Combinatorics question where someone has to be at least one seat away from anyone else? In a doctor’s waiting room, there are 14 seats in a row. Eight people are waiting to be seen.
There is someone with a very bad cough who must sit at least one seat away from anyone else. If all arrangements are equally likely, what is the probability that this happens?
My logic is this: Number the 8 people and let person 8 have the cough. We need to find the number if arrangements that person 8 is next to someone and then subtract this from the total to get the number of arrangements where he sits at least one seat away. 
So treat person 8 and person 7 as one ‘object’. There are 13!*2!/6! distinct arrangements of this (if you say the seats are identical objects). We can repeat this principle 7 more times, pairing up person 8 with person 6, then 5 etc. so we multiply the previous expression by 7 to get 121080960 permutations where the coughing person is next to someone. However, this is way too much.
How would you do this problem?
 A: Here is my first instinct. It shouldn't be too cumbersome to do the necessary calculations:

Either person 8 sits on one of the two end seats, or they sit somewhere in the middle. If they sit on an end seat, there are 12 seats freely available to the remaining 7 people. If person 8 sits somewhere in the middle, there are 11 seats available for the remaining people to sit.

A: There are $8 \cdot \frac{14!}{8! \cdot 6!}$ configurations possible in total, where the added $8$ arises from the fact that person 8 is 'distinguishable'.
If person 8 sits on the edges, this leaves $\frac{12!}{7! \cdot 5!}$ configurations open for the other people to sit ($12 = 14 - 1 - 1$, where $1$ is the edge seat and $1$ the seat next to the edge seat). We multiply this by two, as there are two edge seats.
If person $8$ sits on a non-edge seat, this leaves $\frac{11!}{7! \cdot 4!}$ configurations for the other people to set. We multiply this by twelve, as there are twelve non-edge seats.
The total amount of 'valid' configurations is therefore $ 2 \cdot  \frac{12!}{7! \cdot 5!} + 12 \cdot \frac{11!}{7! \cdot 4!}$.
A: I would try to place the coughing person at all possible 14 places and let the others to choose the places arbitrarily leaving the gaps around the person, so that the total number of arrangements is:
$$
2\binom{12}7{7!}+12\binom{11}7{7!},
$$
where the terms correspond to coughing person sitting either on one of two end seats or on any of 12 other seats, respectively. 
To find the corresponding probability the above number shall be divided by the total number of possible arrangements:
$$
\binom{14}88!.
$$
