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

The way you are asking the question it seems to me that whether it is social workers or clients has no relevance. So basically we have 10 people seated around a round table and 3 specific people must not sit next to each other.
So label the seats 1 through 10 as you suggested: You can choose 3 out of 10 in $\binom{10}{3}=120$ ways.
You can construct all ways to choose 3 non-adjacent seats by choosing $3$ out of $7$, say $i<j<k\in\{1,2,...,7\}$ and map them to $i+1,j+2,k+3$ which points to three seats among $\{2,3,...,10\}$ separated by at least one seat each. This produces $\binom{7}{3}=35$ succesful patterns.
But this does not account for patterns where the first seat chosen is $1$. These paterns can be constructed by fixing the first seat as $i=1$ and then choose $j,k\in\{2,3,...,7\}$ and mapping these to $j+1,k+2$ in $\{3,...,9\}$. This produces another $\binom{6}{2}=15$ succesful patterns.
To sum up we have $35+15=50$ succesful patterns out of $120$ possible patterns. Thus we have
$$
P=\frac{50}{120}=\frac{5}{12}
$$
probability of succes.
