TITLE: The total number of subsets is $2^n$ for $n$ elements
QUESTION [12 upvotes]: In my probability book, it says that to count the total number of subsets of n elements is a process of $n$ stages with binary choice of either adding this element to the subset or not to add it. Therefore, the total number is
$$2^n$$
But, for instance, we have 3 elements, according to this formula, there are 2 to the power of 3 elements, namely 8, which are
$${\emptyset},A,B,C, AB, AC, BC, ABC$$
However, I have a hard time of imagining the process or N stages binary choice that form this many subsets. Can anyone explain/help me to understand it? I mean, ABC, if we are making the choice of A, put it in or do not put it in, exactly which subset are we choosing to put in or not? Thank you.

REPLY [5 votes]: Write out all the binary numbers with $3$ digits. Each digit represents one element ($0$ means exclude, $1$ means include):
$$
\begin{array}{|c|c|}
\text{decimal}&\text{binary}&\text{subset}\\\hline
0&000&\{\}\\
1&001&\{C\}\\
2&010&\{B\}\\
3&011&\{B,C\}\\
4&100&\{A\}\\
5&101&\{A,C\}\\
6&110&\{A,B\}\\
7&111&\{A,B,C\}\\
\end{array}
$$
This is easily extendible to an $n$ element set with $n$ digit binary numbers.