jmc

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 $4+6+4+1=\boxed{15}$.