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Current time:0:00Total duration:4:06

CCSS.Math:

what I want to do in this video is further connect our understanding of the binomial theorem to combinatorics to Pascal's triangle and so just to review the ideas again if we're taking X plus y to the third power and I'm just using this as an example that's a little bit easy to get around get our heads around that's essentially taking three equivalent expressions and multiplying them by each other you're taking X plus y times X plus y times X plus y so let's call this the first X plus y the second X plus y and the third X plus y and so if you're looking at the expansion of it and you're saying well how many ways how many ways if we're how many ways can we construct XY squared or another way to think about it is you have these three expressions you're going to choose two of them to contribute a Y and take the product so for example you could pick expression 1 and expression 2 to contribute the Y you could pick expression 1 and expression 3 to contribute the Y or you could take expression to an expression 3 to contribute the Y and of course the other expression is what's going to contribute to the X so you have 3 expressions you have 3 expressions and you're going to choose to you're going to choose 2 to contribute the Y and so that's why you have this combinatorics idea it's 3 choose 2 it's the exact same mathematical problem if you have 3 friends and you're choosing to to ride in a car with you you don't care what seat they are going to sit in you're just saying which two friends get to ride in my two-seater or get to ride that what are the two friends that I am going to pick same way here out of the 3 friends which are the two that I'm going to pick to contribute a Y to this product and then the third is going to contribute an X now let's go to Pascal's triangle and hopefully see that this is actually a very similar idea so we could go to the same term so the XY squared term so that's right over there and Pascal's triangle I always visualize it as a map so this is a node in the map and I think well what are the different ways that I can get to this node on the map well I could have a I could have a y squared I could have a Y squared and then multiply by an X and since we're multiplying an X that's why I made this line orange or I could have an X Y and I could multiply by a blue so there's there's two immediate things above it that I can get to them but there's two ways to get to this one there's one way to get to this one and so the combined ways to get to this one is two plus one is equal to three and just to make it the connection between what we just said what's really going on here is each at each of these nodes as we pick a path we're essentially saying are we picking an X or Y from each of these expressions so we number the expressions let's number them here so let's say this is expression 1 this is expression 2 this is expression 3 so we said that there's three paths that we could take us here so let's let's mark them out so there's there's this path there's this path right over here that gets us there and that's equivalent to picking the X from expression 1 notice we pick the left hand side then we go to the right hand side we pick the Y from expression 2 and then we pick the Y the Y from expression 3 the last two we pick the right path we picked the blue path so that's essentially one way that we could choose to wise out of these three expressions once again one way to choose two wise out of three expressions but we don't care about just that one way we care about all of the different paths so this was one way another way would be to go like this so to pick a Y from the first one an X from the second one and a Y from the third one or pick a Y from the first expression a Y from the second expression and an X from the third expression so it's really at the at kind of a fundamental mathematical level it's solving the exact same problem so hopefully this gives you a little intuition on why Pascal's triangle is connected to combinatorics and why they're both useful for finding binomial expansions