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={1,2}.
 
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, or{} and {1,2}.
 
Now let us write the elements and its combination of subsets.
 
The singleton subsets are {1} and {2}.
 
Therefore, the subsets of set \(A\) are {},{1},{2}and{1,2}.
 
Thus, the set of power sets of \(A\), P(A)={{},{1},{2},{1,2}}.
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(A)n[P(A)].
Let \(A\) be a set with \(m\) elements. That is, n(A)=m.
  • The number of elements in the power set of \(A\) is given by n[P(A)]=2m.
  • The number of a proper subset of \(A\) is given by n[P(A)]1=2m1.
Example:
Consider the set with three elements B={a,b,c}.
 
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[P(B)]=2m.
 
Substitute the value of \(m\).
 
n[P(B)]=23=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, or{} and {a,b,c}.
 
Now let us write the elements and its combination of subsets.
 
The singleton subsets are {a}{b} and {c}.
 
Let us write the subsets with two elements {a,b},{a,c}and{b,c}.
 
Thus, the subsets of set \(B\) are {},{a},{b},{c},{a,b},{a,c}and{b,c}.
 
Therefore, the power set of \(B\) becomes n[P(B)]={{},{a},{b},{c},{a,b},{a,c},{b,c}}.
 
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.