#1**+1 **

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 ]

CPhill
Aug 31, 2018