probability

The first is the improper subset {1,2,3, 4, 5}

Next...we can choose any 3 of the 4 remaining elements to  include in a set with 5  = 4C3  = 4 more subsets

Then, we can choose any 2 of the remaining elements to include in a set with 5  = 4C2  = 6  additional subsets

Then, we can choose any 1 of the remaining elements to include in a set with 5  = 4 C1   = 4 additional subsets

Lastly...we can  just have a set only including 5   =  1  additional subset

Note....these  are just  the sum of the entries in Row 4 of Pascal's Triangle  = 1 4 6 4 1  =  2^4  =  16  subsets  [if we include the improper subset ]