How to understand Bionomial Theroem and the Recursion Rule? In this video from EDX, the instructor explains the binomial theorem as:
Binomial Theorem:
When you calculate $(a + b)^n = a^n + C(1)a^{(n-1)b} + C(2)a^{(n-2)b^2} + ... + C(n-1)ab^{(n-1) + b^n}$
The coefficients = $C1, C2, ..., C(n-1)$
Coefficient of $(a + b)^4 = 1, 4, 6, 4, 1$
From the formula above, you get after some manipulation:
$(a + b)^n = a^n + (n choose 1) a^{(n-1) b^1} + (n choose 2) a^{(n-2) b^2} + ... + (n choose n-2) a^2 b^{(n-2)} + (n choose n-1) a^1 n^{(n-1)} + b^n$
The coefficients will be the same as pascal's triangle
Is it necessary to know this? I still do not understand how he went from $(a + b)^n$ formula to plugging in $(n choose 1) (n choose 2)...$
Additionally, for the recursion theory:
Recursion Rule:
If we have 52 cards and we select 5 cards we have a collection of 5 cards taken from 52 cards
The order the cards are dealt does not matter
To get the recursion, we will split the hands into hands that contain the ace of spaces with hands that do not contain the ace of spades
For the hand that does not contain the ace of spades, there are 51 cards to choose from
So it is 51 choose 5
For the hand that does contain the ace of spades, then that hand is made from
4 cards in the remaining 51 cards in the deck
So it is 51 choose 4
So since every hand either has ace of spades or does not
We get our recursion
52 choose 5 is = 51 choose 5 + 51 choose 4
But if you actually calculate 52 choose 5 you get 311875200
And if you calculate 51 choose 5 + 51 choose 4 you get 287884800
To me it seems mathematically inaccurate?
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!