SOLUTION: Subsets of 1,2,3,4,5,6 ??

The given set has 6 elements.


Hence, it has  sub-sets in all, including the empty subset and so named improper subset which contains all the elements of the original subset.


It is a general rule/theorem from the set theory:


    Any finite set consisting of n elements, has  subsets including the empty subset and 
    improper subset which contains all the elements of the original subset.

In your case the number os all subsets is  = 64,  and it is not simple task (for me, at least) to list them all.


But I can list some of them to get you an idea what is this:

{1}, {2}, . . . , {6},

{1,2}, {1,3}, . . . , {1,6}

{1,2,3}, . . . , {4,5,6}  

and so on . . . until the last IMPROPER subset 

{1,2,3,4,5,6}.

1)  {}   empty set

2)  {1}, {2}, . . . , {6}          <<<---===  6 subsets consisting of 1 element each;

3)  {1,2}, {1,3}, . . . , {1,6},   <<<---===  15 subsets consisting of 2 elements each;
           {2,3}, . . . , {2,6},   

    and so on 

4)  {1,2,3}, . . . , {4,5,6},      <<<---===  20 subsets consisting of 3 elements each;

    and so on . . . 


and so on . . . until the last IMPROPER subset 

{1,2,3,4,5,6}.