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## Algebra (all content)

### Course: Algebra (all content) > Unit 10

Lesson 26: Understanding the binomial theorem# Binomial expansion & combinatorics (old)

An old video of Sal explaining why we use the combinatorial formula for (n choose k) to expand binomial expressions. Created by Sal Khan.

## Want to join the conversation?

- i did not get the part of having 3 unique permutations and 3 identical combinations @11:20.

what do both of them mean?thanks(4 votes)- Good question, here Sal coloured the three a's and b's differently to make them distinct. Now, among the different arrangements, we've got aab, aba and baa. As we know, in permutations, order matters, so these three are 3 unique permutations. However, for combinations, order doesn't matter, so all these three correspond to the identical combination a^2 b(1 vote)

- Forgive me for asking, what would you guys think I ought to look over if I'm having a slight bit of trouble fully understanding this video?(3 votes)
- Thanks Dylan. I sussed it a while ago now, i was having issues knowing where to start mechanically with all this but i seem to have clicked with the formula. thanks again(2 votes)

- at11:26wouldnt it be 3 different combinations and not indentical combinations as it was mentioned as the color of b and a in each of the possibility of a^2b is different? if not then why is all the different permutations showed in the case of a^2b but not in a^3 or b^3?(2 votes)
- Yes, you're right. He made a minor mistake there.(1 vote)

- how do you do binomial expansions when the powers are rational and not integer?(2 votes)
- what do we call (a+b+c)^n?

is there a formal way of expanding this term?(1 vote)- Yes there is, actually.

http://en.wikipedia.org/wiki/Trinomial_expansion

Let's expand (a + b + c)².

= (a + b + c)(a + b + c)

= a(a + b + c) + b(a+ b + c) + c(a + b + c)

= a² + ab + ac + ab + b² + bc + ac + bc + c²

= a² + b² + c² + 2ab + 2ac + 2bc

(a + b + c)^3

= (a + b + c)(a + b + c)²

= (a + b + c)(a² + b² + c² + 2ab + 2ac + 2bc)

= a^3 + ab² + ac² + 2a²b + 2a²c + 2abc

+ a²b + b^3 + bc² + 2ab² + 2abc + 2b²c

+ a²c + b²c + c^3 + 2abc + 2ac² + 2bc²

= a^3 + b^3 + c^3 + 3a²b + 3a²c + 3ab² + 3b²c + 3ac² + 3bc² + 6abc(3 votes)

- At8:34shouldn't 3a^2b = (3 choose 1)3a^2b?(2 votes)
- So in cases of binomial expansion, we treat permutations and combinations as same things?? Because in expansion of (a+b)^3, there are 3 terms which are a*a*b = a^2b (a squared into b); they are essentially 3 permutations, but just one unique combination.(2 votes)
- Why is it a combination instead of a permutation, if aba, baa and aab are not the same thing? I know it has something to do with the a's and b's being interchangeable, but what exactly is going on there?(1 vote)
- If you think about it, aba, baa and aab
*are*the same thing.

Think of it like this, if a=3 and b=2, then aba=3*2*3=18, baa=2*3*3=18 and aab=3*3*2=18.

Because of this, the order of 'selection' does not matter and therefore it is a combination rather than a permutation.

Hopefully I answered your question there.(3 votes)

- why some videos of sal are poor in quality and look?(1 vote)
- They were made years ago compared to his more recent videos.(3 votes)

- i watched the videos , and understand the concept , but could you explain a question if we had three variables in it, like (a+b+c)^3 , how would we do that question using binomial combinations/theorem?(1 vote)

## Video transcript

In this video, I'm going to
attempt to give you an intuition behind why
multiplying binomials involve combinatorics Why we actually have the
binomial coefficients in there at all. And I'm going to do
multiple colors. The colors will actually be
non-arbitrary this time. Just to give you an intuition. So let's multiply a plus
b to the third power. Well, a plus b to the third
power, that's a plus b times-- and I'm going to keep
switching colors. You're going to have to bear
with me, but it should hopefully be fruitful.-- Times
a plus b, times-- let me pick an appropriately different
color, maybe a blue-- times a plus b. And let's do this as a
distributive property. This equals a times-- go
back in green-- a plus b. With a different green. Want to make sure I use the
right green, just because the colors matter this time. a plus b-- this is tedious,
but it's worth it-- plus b times a plus b. And then all of that
times a plus b again. And let's multiply
this inside part. I'm just going to keep
it and multiply it out. So it's going to be a times
green a-- we know they're all the same a-- plus-- I should do
the pluses in a neutral color, but it's OK-- plus a times--
you might be finding this tedious, but it's going to pay
off in the end-- a times b. Then we have plus b times a,
plus yellow b times green b. And then all of that--
we're almost there. We're almost there. All of that times a plus b. And it's essentially,
we're going to multiply a times everything. This blue a times all of
this, and plus this blue b times all of this. So let's multiply the blue
a times all of this. So the first term will
be yellow a, green a, and then blue a. So it'll be a, a, a. Oh it's a different blue, but
I think you get the point. Plus this times blue a. So yellow a, green b, and then
blue a, plus yellow b, green a, blue a-- hopefully I'm
not confusing you-- baa. b, a, a. We're almost there. Plus yellow b times
green b times blue a. So we did all the
blue a's, finally. There's a blue a. Plus--. Now we're going to do the
blue b times everything. So it's yellow a times green a. Right? Yellow a, green a, then blue b,
times green a, times blue b-- almost there, I know this is
tedious-- times blue b, plus yellow a. Intuition doesn't
come easy, though. Yellow a-- so we're on
this term-- yellow a, green b, blue b. So green b times blue b. Now we're at yellow b-- the
good thing about the colors is it's easy to keep
track of where we are. Plus yellow b times
green a times blue b. And then we're at the last one. Plus yellow b, green
b, times blue b. So this is the expansion
of a plus b to the third power, right? We haven't simplified
it at all, and I did that for a reason. Because you see that every term
here-- what's happening here? Every term has exactly
one-- it's 3 numbers being multiplied, right? Every term is 3 numbers
being multiplied. And it's one of, you know, the
yellow number comes from the first-- from this
yellow a plus b. The green number-- the
middle number-- comes from this middle a plus b. And then the blue number comes
from this right hand a plus b. And you saw me, I went all
the way through it, right? So hopefully you
believe this point. So let's think about it a
couple of different ways. To generate each of these terms
in the expansion of a plus b to the third, we're picking either
a or b from-- from the yellow a plus b, we're picking
either a or b. Right? We picked an a here,
we picked an a here. We picked a b here, a b here,
an a here, an a here, a b--. And the group from the green
a plus b, we're picking either an a or a b. And then from the blue a
plus b, we're picking either an a or a b. Right? So essentially the expansion,
if you think about it, this expansion-- we've essentially
done every way of choosing 3 different things. Every way of picking either
an a or a b, from these 3 different terms. And that ends up with
these-- what is this? 1, 2, 3, 4, 5, 6, 7, 8 terms. Right? Now let's make it a little
bit-- let's give you a little bit more intuition
of what's going on. And I think if you're starting
to realize why this is dealing with permutations and
combinations, in particular. Once we simplify
it, what happens? This is a to the cubed, right? That's the only a cubed term. This is a squared b. What are the other
a squared b terms? Let's see, a squared b. This is also a squared b. So let me write down
all the a squared. So let's see-- let's
see how many a squared b terms there are. I'll do it in a neutral color. So this is a squared b. This is ba squared, but
that's also a squared b. What's another a squared b? This is also a squared b,
right? a times a times b. So there was 3 ways
to get a squared b. And that's why when we
eventually write the expansion, it's going to-- we know that
the coefficient in front of it is 3a squared b, right? The coefficient on the a
squared b term, when we actually multiply it out. And we've done that several
times already, when we did the binomial theorem, which we
took to the third power. So where did this 3 come from,
and why is that the same thing as when we learned the
definition of the binomial theorem? Why is, you know, does it just
happen to be the case that that's the same thing as 3
choose 2 times a squared b? Well, no. Think of it this way. We already know that every
term here, every term in the expansion, we're essentially
picking either an a or a b from each of these, right? We have to pick one term
from each of these. So the way you could think of
it, for a squared b-- to get a squared b-- we have to
essentially say, well, how many combinations? And that's the key word. How many combinations are
there, where out of these 3 a plus b terms, I'm
choosing the a term? I'm choosing 2 a terms. Because to get a squared, I
have to pick an a term twice. And so that's where I have
to pick 2 a terms out of the three times I pick. So I'm picking three times. Two of the times, I'm
picking an a term. So out of three times,
I'm choosing 2. And that's where 3 choose
2 comes from, for the a squared b term. And so you could, for the ab
squared term, you could say, well, I'm picking an a once. How many ways are there to
choose a once when I'm taking from 3 things? So it could be 3 choose 1. But that's also the same thing,
or should be equal to-- you could also say, well, I'm
picking-- how many ways are there to pick b twice if
I'm picking three times? Well, if it's a b squared,
I'm picking b twice. So that should be equal to
3 choose 2 ab squared. And if you work these out, you
will find that, yes, these both turn out to be 3. And actually, that's why
there's some symmetry there, and the combinations
all work out. But hopefully, that's
giving you an intuition. Essentially, when you're
doing the-- so the binomial expansion of this--. Let me just rewrite it again. You know, this is a
plus b to the third. It's 3 choose 0 of a cubed b to
the zero, plus 3 choose 1 of a squared b to the
one, we could say. Plus 3 choose 2 of ab squared,
plus 3 choose 3 of a to the zero, b cubed. So what's this saying? What's 3 choose 3 saying? How many ways-- if I'm choosing
from 3 different things-- how many ways can I pick
exactly 3 b's? That's how you can view it. How many ways can I pick 3 b's? I'm either picking
an a or a b, right? You could say it's either a
heads or tails, or you know, red or black or white,
but it's either a or b. How many ways can I choose b
three times from 3 things? Well, when you evaluate this,
you get this to b one, and that makes sense. Because it's 1b cubed. And similarly, I mean, you
could view this as how many ways, when I'm picking
out of 3 things, can I pick exactly 0 b's? That's 3 choose [? b. ?] How many ways can I
pick exactly 0 b's? Well, that's the same thing
as how many ways can I pick exactly 3 a's? This is also 1. There's only 1 way to do it. And that's the way. Once again, there's only
1 way to do this one, and this was the way. There's 3 ways to
get a squared b. There's 3 combinations. There's 3 possible-- actually,
well, unique permutations. But they're all the
same combination. So there's 3 identical
combinations for picking 2 a's and a b, and those are this
one, this one, and this one. So hopefully, I
didn't confuse you. And hopefully at minimum, that
gives you a glancing intuition of why combinations are even
involved in the binomial theorem, or whether they're
even involved when you're expanding a binomial
to some power. And at best, I really hope that
I've given you a deep intuition for why this happens. I will see you in
the next video.