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Binomial expansion & combinatorics

Video transcript
Voiceover:What I want to do in this video is hopefully give more intuition as to why the binomial theorem or the binomial formula involves combinatorics. Let's just think about what this expansion would be. I'm just using a particular example that's pretty simple, x plus y to the third power which is x plus y, times x plus y, times x plus y. We already know what the terms look like, we could pick an x from each of these. If we pick exactly an x from each of these that's going to be our x to the third term, and we essentially picked no y [zen] because we picked an x from each of this when we multiplied out to get this term. How many ways are there to construct that? How many ways are there, if we're picking from three things. We're picking from three things, how many ways can we choose exactly zero y's of not picking a y? There's only one way to do that, there's only one possible way of doing that. By not picking a y from each of them or you can say by picking an x from each of them. The coefficient right over here is one x to the third. There's only one way, when you take this product out, only one of the terms initially is going to have an x to the third in it. Now, what about the next term? So plus. Now we're going to pick from three buckets but we want to pick one y. Another way of thinking about it you have three people or we have three people, this person, this person, and this person. One of them you're going to choose to be I guess be your friend that's another way of saying which of these are you going to pick the y from? This is the exact same thing as a combinatorics problem, from three things you are choosing one of them to be y. Of course three choose one, this is equal to three. If you're picking one y that means you're picking two x's, and you're taking the product of everything so it's x squared times y to the one and we could keep doing that logic, so plus. Now we're going to think about the situation where we are picking two y's out of three things. It's x to the first, y squared. We're starting with three things and we're going to pick two to two of this y's so it could be this y and this y to construct the product y. It could be y times y times that x or it could be y times y, times that x or it could be y times x, times y. This is three choose two, out of three things what are the different ways, what are the combinations of picking two things. If you have three friends how many combinations can you put two of them in your two seater car where you don't care about who sits in which seat. You're just thinking about how many different combinations can you pick from that and once again this evaluates to three and then finally, how many different ways from a set of three things, from three different things, so from each of these expressions, how many ways can you pick exactly three y's? Well there's only one way to do it, you pick this y, this y and that y. If you're picking three y's that means you picked zero x's and you have picked three y's. That's why we're dealing with combinatorics here. Three choose three. From three expressions, your three binomials, right over here that you are taking the product from you need to choose three y's, so you have to pick the second term in all of them. There's only one way to do that, here you want to pick the second term, the y term, in two of them, exactly two of them. Here you want to pick the y term in exactly one of them, there's one way to do this, three ways to do this, so there's three ways to do this, three ways to do this, one way to do that, one way to do that. This is one, this is three, when you evaluate, when you apply the formula for three choose and choose k, three and this is one.