If we had fifteen distinct objects, we could arrange them in a row in $15!$ ways. In a circular permutation, only the relative order of the objects matters. There are two ways to handle this:
- Pick one of the objects as a reference point. Relative to this object, the other $14$ objects can be arranged in $14!$ ways as we proceed clockwise around the circle.
- Break a circular arrangement between two objects. Now place the objects in a line as we proceed clockwise around the circle from the point where we made the break. There are $15$ ways we could do this, one to the left of each object. Hence, there are $15$ linear arrangements corresponding to each circular arrangement. Since there are $15!$ linear arrangements of $15$ objects, there are
$$\frac{15!}{15} = \frac{15 \cdot 14!}{15} = 14!$$
circular permutations of $15$ distinct objects.
In general, there are $(n - 1)!$ circular permutations of $n$ distinct objects.
In this case, we have ten indistinguishable large chairs and five indistinguishable small chairs to arrange in a circle. As you observed, there are
$$\binom{15}{5}$$
ways to arrange the objects in a line. If we divide by $15$ to obtain the number of circular arrangements, we do not get an integer because as @lulu pointed out in the comments, the circular arrangement in which there are exactly two large chairs between each pair of small chairs has just three corresponding linear arrangements:
$$LLSLLSLLSLLSLLS$$
$$LSLLSLLSLLSLLSL$$
$$SLLSLLSLLSLLSLL$$
Therefore, when we divide by $15$, we count this particular arrangement $3/15 = 1/5$ times. We want to count it once, so we must add $4/5$ to your answer. Hence, the number of distinct circular arrangements of $10$ indistinguishable large chairs and $5$ indistinguishable small chairs is
$$\frac{1}{15}\binom{15}{5} + \frac{4}{5} = 201$$
