Total number of subsets in a set I was reading about subsets, in that, the article suggests the total number of subsets in a set is $2^n$, where $n$ is the number of elements in the set. For example - $\{1, 2, 3, 4, 5\}$ the total number of subsets is $32$ because $n$ is $5$ and $2^5$ is 32 by multiplicative principle.
But the multiplicative principle is that if m events can happen in n ways then the possible outcomes are $m \times n$. So in the subsets problem if every element has $2$ possibilities of it being in set or not being in set why is it not $2 \times 5$ and $2 ^ 5$? I know that the $2 ^ 5$ is correct but not able to visualize it.
 A: It is not true that if $k$ events are independent and each have $m$ different outcomes, then there are $k\times m$ possibilities. The true number is actually $m^k$, which immediately gives us a formula for the number of subsets here.
The $m\times n$ expression you give is from a different scenario: there are $2$ independent events; the first event has $m$ possibilities; the second event has $n$ possibilities. We can see how the $m^k$ rule arises from repeated application of this rule (until we have broken the situation into $k$ independent events) in the special case where $m=n$.

Some examples of each rule:
$2\times 5$ in the "multiplicative principle" case represents an event that can be divided into $2$ independent choices: for the first choice, there are $2$ possible outcomes; for the second, there are $5$. For instance, in choosing a digit from $\{0,1,2,\dots,9\}$ you can split the process into:

*

*Choosing "odd" or "even".

*Choosing a number out of the five remaining options ($\{0,2,4,6,8\}$ or $\{1,3,5,7,9\}$). (An alternate way to frame this: choose $n$ from $\{0,1,2,3,4\}$ and then your number is $2n+1$ if you chose "odd" and $2n$ if you chose "even".)

And of course, there are $10=2\times 5$ digits to choose from, so our formula holds.
In the case of subsets of $\{1,2,3,4,5\}$, we can split the event into five independent choices (not $2$), so we're going to have a multiplication of five different numbers. The choices are about whether to include or exclude $1,2,3,4$ and $5$. In each choice, there are $2$ options: include or exclude. So our multiplication is $2\times 2\times 2\times 2\times 2=2^5=32$.
A: Let $\Omega$ be a finite set.
Then you can think this way: for each subset $S\subseteq\Omega$, each element $\omega_{i}\in\Omega$ belong or do not belong to such set  $S$, where $1\leq i \leq n$.
So the first decision has two possibilities: either $w_{1}\in S$ or $w_{1}\not\in S$.
Once you have made the first decision, there are two possibilities for the second decision: either $w_{2}\in S$ or $w_{2}\not\in S$.
This process goes on until the last element $\omega_{n}$ is considered.
Since there are $n$ elements and two possibilities associated to each decision, we conclude $|\mathcal{P}(\Omega)| = 2^{n}$.
Hopefully this helps !
A: It might help if you visualize your 5-element set as a bank of 5 bits, or electric light switches.

To form a subset, you need to go through the series and make a decision for each element whether it's in or out of the subset. In the bit representation, that's indicated by setting each switch to either on or off.
The multiplicative principle is that each event generates a separate factor in the count of total ways to make a selection. In this case, each switch is a separate simple event. There's 2 ways to set the first switch, times 2 ways to set the second switch, etc., for a total of $2^5$ ways to set the 5 switches.
As an exercise, consider writing out all the different possibilities in the form of 5-bit strings. There's 00000 (all switches off, i.e., all elements left out of the subset, i.e., the empty set), 00001 (only the rightmost switch on), 00010, 00011, etc.
