cubed.

So that was pretty fast.

And we can keep going down
the Pascal triangle.

So let's do the next one.

So we can bring down a 1.

1 plus 3 is 4.

3 plus 3 is 6.

And this is neat.

I mean, just very simply, you
can actually generate binomial

coefficients without
having to compute them.

Very simple, I guess you
could call it an algorithm.

Or drawing.

And it's symmetric, just as
you would expect it, right?

Because you could
easily switch b and a.

a plus b is the same thing
as b plus a, so you

should essentially
get the same answer.

And so, we just-- very quickly
we figured out the binomial

coefficients for a
plus b to the fourth.

Which was a lot faster than
we did in the last example.

a plus b to the fourth.

So then we-- I think
you get the point.

But, so it's 1-- let me write
in a different color-- 1a

to the fourth, b to the
zero, plus 4a cubed b

squared-- b to the one.

Plus 6a squared b squared.

Which makes sense that, you
know, this is a middle number

and they both-- and a and b
have the same exponent

at this point.

And then plus 4a-- we
decrement it-- b cubed.

Plus b to the fourth.

1b to the fourth, right? a
to the zero, so that's what

we didn't write there.

So 1b to the fourth.

And that was very fast compared
to what we had to do at

the end of the last video.

We could just keep going.

You know, for 5.

So 1 plus 4 is five.

4 plus 6 is 10.

6 plus 4 is 10.

4 plus 1 is 5.

Bring down the 1.

So these are the coefficients
for the expansion of a

plus b to the fifth power.

And so this is a reasonably
fast way of doing it, although

it can get-- one, it will
take a lot of space.

And it can work reasonably
well, you know, for up to a

power of eight or nine or ten.

Even then it starts to get
pretty big and cumbersome.

But you know, for powers
up to seven or eight or

nine, you could do this.

You could draw it out really
fast and do this, and it's

probably faster than actually
computing each of the

binomial coefficients.

Although you might be pretty
fast at computing n choose k,

in which you don't
have to do this.

So with that out of the way,
let me show you an even faster

way of doing it, short
of memorizing it.

And this will allow you to
really calculate a plus b

to the nth-- you know, to
the twentieth power--

almost in your head.

Depending on how good you are
at arithmetic in your head.

So here is the trick.

And I encourage you to
experiment for why it

works, but it does work.

And I mean it's
not even a trick.

It's just-- and this Pascal's
triangle isn't even a trick.

Pascal's triangle is just an
alternative way to generate

binomial coefficients, and what
I'm about to show you is just

another way of essentially
generating the binomial

coefficients.

Although it's probably a
faster way to compute them.

And it's a good project for you
to think about why this works.

So I'm just going to start
with a very concrete example.

Instead of a plus b, let
me just do x plus y.

Just because you might
see the binomial theorem

written that way.

So let's say x plus y
to the tenth power.

This would take me all
day if I was to actually

multiply it out.

It would take me probably 20
to 30 minutes, without

making careless mistakes, to
actually figure out all the

binomial coefficients.

Maybe not that long, but
it would take me a while.

And to draw Pascal's triangle
would fill up a whole page,

and I'd still probably
make a careless mistake.

So how can I do this?

So what you do is-- so
one thing you know.

This is going to have
11 terms, right?

Because you're going to
start with x to the

tenth, y to the zero.

And you're going to go all
the way to y to the tenth.

So if you start at 0 and you
go to 10, that's 11 terms.

So it has 11 terms.

What I want you to do
is just write down the

first, just the numbers.

You know, you can almost
count the terms.

You don't have to go all the
way to 11, and I'll show you.

But actually, let's write
all the way to 11.

So 1, 2, 3, 4, 5, 6,
7, 8, 9, 10, 11.

Just squeezed it in.

And you'll see, you don't
actually have to go

all the way to 11.

You could probably
just stop at 6.

So here's the trick.

We know that the first
term is going to be x

to the tenth, right?

We know that x to the tenth.

Actually, we know that it's
going to be x to the tenth.

The second term is going
to be x to the ninth.

Then it's going to
be x to the eighth.

It's going to be x
to the seventh.

A little tedious.
x to the sixth.

x to the fifth.

x to the fourth.

x to the third.

x squared.

x.

And then there's going to be
x to the zero, or just one.

Let me just do the y's.

So this was x to the tenth.

That's not bright
enough, this color.

So this is y to the zero.

So we don't have to
write it there.

But then we have a y.

y to the first.

y squared.

y to the third.

y to the fourth.

y to the fifth.

Which makes sense, this
is the middle term.

y to the sixth.

y to the seventh.

y to the eighth.

y to the ninth.

I don't want you to get
confused, each of these

is a separate term.

I don't want you to think
I'm multiplying them all.

And then, we just have to
figure out the coefficients

on each of these terms.

Those are divider lines
I attempted to draw.

I wasn't trying to
confuse you more.

I just wanted to-- because they
seemed to be running together,

each of the terms
that I'm writing.

But I think you know
what I'm doing.

So now we have to figure
out the coefficients.

And then this is the neat part.

So we know that the coefficient
on the first term-- let me draw

a dividing line here and here--
the coefficient on the first

term is always 1, right?

So, the coefficient is 1.

So the coefficient on the
second term is going to be the

exponent on the first term
times its coefficient-- so 10

times 1-- divided by
the term that it is.

So it's going to be 10
times 1, divided by 1.

So it's going to be 10.

The third term's coefficient
is going to be the

exponent on the x, right?

So it's 9 times its
coefficient-- which is 10-- so

it's going to be 9 times 10--
divided by the term it is.

So it's going to be 9 times its
coefficient, 10, divided by 2.

So what's 9 times 10?

That's forty-- it's 90
divided by 2 which is 45.

And you keep going.

The fourth term is going to be
the third term's exponent-- so

it's going to be 8 times-- let
me write this down in a

different color-- it's going to
be 8 times its coefficient,

times 45 divided by
which term it is.

So it's the third term.

Divided by 3.

Well, that's just 8 times 15.

And we'll see,
that's 80 plus 40.

So that's equal to 120.

So that is the fourth term.

And so then let me just
draw these dividers.

I know it's getting a
little complicated.

And I'm writing it all out like
this, but if you practice

this enough, you can actually
just write it straight out.

And so the fifth term.

What is the fifth term?

Well, you take the
exponent on the x.

So 7 times the fourth
term's coefficient--

times 120-- divided by 4.

Right?

Divided by the
previous term, by 4.

Well, that's just 7 times
30, which equals 210.

That's the fifth coefficient.

What's the sixth coefficient?

Well, it's 6 times-- you
know the exponent on

the x-- times 210.

Times its coefficient-- times
the fifth term's coefficient--

divided by 5, for
the fifth term.

Well, 5 goes into
210 how many times?

42 times, right?

So it's 6 times 42--
that's 240 plus 12.

That's 252.

And then once you're at the
middle point-- the sixth term

is the middle term-- you'll
see that, you know, you start

going back the other way.

And we learned one from the
Pascal's triangle, or even the

definition of the binomial
theorem, that the

coefficients are symmetric.

So we know that the next one is
going to be the same-- this was

the middle one, right?-- so we
know the next one is

going to be 210.

And you could calculate it
using the same system.

This is just a quick
way of doing it.

This one's going to be 120.

This one's going to be 45.

And this one is going to be
the tenth coefficient--

this is going to be 10.

And then of course the last
coefficient is just 1.

1y to the tenth power.

So if I were to write this out,
the answer is-- and if you

practice this, you'll find that
you can do quite fast-- it's x

to the tenth, plus 10x to the
ninth y, plus 45x to the eighth

y squared, plus 120x to the
seventh y to the third, plus

210x to the sixth y to the
fourth, plus 252-- already at

the middle term-- x to the
fifth y to the fifth, plus 210x

to the fourth y to the sixth--
I'm running out of space.

But you can you can hopefully
extrapolate what I'm doing,

and it makes sense to you.

And hopefully you have an
appreciation that if you

actually had to multiply out x
plus y to the tenth, it would

have taken you all day.

Maybe I'll do one more video
with a smaller example to show

you that it's a little less
complicated when you do, say,

x plus y to the sixth power.

See you soon.