Raymond B. answered 02/27/23
Math, microeconomics or criminal justice
A is an improper subset
for proper subsets subtract 1
6 elements of the set
the total subsets = 2^6 = 64
proper subets = 64-1 = 63
total subsets includes the set A itself
there is one improper subset with 6 elements, A
there are 6 subsets with 5 elements
10 subsets with 4 elements
15 with 3 elements
10 with 2 elements
6 with 1 element each
1 with 0 element, the null set
proper subsets includes any subset except A
add them to get 1+6+10+15+20+15+10+6+1 = 64
add the 7th row of Pascal's triangle