A round table probability question Hi guys I am writing my P exam for the second time and I remembered two question that confused me when writing the exam. I asked my prof. but it confused him as well. For simplicity I will ask one question here and post the other one after.
So if you could please help me I would really be greatful.
Question:
You have 3 Actuaries, 4 Social Workers, and 3 clients. They all sit at a ROUND table. What is the probability that NO actuaries sit beside each other.
My attempt:
I labelled 3 A, 4S, 3C. Then I labelled 1 to 10 on a circle. I placed A in one spot and I knew no other actuaries sit beside each other so we have 7 spaces left. so now how can we place those other 2 actuaries so they dont sit beside each other.
Then I got confused.Please help out.
Thank you

Let's think of putting all $10$ people in a line arrangement and then having the end points meet to make the table arrangement.  For starters there are $10!$ ways to do this. This will be our denominator. 
First we line up the $7$ non-actuaries. There are $7!$ ways to do this. Now we seat the $3$ actuaries. We can seat them in any of the $6$ spots between the non-actuaries plus one of the end seats (but not both since in the table arrangement they'd be next to each other). So there are $7$ choices for the first actuary, $6$ for the second and $5$ for the third.  In total we have:
$$7!\times7\times6\times5$$ 
ways to do this and likewise when allowing one of the actuaries to sit at the other endpoint. Summing these up we get twice our number above. But then in total this is double counting all of the arrangements when the actuaries are all on the inside (i.e. none are at the end seats in the line arrangement). The count of these ways is $7!\times6\times5\times4$. So in total we have the probability as:
$$\frac{2\times7!\times7\times6\times5-7!\times6\times5\times4}{10!}$$
$$=\frac{7!\times6\times5\times2(7-2)}{10!}$$
$$=\frac{300}{720}$$
$$=\frac{5}{12}$$
Good luck on your exams. 
