Hello ulit, haha. So, let's look at all the possible subsets of A.
First instance, the subset is a null set.
Second instance, the subsets has one element each.
Third instance, the subsets has two elements each.
...
Last instance, the subset has five elements
So for the first instance, C(5,0) = 1
second instance, C(5,1) = 5
third instance, C(5,2) = 10
fourth instance, C(5, 3) = 10
fifth instance, C(5,5) = 1
Adding those up, the answer is 32 :)
If you're not aware of what combination is, you might want to research about combinatorics and probabiliry. and the number theory, too.
or, if you don't want to, well, just have patience in counting all of them manually.
lastly, if you would notice 1, 5, 10, 10, 5, 1 is a pattern that you would see in the Pascal's triangle so you might want to check that out too ;)
