Do not understand the proof of $|P(A)|=2^{|A|}$ (numbers of subsets of a A) The proof uses the multiplication rule successively. Given a set $A = \left\{ a_1,a_2,...,a_n \right\} $. It says for each element we can choose to include or not. Therefore $|P(A)|=2^{|A|}$.
I really need a more illustrative way of doing it (draw the proof, that's what I mean. Not a hard mathematical way for example by induction). I simple do not understand it. 
I'm taking an introduction course in probability and statistics.
 A: The way I think of this is as follows: for each element $a \in A$, we have two choices: $\{ \texttt{PICK}, \texttt{DROP} \}$. Every subset $S \subseteq A$ can be described by a function $I_S:  A \rightarrow \{ \texttt{PICK}, \texttt{DROP} \}$ $I_S(a) \equiv \begin{cases} \texttt{PICK} & a \in S \\ \texttt{NOPICK} & a \not \in S \end{cases}$.
The total number of functions of the form $A \rightarrow \{ \texttt{PICK}, \texttt{DROP} \}$ is:  $|\{ \texttt{PICK}, \texttt{DROP} \}|^{|A|} = 2^{|A|}$
Let's consider a set $A \equiv \{\#, \star\}$. Now, to describe the subset $B \equiv \{ \star \}$, we create the function $I_B$: 
$$
I_B(\#) \equiv \texttt{NOPICK} \qquad I_B(\star) \equiv \texttt{PICK}
$$
To understand why there are $2^{|A|}$  such functions, notice that we can think of the above $I_B$ as a tuple $I_B \sim(\#: \texttt{PICK}, \star: \texttt{NOPICK})$. How many such tuples $(\#: \_\ , \star: \_ \ )$ exist? For each $\_$, we have two choices, and we have $|A|$ such choices. So the total size is $2^{|A|}$.
A: Say $A= \{1,2,3,4,5,6,7\}$. Each subset $X$ of $A$ we map to a string made of $0$ and $1$. Write $1$ on a place if the number of a place is in $X$ else write $0$. Example:
\begin{align}\{1,2,5\} &\mapsto 1100100\\
\emptyset &\mapsto 0000000\\
A &\mapsto 1111111
\end{align}
Clearly this map is a bijection on a set of all strings of lenght $7$ made of $0$ and $1$. So you have to count the number of those strings which is $2^7$
A: Let $A=\{a_1,a_2,\dots,a_n\}$, thus $|A|=n$. Imagine you have to construct a subset of $A$, you would have to pick and choose from the elements of $A$. Thus for each element $a_j\in A$ you have two choices; 1. keep it in the subset, 2. leave it out of the subset. Thus starting from $a_1$ you have 2 choices, keep $a_1$ in the set or leave it out. This gives two possibilities of subsets. Then you move to $a_2$, keep it or leave it. This gives you two possibilities for $a_1$ and two for $a_2$. Since you know of the multiplication rule, you know that the number of possibilities then would be $2*2$. Continue this with all the remaining $a_j$ up to $a_n$ and see that you would have written the number of possibilities of subsets as $n$ products of $2$. Of course the rigorous way of performing the last step would be by induction.
If you want an illustrative way of showing that the multiplication rule works in this context, you can grab a piece of paper and write down the subsets of a set of $0,1,2,3,\dots$ elements.
