role of nCr factorial in binomial expansion I have been looking for a concrete answer on this matter since leaving secondary school and thought now to ask online since YouTube and Wikipedia seem too convoluted. 
I remember watching this video on the proof behind the nCr factorials and I understood the logical proof clearly. 
However, I remember when asked

Expand $(x+y)^3$

I would, automatically, draw out the following line:

${3 \choose 0}x^0y^3 + {3 \choose 1}x^1y^2 + {3 \choose 2}x^2y^1 + {3 \choose 3}x^3y^0$

and then use my calculator to get

$y^3 + 3x^1y^2 + 3x^2y^1 + x^3$

and further to my disappointment, my teacher would speak about Pascal's triangle and say that the coefficients (the numbers infront of my $x$ and $y$ terms) came from this so-called triangle. There was no explanation given for the purpose of factorials or where this Pascal's triangle came from. 
After looking at the video above and looking at the Pascal's triangle, I'm just confused as to why I would need ${n \choose r}$ for expanding the equation above. 
What is the link between using this factorials formula:
$$\frac{n!}{r!(n-r)!}$$
and applying it to expanding algebraic expressions as above?
I understand if it seems trivial to most people here, but after my first year as a chemistry student, I'm still interested in knowing the link between the two!
EDIT - I should add I understand the video where he describes how many words can be made but I can't apply this understanding for an algebraic perspective
 A: Expanding on Don Thousand's comment, as well as the video's way of explaining it, you can think of each term in the expansion of $(x+y)^n$ as a word using $x$ and $y$. For example, $$(x+y)^3=xxx+xxy+xyx+xyy+yxx+yxy+yyx+yyy.$$ Algebraically, many of these terms are the same, of course, e.g. $xxy=xyx$. In fact, for each such word, the only thing that matters is the number of $x$s (or, conversely, the number of $y$s). So asking, "What is the coefficient of $xy^2$ in the expansion of $(x+y)^3$?" is the same as asking, "How many words of length $3$ can be made using $x$s and $y$s such that there is exactly one $x$?" And both of these questions are equivalent to asking, "How many size $1$ subsets are there of a size $3$ set?" All of these are answered with $3\choose 1$, or, in general, $n\choose k$.
A: The idea is this: whenever we expand something of the form $(x+y)^2=(x+y)(x+y),$ what we're doing is pairing up every term in one bracket with every term in the other, to get $xx+xy+yx+yy,$ in order.
We may have more than two brackets, and that only means selecting one letter from each of the brackets, in such a way that all possible combinations are made. So, for five brackets, or $(x+y)^5,$ we get distinct terms $xxxxx+xxxxy+xxxyy+xxyyy+xyyyy+yyyyy.$ However, the mixed terms can appear in more orders -- e.g., in the second term, there are a total of five places for the $y$ to be, so we take this term $5$ times. In general, we can find the total number of possible appearances by focusing on only one of the letters (say $y$). Then the problem reduces to counting the number of ways of selecting, say two, spaces out of a total number of $5,$ where we put the $y$'s. For the fourth term, for example, we calculate $\binom{5}{3},$ and so on.
In general, we have for the expansion of $(x+y)^n,$ each term containing $n$ letters, the first of which contains no $y,$ then the next, one $y,$ etc. Thus, the first appears $\binom{n}{0}$ times, the second $\binom{n}{1}$ times, the third $\binom{n}{2}$ times, and in general the $r+1$th appears $$\binom{n}{r}$$ times. These are the coefficients of the terms of the expansion.
So, when we expand $(x+y)^n,$ first we have all $x$'s, so that the first term is $x^n.$ Then we have one $y.$ Thus, we count the number of selections of one of $n$ spaces, this is $\binom{n}{1}=n.$ Thus the second term is $nx^{n-1}y.$ The third term has two $y$'s, so we count the number of selections of $2$ spaces from $n,$ to get $\binom{n}{2}=n(n-1)/2!,$ and so on. So the development may immediately be written down as $$x^n+nx^{n-1}y+\frac{n(n-1)}{2!}x^{n-2}y^2+\frac{n(n-1)(n-2)}{3!}x^{n-3}y^3+\frac{n(n-1)(n-2)(n-3)}{4!}x^{n-4}y^4+\cdots.$$
