Arrangement of $10$ people given $3$ cannot be next to each other. Question: How many ways can you arrange a group of $10$ students given that $3$ of them cannot sit next to each other (they cannot sit as a group of $3$ nor can $2$ of them sit next to each other $-$ the three students need to be separated completely).
My attempt: Naturally, I went to counting possible cases and I decided to visualise different combinations if they were arranged in a line. I ended up with $15$ combinations that they could sit in a line such that they are unique and follow the condition if each line was a circle. From there it was pretty easy to see that $\dfrac{15 \times 3! \times 7!}{10}$ was the resulting answer. I have tried to justify the $15$ without counting situations, although I have been unsuccessful. Is this answer correct? If it is, how can I properly justify it? If it isn't, how would I approach a problem like this?
Any guidances or help would be greatly appreciated!
EDIT: I know for a fact $15$ is incorrect. I feel like it may be $12$ but still I'm not entirely sure.
(1) Assuming they in a line, not a circle, then imagine $7$ stars a row (the &quot;other&quot; students). Among the $8$ gaps in between or outside of those stars, choose $3$ for the locations of the special $3$ students. This can be done in $C(8,3)=56$ ways.
(2) Now order (say, left to right) both of the two subsets of students arbitrarily, this can be done $(7!)(3!)$  ways.
Each admissible arrangement of students corresponds to a unique sequence of choices in (1) and (2). So there are $56(7!)(3!)$ ways to do it.
The assumption in crush3dice's answer that the students sit in a circle is just as reasonable as the assumption that I am going to make, that the students sit in a row of chairs numbered, from left to right as 1, 2, ..., 10.
My approach is to favor brute force over elegance of thought in this one-off (small) setting.  If instead the # of chairs was (for example) 1000, then I would look for a pattern and then try for elegance.
Suppose that you select chairs 1, 3, 5, and decide that the 3 troublesome students (refer to them as a, b, and c) will sit in those 3 chairs.  There are 3! ways of permuting these 3 students among these 3 chairs and 7! ways of permuting the other 7 students among the other 7 chairs.
Therefore, the exact answer is $3! \times 7! \times T$, where $T$ represents the total # of ways of selecting 3 chairs out of 10 so that none of the three chairs is next to each other.
Now for the brute force:
Rather than try for elegant analysis, why not manually enumerate $T$.
Let $L$ = the # of the lowest of the three chairs. 
Let $M$ = the # of the middle of the three chairs. 
Let $N$ = the # of the largest of the three chairs.
Clearly, $1 \leq L \leq 6.$
That is if $L = 7$, there is no way of selecting 2 chairs among chairs 8, 9, and 10 so that the 2 chairs are apart both from each other and chair 7.
If $L$ and $M$ are set then $N$ must be in $\{(M+2), \cdots, 10\}.$
If $L$ is set, then $M$ must (similarly) be in $\{(L+2), \cdots, 8\}.$
Since you know that $L$ is in $\{1,2,3,4,5,6\},$ then all you have to do to enumerate $T$ is to count.
Addendum : elegance rears its ugly head
Since the following idea just came to me, I might as well share it.  There is a shortcut to enumerating $T.$
As analyzed, $M$ must range from 3 through 8, inclusive.
For any specific $M$ in this range, $L$ must range from 1 through $M-2$ and $N$ must range from $M+2$ through 10.
Therefore, $T = \sum_{M=3}^8 \,[(M-2) \times (9-M)].$
Addendum-2 : A symmetry argument.
By symmetry, the # of arrangements with $M=3$ equals the # of arrangements with $M=8.$  A similar argument relates $M=4$ to $M=7$ and $M=5$ to $M=6.$
Therefore, $T = 2 \times \left\{\sum_{M=3}^5 \,[(M-2) \times (9-M)]\right\}.$