The number of elements in the power set $P(S)$ of the set $S=\{\{\phi\}, 1, \{2, 3\}\}$ is:
Would we not consider that an element itself has two elements (2,3) which itself have its own power set of size = 2^{2 }?
@BharathiCH what do u mean by no of power set? no of element in power set is 2^(set size)
in {∅,1,{2,3}} there are 3 element i.e phi,1,{2,3} [Note 2,3 single element] now power set 2^3=8
Note ∅ and {∅} not same. first means no element whereas second means one element which is ∅.
@Abhisek Tiwari 4
Then the elements will be as
{∅,{∅},{1},{{2,3}},{∅,1},{1,{2,3}},{∅,{2,3}},{∅,1,{2,3}}}..
Then {∅,1} and {1} are they not equal?
@BharathiCH
Yes {1} and { {∅},1} different. [Note it should be {{∅},1} instead of {∅,1}
no of elements in power set is = $2$^{(no of elements in the set)} = $2^{3}$ = $8$ Elements are {∅,{{∅}},{1},{{2,3}},{{∅},1},{1,{2,3}},{{∅},{2,3}},{{∅},1,{2,3}}}
Hence, Option is $(C)$ 8.
S={{ϕ},1,{2,3}}
cardinality of S= 3
then cardinality of Power set of S = 2^3 =8