Question 663055
each element has 2 states, there or not there, so there are

{{{2^100}}} different ways of choosing sets there or not there, including none of them there (the empty set)

so there are

{{{2^100 - 1}}} non empty subsets


{{{100C1+100C2}}} have {{{2}}} or {{{less}}} elements

{{{2^100 - 1-100C1-100C2 = 2^100 -1-100-4950 = 2^100- 5051}}}