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24 votes
24 votes

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

  1. $2$
  2. $4$
  3. $8$
  4. None of the above
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6 Answers

Best answer
34 votes
34 votes
Number of elements in power set  $= 2^{\text{(no of elements in the set)}} = 2^{3} = 8$

Elements are $\{∅,\{\{∅\}\},\{1\},\{\{2,3\}\},\{\{∅\},1\},\{1,\{2,3\}\},\{\{∅\},\{2,3\}\},\{\{∅\},1,\{2,3\}\}\}$

Hence, Option is $(C): 8.$
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5 votes
5 votes

S={{ϕ},1,{2,3}}

cardinality of S= 3

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

Hence ,Option (C)8  is the correct answer.

1 votes
1 votes
Since there are three elements in the set S , so the power of this set  P( S)= 2 ^3= 8.

Hence the option C is correct.
0 votes
0 votes

S = {(φ), 1, (2, 3)}
P(S) = {φ, {{φ}}, {1}, {{2, 3}}, {{φ}, 1}, {1, {2, 3}}, {{φ}, 1, {2, 3}}}
In P(S) it contains 8 elements.

Answer:

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