How many numbers can you get by multiplying two or more distinct members of the set $\{1,2,3,5,11\}$ together? Apparently the answer is 15,  but I got 26.
My process:
The set $\{1,2,3,5,11\}$ has five numbers.
Ways to choose two members: $_5C_2$
Ways to choose three members: $_5C_3$
Ways to choose four members: $_5C_4$
Ways to choose five members: $_5C_5$
Total: $_5C_2 + {}_5C_3 + {}_5C_4 + {}_5C_5 = 26$
Is there a detail I have misunderstood in this question?
 A: The people in the comments have helped me realize that multiplying by 1 results in duplicates of the previous case. So I worked my way to the correct answer:
Case 1: Ways to choose two members: $_5C_2$
Case 2: Ways to choose three members: 
$_5C_3$ initially, but there are duplicates. If 1 is chosen, then it will be duplicates of some combinations of Case 1.
If one of the numbers is 1 then
1 _ _ 
Four numbers in the main set that are not 1; how many ways to choose 2 of them is $_4C_2$, so the total number for Case 2 is $_5C_3{} - {}_4C_2 = 4$
Case 3 Ways to choose four members: 
$_5C_4$ initially, but there are duplicates. If 1 is chosen, then it will be duplicates of some combinations of Case 2.
1 _ _ _
Four numbers in the main set that are not 1; how many ways to choose 3 of them is $_4C_3$, so the total number for Case 2 is $_5C_4{} - {}_4C_3 = 1$
Case 4 Ways to choose five members: $_5C_5$, but that entire combination is a duplicate of a combination from Case 3. So ignore.
--
Total number: 10 + 4 + 1 = 15
Thank you for those who have helped.
A: Adding $1$ into a set doesn't change the product.  It does let you choose only one of the other numbers.  You have $4$ choices of $1$ and one other, then ${4\choose 2}+{4\choose 3}+{4\choose 4}=6+4+1=11$ combinations with at least two of the others, so there are $15$ products that can be obtained.  They are $2,3,5,11,6,10,15,22,33,55,30,66,110,165,330$
A: Note that, because two or more members can be multiplied, multiplying by $1$ will only make a difference if it is one of two numbers. Thus, multiplying by $1$ adds four potential numbers.
Now, we only need to consider the number of combinations that can be made from $2$, $3$, $5$, and $11$.
Choosing two from this set offers six possiblities: $2 \cdot 3$, $2 \cdot 5$, $2 \cdot 11$, $3 \cdot 5$, $3 \cdot 11$, and $5 \cdot 11$.
Choosing three offers four possibilities: $2 \cdot 3 \cdot 5$, $2 \cdot 3 \cdot 11$, $2 \cdot 5 \cdot 11$, and $3 \cdot 5 \cdot 11$.
Finally, there is one possibility with four chosen: $2 \cdot 3 \cdot 5 \cdot 11$. Thus, there are $4 + 6 + 4 + 1 = \boxed{15}$.

