The numbers $1, 2, 3, 4, 5$ are to be arranged in a circle. An arrangement is $\textit{bad}$ if it is not true that for every $n$ from $1$ to $15$ one can find a subset of the numbers that appear consecutively on the circle that sum to $n$. Arrangements that differ only by a rotation or a reflection are considered the same. How many different bad arrangements are there?
$\textbf {(A) } 1 \qquad \textbf {(B) } 2 \qquad \textbf {(C) } 3 \qquad \textbf {(D) } 4 \qquad \textbf {(E) } 5$
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We see that there are $5!$ total ways to arrange the numbers. However, we can always rotate these numbers so that, for example, the number $1$ is always at the top of the circle. Thus, there are only $4!$ ways under rotation. Every case has exactly $1$ reflection, so that gives us only $4!/2$, or $12$ cases, which is not difficult to list out. We systematically list out all $12$ cases. 
Now, we must examine if they satisfy the conditions. We can see that by choosing one number at a time, we can always obtain subsets with sums $1, 2, 3, 4,$ and $5$. By choosing the full circle, we can obtain $15$. By choosing everything except for $1, 2, 3, 4,$ and $5$, we can obtain subsets with sums of $10, 11, 12, 13,$ and $14$. 
This means that we now only need to check for $6, 7, 8,$ and $9$. However, once we have found a set summing to $6$, we can choose all remaining numbers and obtain a set summing to $15-6=9$, and similarly for $7$ and $8$. Thus, we only need to check each case for whether or not we can obtain $6$ or $7$. 
We can make $6$ by having $4, 2$, or $3, 2, 1$, or $5, 1$. We can start with the group of three. To separate $3, 2, 1$ from each other, they must be grouped two together and one separate, like this.

Now, we note that $x$ is next to both blank spots, so we can't have a number from one of the pairs. So since we can't have $1$, because it is part of the $5, 1$ pair, and we can't have $2$ there, because it's part of the $4, 2$ pair, we must have $3$ inserted into the $x$ spot. We can insert $1$ and $2$ in $y$ and $z$ interchangeably, since reflections are considered the same.

We have $4$ and $5$ left to insert. We can't place the $4$ next to the $2$ or the $5$ next to the $1$, so we must place $4$ next to the $1$ and $5$ next to the $2$.

This is the only solution to make $6$ "bad."
Next we move on to $7$, which can be made by $3, 4$, or $5, 2$, or $4, 2, 1$. We do this the same way as before. We start with the three group. Since we can't have $4$ or $2$ in the top slot, we must have one there, and $4$ and $2$ are next to each other on the bottom. When we have $3$ and $5$ left to insert, we place them such that we don't have the two pairs adjacent.

This is the only solution to make $7$ "bad."
We've covered all needed cases, and the two examples we found are distinct, therefore the answer is $\boxed{\textbf {(B) }2}$.