Question 1119390
{}
{g},{e},{o},{m}
{g,e},{g,o},{g,m},{e,o},{e,m},{o,m}
{g,e,o},{g,e,m},{g,o,m},{e,o,m}
{g,e,o,m} 
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Counting each row, that's 1+4+6+4+1 = 16 subsets (notice it forms a row from Pascal's triangle)
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For n elements, there are {{{2^n }}} subsets, including the improper subset of all elements. 
This is so because each element is either in the subset or it is not, and that leads to  2 choices for the 1st element * 2 choices for the 2nd, *…* 2 choices for the {{{n^(th)}}} element or {{{2^n}}} in all.