# Solution gives wrong answer to probability problem

Great Northern Airlines flies small planes in northern Canada and Alaska. Their largest plane can seat 16 passengers seated in 8 rows of 2. On a certain flight flown on this plane, they have 12 passengers and one large piece of equipment to transport. The equipment is so large that it requires two seats in the same row. The computer randomly assigns seats to the 12 passengers (no 2 passengers will have the same seat). What is the probability that there are two seats in the same row available for the equipment?

This problem was posed on Brilliant last week, and now that the official solution is posted, I would like to know why my solution gives wrong result. Here is my solution:

The number of ways we can seat 12 (identical) people on 16 seats is $16\choose 12$. Now the equipment can occupy any of the rows (8 possibilities), and the 12 people must be seated on the remaining 14 seats ($14 \choose12$ ways). Thus the desired probability is $$\frac{8 {14\choose12}}{16\choose12}=\frac25$$ The official solution gives $\frac5{13}$, but I don't understand what is wrong with mine.

-
Note that the link you provided is unique to you. You need to use the "Share this problem" link for others to view the problem properly. – Calvin Lin May 10 '13 at 0:55
@CalvinLin Oh, I'll keep that in mind. – Dave May 10 '13 at 12:02

Think of it this way - you can place four seats, with the restriction that at least one pair must be together in a row. So subtract off the number of ways you can have no pairs from the total number of combinations $${16\choose4}-{8\choose4}\cdot2^4 = 700$$ as each of the four rows containing an empty seat can have it in either of the two seats. So the probability is $$\frac{700}{1820}=\frac{5}{13}$$ (note that this means that you were double-counting 28 combinations... which makes sense, as $28 = {8\choose2}$, and that's the number of ways you can have two pairs of empty seats)