jmc

Solution 1: For each of 1, 2, 3, and 4, we can either choose to include the number in the set, or choose to exclude it. We therefore have 2 choices for each of these 4 numbers, which gives us a total of $2^4=\boxed{16}$ different subsets we can form.

Solution 2: We can have either 5 by itself, 5 with one other number from the four, 5 with two other numbers, 5 with three other numbers, or 5 with all four other numbers. The number of ways to form a subset with 5 by itself is $\left(\genfrac{}{}{0ex}{}{4}{0}\right)$. The number of ways to form a subset with 5 and one other number is $\left(\genfrac{}{}{0ex}{}{4}{1}\right)$. Similarly, the number of ways to form a subset with 5 and two other numbers is $\left(\genfrac{}{}{0ex}{}{4}{2}\right)$, with three other numbers is $\left(\genfrac{}{}{0ex}{}{4}{3}\right)$, and with all four other numbers is $\left(\genfrac{}{}{0ex}{}{4}{4}\right)$. Thus, our answer is $1+4+6+4+1=\boxed{16}$.