SOLUTION: Suppose you are going to list all the subsets,how many subsets are in {3,6,9,12,15,18 ]?

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Question 756454: Suppose you are going to list all the subsets,how many subsets are in {3,6,9,12,15,18 ]?
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
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 %28matrix%282%2C1%2C6%2C2%29%29=6%2A5%2F1%2F2=15 sets of 2 elements,
%28matrix%282%2C1%2C6%2C3%29%29=6%2A5%2A4%2F1%2F2%2F3=20 sets of 3 elements, and
%28matrix%282%2C1%2C6%2C4%29%29=6%2A5%2A4%2A3%2F1%2F2%2F3%2F4=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%2A5=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%2F2=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%2A5%2A4%2F%283%2A2%29 because you could make 6%2A5%2A4=120 ordered triples, but that would count each set of 3 elements 3%2A2=6 times, because there are 3%2A2 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%2B6%2B15%2B20%2B15%2B6%2B1=highlight%2864%29
The fancy way to add would be to add the combinations as a power of a binomial