Like when you expand (x+y)²= (1,1)x² + (2,1)xy + (1,1)y². Its obvious in this case but when the calculation gets complicated its hard to fit in your mind
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Google and Wiki can cover. It follows a pattern.
[https://en.wikipedia.org/wiki/Binomial_theorem](https://en.wikipedia.org/wiki/Binomial_theorem)
Imagine you have a magic box with two switches on it, one labeled “x” and the other “y”. When you flip the switches, something amazing happens: the box gives you either “x” or “y”. Now, let’s see what happens when we use this magic box twice.
First, you flip the “x” switch, and the box gives you “x”. Then, you flip the “y” switch, and the box gives you “y”. So, the box gave you “xy”.
Now, let’s reverse the order. First, you flip the “y” switch, and the box gives you “y”. Then, you flip the “x” switch, and the box gives you “x”. So, the box gave you “yx”.
But wait! In math, the order doesn’t matter when we’re just adding things up. “xy” and “yx” are like two friends who like to stand together, so we can say they’re the same thing. That’s why we call them “xy” and don’t worry about the order.
Now, let’s put all this together. When you multiply (x + y) by itself, you’re using the magic box twice. So, you can get “xy” or “yx” as we talked about earlier. And that’s why you have the “xy” term in the binomial expansion.
The (1,1) thing you mentioned is like the number of ways you can choose “x” and “y” from the magic box. The numbers in the parentheses show how many times each switch was flipped. For example, (2,1) means you flipped the “x” switch twice and the “y” switch once, which gives you “xxy”. It’s like saying you got “x” twice and “y” once from the magic box.
So, when things get more complicated, just think of the magic box with its switches, and remember that “xy” and “yx” are like buddies who are the same. That should help you understand how combination works in the binomial expansion!
You can think of expanding binomials in terms of a probability tree/tree diagram.
If you have:
> (x + y)^n
if you write it out in full, you get *n* brackets of *(x + y)*, and when you multiply it you need to pick one term (*x* or *y*) from each bracket.
So if you want the *x^r y^(n-r)* term, you need to know how many ways you have got of picking *r* “x”s and *(n-r)* “y”s from a *n* things. Because you only have 2 options (x or y) that is the same as picking just the *r* “x”s (which is why the choose functions are symmetric – *nCr = nC(n-r)* – picking the *r* “x”s is the same as picking the other *(n-r)* “y”s).
And this is just the “choose” function; *nCr* gives you how many ways are there of picking “r” things from “n” options.
For example, if we have *(x+y)^4* we get:
> (x+y)(x+y)(x+y)(x+y)
Our choices are to pick the “x” from all for brackets (only one way to do that, “xxxx”), to pick 3 “x”s and one “y” (four ways for that: xxxy, xxyx, xyxx, yxxx), two of each (6 ways for this: xxyy xyxy xyyx yxxy yxyx yyxx), and then by symmetry, 4 ways to pick one “x” and three “y”s (the reverse of three “x”s), and finally one way to pick all four “y”s (yyyy). So we get:
> (x+y)^4 = x^4 + 4 x^3 y + 6 x^2 y^2 + 4 x y^3 + y^4
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If you want to know why the *nCr* formula works…
For the first of your *r* “x”s you have *n* to choose from (so n options, could be the first bracket, second bracket all the way to the final bracket). For the next you now have *n-1* options (as one bracket is already fixed). Then *n-2* for the third, and so on, all the way to the *r*th “x” for which you will have *n-r+1* options. Giving you:
> n (n-1) (n-2) … (n-r+1)
But the order in which you pick doesn’t matter. Picking the first “x” in the first bracket and the second “x” in the second bracket gives you the same result as picking the first “x” in the second bracket and the second “x” in the first bracket. For each option we need divide by the number of ways of re-arranging all those “x”s. Which is just how many ways we can place each “x” in any given pattern. For example, in xxxy, the first “x” can go in position 1, 2 or 3. Once we’ve fixed the first, the second “x” has two options. The third “x” is then fixed. Generalising to *n*, our first “x” has *r* choices (as there are *r* slots to fill), the next has *(r-1)* and so on, all the way down to 1. Meaning we have to divide by:
> r (r-1) (r-2) … 1
Giving us our formula:
> nCr = n (n-1) (n-2) … (n-r+1) / 1 2 3 … (r-1) r = n! / (n-r)! r!
Also gives a neat way to work out the next nCr value. nC0 = 1 for all n. nC1 = n for all n. Then for the next you multiply by the next number down (so *(n-1)*) and divide by the next number up (so 2). And so on.
For example, the “10Cr”s will give us:
> 1, 10, (10 * 9 / 2) = 45, (45 * 8 / 3) = 120, (120 * 7 / 4) = 210, (210 * 6 / 5) = 252, (252 * 5 / 6) = 210 … (and we can see how the pattern repeats.
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If you want to explore this further and check you really understand it, try multiplying out a “trinomial”, something like (x + y + z)^n
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