Question 756454
The set {3,6,9,12,15,18} has 6 elements.
 
NOTE:
There is an issue with the interpretation of what constitutes a subset.
I consider the set {3,6,9,12,15,18} to be a sunset of itself, and I consider the empty set to be a subset too. 
So I would have to count all the sets containing 0, 1, 2, 3, 4, 5, and 6 elements from that set. With a different interpretation, the total number of subsets would be less.
 
There is 1 set with 6 elements, and 1 set with 0 elements.
There are 6 sets with 1 element and 6 sets with 5 elements.
 
IF YOU KNOW AND UNDERSTAND COMBINATION FORMULAS:
The number of sets of 2, 3, and 4 elements can be calculated as combinations.
There are {{{(matrix(2,1,6,2))=6*5/1/2=15}}} sets of 2 elements,
{{{(matrix(2,1,6,3))=6*5*4/1/2/3=20}}} sets of 3 elements, and
{{{(matrix(2,1,6,4))=6*5*4*3/1/2/3/4=15}}} sets of 4 elements.
(The same combinatorial formulas could have been used to calculate the number of sets of 0, 1, 5, and 6 elements too).
 
WITHOUT USING COMBINATION FORMULAS:
When assembling sets of 2 elements, you have 6 choices for the first element, and then 5 elements to chose for the second element, for a total of {{{6*5=30}}} possible ordered pairs. However, that would count each set twice, with tje elements in different order. For example, you would have counted the ordered pairs (3,6) and (6,3) as 2 ordered pairs, but they represent just one set, the set {3,6}. So there are only {{{30/2=15}}} sets of 2 elements.
The number of sets of 4 elements is also 15, because each set of 2 elements leaves behind a set of 4 elements.
The number of sets of 3 elements is {{{20=6*5*4/(3*2)}}} because you could make {{{6*5*4=120}}} ordered triples, but that would count each set of 3 elements {{{3*2=6}}} times, because there are {{{3*2}}} different ways to arrange 3 elements (3 choices for which element is first times 2 choices for which one to place second).
 
ADDING UP:
You can add up the numbers of sets with 0, 1, 2, 3, 4, 5, and 6 elements the simple way:
{{{1+6+15+20+15+6+1=highlight(64)}}}
The fancy way to add would be to add the combinations as a power of a binomial
{{{(matrix(2,1,6,0))+(matrix(2,1,6,1))+(matrix(2,1,6,2))+(matrix(2,1,6,3))+(matrix(2,1,6,4))+(matrix(2,1,6,5))+(matrix(2,1,6,6))=(1+1)^6=2^6=highlight(64)}}}