How many different subsets, including the empty set, of the set {0, 1,

Official Solution:

How many different subsets, including the empty set, of the set {0, 1, 2, 3, 4, 5} do not contain 0?

A. \(16\)
B. \(27\)
C. \(31\)
D. \(32\)
E. \(64\)


Consider the set without 0: {1, 2, 3, 4, 5}. Each of the 5 elements in the set {1, 2, 3, 4, 5} has two possibilities: either to be included in a subset or to be excluded. Therefore, the total number of subsets for this set is \(2^5 = 32\). These 32 subsets include the empty set, single-element subsets like {1}, {2}, {3}, {4}, and {5}, and multi-element subsets like {1, 2}, {1, 2, 3}, {1, 3, 4, 5}, and so on. Each of these subsets will also be a subset of {0, 1, 2, 3, 4, 5} and will not contain the element 0.


Answer: D

Similar question: https://gmatclub.com/forum/fresh-meat-1 ... l#p1215323

Hope it helps.

The solution offered counted it differently.

So I just want to clear it up whether I am also thinking in the right direction.

Number of subset

Since we have 5 digits other than 0, we can take any numbers from the set of 5 to make a subset. Also it is a matter of selection and not arrangement.So we will consider combinations.

5c1+5c2+5c3+5c4+5c5=31

And one set is the NULL set having no elements in it so

31+1=32

Please confirm my reasoning of NULL set.

Rgds,
TGC!

Is null set a subset of all the sets ?

Yes, an empty set is a subset of all sets.

Total number of subsets: 6 + 6C2 + 6C3 + 6C4 + 6C5 + 6C6 = 2^6 - 1 = 64 - 1 = 63

Number of subsets with zero = 1 + 1 x 5C1+ 1 x 5C2 + 1 x 5C3 + 1 x 5C4 = 1 + 5 + 10 + 10 + 5 = 31

Number of subsets without zero = 63 - 31 = 32

The total no of subsets that can be formed is given by2^n
Now,
Let us consider a set without 0 (1,2,3,4,5) in total there are 5 elements so possible no of sets with these 5 elements is 2^5=32
Hence
There are 32 sets which do not contain the number 0.

Please consider a kudos if this was helpful

I did not know the subset formula when trying to solve this initially - clearly it is a very simple and easy to remember formula that renders the problem easy at a mechanical level, if you understand the formula - but I am struggling to see why the alternative logic I used initially produced an incorrect answer. Can anyone explain what element is missing?

Sets consisting of 6 numbers: 0 (because one of the numbers must be excluded)
Sets consisting of 5 numbers: 1 (1,2,3,4,5)
Sets consisting of 4 numbers: 5 (combination formula; 5 choose 4)
Sets consisting of 3 numbers: 10 (combination formula; 5 choose 3)
Sets consisting of 2 numbers: 10 (combination formula; 5 chose 2)
Sets consisting of 1 number: 5 (1 or 2 or 3 or 4 or 5)

All of which sums to 31. What subset is not being picked up?

You are missing an empty set, which is a subset of all sets.