$n-$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.

Comment of @Christian Blatter might be helpful to you in counting the missing cases. However, here's an alternate method.
Let the three people be $P_1,\ P_2,\ P_3$ and the number of people between $P_1P_2,\ P_2P_3,\ P_3P_1$ be $x_1,\ x_2,\ x_3$ respectively. Now, we've to find number of positive integral solutions of the equation $$x_1+x_2+x_3=n-3$$ Also, the $n-3$ people can arrange themselves in $(n-3)! $ ways and the three people in $2! $ ways.
The total no. of ways are $\displaystyle{{n-4}\choose 2}×(n-3)!×2! =(n-3)!(n-4) (n-5) $.
