How to understand Bionomial Theroem and the Recursion Rule?
Firstly, this is probably best to ask on the math stackexchange, but I'll have a crack at it anyway.

Let's start by establishing something you already know: $(a+b)^n$ will expands to a sequence of terms with general form $c_ia^xb^y$ such that $n=x+y$. We also know that the expansion process hits every positive integer combination of $x$ and $y$ that sum to $n$ (of course assuming $n$ is a positive integer).
I'd also like to mention that you can notate the expansion entirely with $choose$ coefficients: the first and last terms have coefficients given by $n\choose{0}$ and $n\choose{n}$. Below, we'll see why this is important.

Let's do an example where $n=4$.
Let's start by solving for the first term where $x=n$ and $y=0$. We can ask ourselves: Given the first $(a+b)$ term, how many paths can we take, by multiplying out the rest of the terms $(a+b)(a+b)(a+b)(a+b)$, to reach a term with $x=n$ and $y=0$?

For the first term, it is of course one path: you need to multiply the first $a$ with the second and third and fourth: $aaaa$. Given only one path, the first coefficient is $c_1=1$.

Now let's solve for the second term where $x=n-1$ and $y=1$. We can ask ourselves the previous question with the modified values of $x$ and $y$. We find, stepping from one $(a+b)$ to the next, that starting with $a$, there are three ways to get to $a^3b$ and starting with $b$, there is only one way to get to $a^3b$. Namely, starting with $a$, we could take the following paths: $abaa$,$aaba$, or $aaab$. And starting with $b$, we can only take one path: $baaa$. So $c_2=4$.

Solving for the third term where $x=n-2$ and $y=2$, we can find three paths starting from $a$ and three paths from $b$. Namely: $aabb$,$abab$, and $abba$ for starting with $a$ and $bbaa$,$baba$, and $baab$ for starting with $b$. So in total, $c_3=6$.


We can see a pattern of solving for $c_i$ whereby we ask ourselves How many ways can we construct a sequence of $a$'s and $b$'s such that the number of $a$'s is $x$ and the number of $b$'s is $y$?
This question is precisely solve via the choose operator!
The first coefficient, $c_1$, can be solved by asking, How many ways can we construct $4$ elements consisting of $a$'s and $b$'s such that the number of $a$'s is $4$? This pretty much analogous to the choose operator's definition: ${4\choose4}=1$. Similarly, $c_2$ asks how many ways we can construct the sequence such that there are $3$ $a$'s and $1$ $b$, which is, by definition, ${4\choose3}=4$.

I hope this answers your question about the binomial expansion theorem. Unfortunately, I do not understand what you are trying to ask in your second question, but again, it probably should be asked on the math stack exchange instead. Good luck!