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 }?
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
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