$n-$circular arrangement problem Find the number of ways to arrange $n$ people in a circle so  that $3$ people are separated.
My approach: The number of ways to arrange $n$ people in a circle is $(n - 1)!$. If the $3$ people are together, the number of arrangement is $(n - 3)!$. The $3$ people can rearrange themselves in $3!$ ways, the number of ways for the $3$ people together is $3!(n - 3)!$. Therefore, the number of ways so that none of the $3$ people is seated together is $(n - 1)! - [3!(n - 3)!]$. Is that correct? If not, where did I go wrong? For instance, 4 girls and 3 boys to be arranged in a circle so that none of the boys is together. In this case, we have $(7 - 1)! - [3!(7 - 3)!] = 576$.
Any help is appreciated.

Since 3 people must not sit together the no.of of ways to choose 3 people from $n$ is  here $\binom{n}{3}$ and since we can arrange them in 2! ways (in a circle) it will be $\binom{n}{3} 2!$ . And the remaining no.of of people are $n-3$ and since there are 3 spaces between 3 people (on a circle) so no.of solutions of $x_1 + x_2 + x_3 = n-3 , x_1 , x_2 , x_3 \geq 1 \Rightarrow $ solutions of $ y_1 + y_2 + y_3 = n-6$ viz. $\binom{n-4}{2}$ and the number of ways to arrange $n-3$ objects in a row is $(n-3)!$. So finally we get $ \binom{n}{3} 2! (n-3)! \binom{n-4}{2}$.
I hope this answer is fine. Any tiny errors, correct me!!
