Q:

Basic question about coefficients in external algebra

The exterior algebra formed from a vector space $V$ with vector basis $\{e_1,e_2,e_3\}$ will have basis
$$\begin{align}
&\Delta^0 V=\langle 1 \rangle\\
&\Delta^1 V=\langle e_1,e_2,e_3 \rangle\\
&\Delta^2 V=\langle e_1\wedge e_2, e_1 \wedge e_3, e_2 \wedge e_3 \rangle\\
&\Delta^3 V=\langle e_1\wedge e_2 \wedge e_3 \rangle\\
\end{align}$$
I presume...
Mathworld states

The alternating algebra, also called the exterior algebra, $\Delta^* V$ is a $2^n$ dimensional algebra. In Wolfram Language, an element of the alternating algebra can be represented by an $n$-nested binary list. For example,

$$\{\{\{\color{blue}1,\color{red}2\},\{0,0\} \},\{\{\color{orange}3,0\},\{\color{magenta}4,\color{brown}5\}\}\}$$

represents

$$\begin{align}
\color{blue} 1 \;e_1 \wedge e_2 \wedge e_3\quad+\\
\color{red}2\; e_1 \wedge e_3\quad+\\
\color{orange}3\;e_2\wedge e_3\quad +\\
\color{magenta}4\; e_3\quad +\\
\color{brown}5\; 1\quad \quad
\end{align}$$
The color-coding and $\LaTeX$ transcription is mine in the quote, trying to reflect my problem understanding this. 

The actual formulation in the quote is:
$$\{\{\{1,2\},\{0,0\} \},\{\{3,0\},\{4,5\}\}\}$$

represents

$$e_1 \wedge e_2 \wedge e_3\,+\,
2\; e_1 \wedge e_3\, +\,
3\;e_2\wedge e_3\, +\,
4\; e_3\, +\,
5
$$

For instance, $\{\color{blue}1,\color{red}2\}$ (iff the color-coding makes sense) combines the coefficient $1$ of $e_1\wedge e_2\wedge e_3,$ composed of the maximum combination of basis vectors from $V$ with the coefficient for a wedge of just $2$ basis vectors, i.e. $e_1 \wedge e_3.$ Likewise, I don't see, either, what the next binary element $\{0,0\}$ corresponds to.

What is the equivalence between the WL notation and the different possible wedge products, and how are the coefficients usually "stored" (presumably a regular matrix, provided they are not functions (?))?

A:

I think they've made a mistake, writing $e_1 \wedge e_3$ when they meant $e_1 \wedge e_2$. At least, that's the only way I can make sense of the pattern.
Then the coefficients are leaves in the binary tree obtained by successively making the choices (“$e_1$ or $1$”) $\wedge$ (“$e_2$ or $1$”) $\wedge$ (“$e_3$ or $1$”). So the first number in the (flattened) list is the coefficent of $e_1 \wedge e_2 \wedge e_3$, next comes the coefficent of $e_1 \wedge e_2 \wedge 1 = e_1 \wedge e_2$, and so on, until you get to the last number which is the coefficient of $1 \wedge 1 \wedge 1 = 1$.

