### Theory:

The set of all subsets of a set $$A$$ is said to be power set $$A$$. It is written as $$P(A)$$.
Example:
Consider the set with two elements $A\phantom{\rule{0.147em}{0ex}}=\phantom{\rule{0.147em}{0ex}}\left\{1,2\right\}$.

By the concept, 'Empty set is a subset of every set' and 'Every set is a subset of itself' the two obvious subsets of the set are the empty set and the whole set itself.

That is, $\varnothing \phantom{\rule{0.147em}{0ex}}\mathit{or}\phantom{\rule{0.147em}{0ex}}\left\{\right\}$ and $\left\{1,2\right\}$.

Now let us write the elements and its combination of subsets.

The singleton subsets are $\left\{1\right\}$ and $\left\{2\right\}$.

Therefore, the subsets of set $$A$$ are $\left\{\right\},\left\{1\right\},\left\{2\right\},\phantom{\rule{0.147em}{0ex}}\left\{1,2\right\}$.

Thus, the set of power sets of $$A$$, $P\left(A\right)\phantom{\rule{0.147em}{0ex}}=\phantom{\rule{0.147em}{0ex}}\left\{\left\{\right\},\left\{1\right\},\left\{2\right\},\left\{1,2\right\}\right\}$.
Important!
The number of elements in set $$A$$ is always less than the number of elements in the power set of $$A$$. That is $n\left(A\right)\phantom{\rule{0.147em}{0ex}}\le n\left[P\left(A\right)\right]$.
Let $$A$$ be a set with $$m$$ elements. That is, $n\left(A\right)\phantom{\rule{0.147em}{0ex}}=\phantom{\rule{0.147em}{0ex}}m$.
• The number of elements in the power set of $$A$$ is given by $n\left[P\left(A\right)\right]={2}^{m}$.
• The number of a proper subset of $$A$$ is given by $n\left[P\left(A\right)\right]-1={2}^{m}-1$.
Example:
Consider the set with three elements $B\phantom{\rule{0.147em}{0ex}}=\phantom{\rule{0.147em}{0ex}}\left\{a,b,c\right\}$.

Now we are going to find the number of elements in the power set of $$B$$.

Set $$B$$ has three elements. That is, $$m = 3$$.

The formula for this is given by $n\left[P\left(B\right)\right]={2}^{m}$.

Substitute the value of $$m$$.

$n\left[P\left(B\right)\right]={2}^{3}=8$

Let us verify by the actual method:

By the concept, 'Empty set is a subset of every set' and 'Every set is a subset of itself' the two obvious subsets of the set are the empty set and the whole set itself.

That is, $\varnothing \phantom{\rule{0.147em}{0ex}}\mathit{or}\phantom{\rule{0.147em}{0ex}}\left\{\right\}$ and $\left\{a,b,c\right\}$.

Now let us write the elements and its combination of subsets.

The singleton subsets are $\left\{a\right\}$$\left\{b\right\}$ and $\left\{c\right\}$.

Let us write the subsets with two elements $\left\{a,b\right\},\phantom{\rule{0.147em}{0ex}}\left\{a,c\right\},\phantom{\rule{0.147em}{0ex}}\left\{b,c\right\}$.

Thus, the subsets of set $$B$$ are $\left\{\right\},\left\{a\right\},\phantom{\rule{0.147em}{0ex}}\left\{b\right\},\phantom{\rule{0.147em}{0ex}}\left\{c\right\},\left\{a,b\right\},\phantom{\rule{0.147em}{0ex}}\left\{a,c\right\},\phantom{\rule{0.147em}{0ex}}\left\{b,c\right\}$.

Therefore, the power set of $$B$$ becomes $n\left[P\left(B\right)\right]=\left\{\left\{\right\},\left\{a\right\},\phantom{\rule{0.147em}{0ex}}\left\{b\right\},\phantom{\rule{0.147em}{0ex}}\left\{c\right\},\left\{a,b\right\},\phantom{\rule{0.147em}{0ex}}\left\{a,c\right\},\left\{b,c\right\}\right\}$.

Note that the number of elements in the power set of $$B$$ is $$8$$.
Important!
The number of subsets of a set is equal to the number of elements in the respective power set.