In a doctor's waiting room there are 14 seats. 8 people are waiting but there is someone with a bad cough In a doctor's waiting room there are fourteen seats in a row. Eight people are waiting to be seen, but there is someone with a very bad cough who must sit at least one seat away from everyone else. How many ways can this happen?
I am first supposing that the sick person is going to be at either one end of the row. Thus, there would only be twelve seats left (as the seat next to the sick person must be empty). Of these twelve seats available, we are going to pick seven for the remaining people, and we are going to do this twice (once for each end side of the row).
12!/7!5! + 12!/7!5!= 1 584
However, the sick person is not necessarily at one end, he could be in the middle too.
I choose the seat for the sick one first (of all the seats that are not at either end) and then subtract three from the remaining seats because the sick person must have one seat empty at each side, and of this quality (9) pick 7 and permute this 7 as order matters.
12!/1!11! x 9!/7!2! x 7!= 2 177 280
And them add both results: 1 584 + 2 177 280 = 2 178 864
However, this is not the correct result. The answer is 27 941 760 and I don't know how to get to this :( please help
 A: Your strategy is correct, but you made a couple of errors in executing your solution.
The person with the cough sits at an end of the row:  There are two ways to select the end of the row where the person sits.  This eliminates two seats, the place where the person with the cough sits and the seat next to it.  That leaves $14 - 2 = 12$ seats available to the remaining seven patients.  We can select seven of these $12$ seats in $\binom{12}{7}$ ways, and arrange those seven people in the selected seats in $7!$ ways, giving
$$\binom{2}{1}\binom{12}{7}7!$$
such seating arrangements.
The person with the cough does not sit at an end of the row:  There are $14 - 2 = 12$ places where the person with the cough can sit.  This eliminates three seats, the place where the person sits and the two seats adjacent to it, so there are $14 - 3 = 11$ seats available to the other seven people. We can select seven of these $11$ seats in $\binom{11}{7}$ ways, and arrange those seven people in the selected seats in $7!$ ways, giving
$$\binom{12}{1}\binom{11}{7}7!$$
such seating arrangements.
Total:  Since the two cases above are mutually exclusive and exhaustive, the number of admissible seating arrangements is
$$\binom{2}{1}\binom{12}{7}7! + \binom{12}{1}\binom{11}{7}7!$$
