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The number of elements in the power set $P(S)$ of the set $S=\{\{\phi\}, 1, \{2, 3\}\}$ is:

1. $2$
2. $4$
3. $8$
4. None of the above
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0

Would we not consider that an element itself has two elements (2,3) which itself have its own power set of size = 2?

0
NO. {2,3} is considered as one element in set S. if you were confused replaced that with x.

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.

edited

S={{ϕ},1,{2,3}}

cardinality of S= 3

then cardinality of Power set of S = 2^3 =8