[tex]S=\{\varnothing,d,\{d\},\{\{d\}\},\{\{\{d\}\}\}\}=\{v,w,x,y,z\}[/tex]
A set containing [tex]n[/tex] elements has a power set containing [tex]2^n[/tex] elements; here, [tex]|S|=5[/tex] so [tex]|\mathscr P(S)|=2^5=32[/tex].
The elements of the power set are all possible combinations of up to 5 of the total 5 elements to choose from:
0 choices (1): [tex]v[/tex] (since [tex]v=\varnothing[/tex])
1choice (5): [tex]\{v\},\{w\},\{x\},\{y\},\{z\}[/tex]
2 choices (10): [tex]\{v,w\},\{v,x\},\{v,y\},\{v,z\},\{w,x\},\{w,y\},\{w,z\},\{x,y\},\{x,z\},\{y,z\}[/tex]
3 choices (10): [tex]\{v,w,x\},\{v,w,y\},\{v,w,z\}[/tex] and so on
4 choices (5); [tex]\{v,w,x,y\},\{v,w,x,z\}[/tex] and so on
5 choices (1): [tex]\{v,w,x,y,z\}[/tex]
(1 + 5 + 10 + 10 + 5 + 1 = 32)
