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