What does $C_n(X)$ in a simplicial complex really mean? My Algebraic Topology textbook says:

Let $X$ be a simplicial complex. The group $C_n(X)$ of oriented $n$-chains of $X$ is the free abelian group generated by the oriented $n$-simplexes of $X$. 

For example, an element of $C_2(S)$ would be $9P_2P_3P_4-4P_1P_3P_4-5P_1P_2P_3$. 
What does $9P_2P_3P_4-4P_1P_3P_4-5P_1P_2P_3$ even mean?? What does it mean to multiply a 2-simplex with a scalar? Do we expand the area of the simplex by $9$ times? Is area even defined in a simplicial complex?
Thanks in advance!
 A: When they write $5\cdot S$ where $S$ is a simplex they do not mean to actually do anything with the simplex $S$, it is just a formal expression with which you can calculate. 
$S$ becomes nothing but an abstract symbol for symbolic calculation. So $5\cdot S+3\cdot S=8\cdot S$, for example. This is not a way of writing, $5$ times the area of $S$ plus $3$ times the area of $S$ is $8$ times the area of $S$. Instead, it just means $5$ times of an object called $S$ plus $3$ times an object called $S$ makes $8$ times an object called $S$.
Given any set whatsoever, you can form a free abelian group on it just looking at formal linear combinations of its elements.
Example:
Suppose your set of simplices is $X=\{A,B,C\}$. Then the free abelian group over $X$ is the set given by all the linear combinations
$$x_1A+x_2B+x_3C$$
The $+$'s in this expression do not mean any transformation to be done with the simplices. Imagine the $A,B,C$ being the basis of a vector space of dimension $3$. You can just as well write the vector $x_1A+x_2B+x_3C$ in the form $(x_1,x_2,x_3)$.
Now you make this into a group by saying that addition is defined componentwise:
$$(x_1A\color{red}{+}x_2B\color{red}{+}x_3C)\color{red}{+}(y_1A\color{red}{+}y_2B\color{red}{+}y_3C):=(x_1+y_1)A\color{red}{+}(x_2+y_2)B\color{red}{+}(x_3+y_3)C$$
This makes the set of these combinations into an abelian group. To avoid confusion, I drew the $+$'s between elements of the free group in red and the $+$'s of integers in black.
Entirely equivalently, if we had decided to denote the elements of our free abelian group as $(x_1,x_2,x_3)$ instead, the definition of addition would look like
$$(x_1,x_2,x_3)+(y_1,y_2,y_3)=(x_1+y_1,x_2+y_2,x_3+y_3)$$
We can connect these notations by defining $A:=(1,0,0)$, $B:=(0,1,0)$ and $C:=(0,0,1)$. Indeed, then
$$x_1A+x_2B+x_3C=(x_1,x_2,x_3)$$
So whenever we make the free abelian group over a set with $k$ elements, say, then that means we are taking all the $k$-tuples of integers with componentwise addition. Naturally this also works with infinitely large sets, but then we consider only finite linear combinations.
As always in algebra, what matters is not how you name your elements or where they come from, the point is only what you do with them. The structure of the algebraic operations is important and nothing else.
