Eight girls are seated around a carousel. In how many ways can they change seats so that each has a different girl in front of her?
I thought the answer was as follows: For $i = 2,\dots, 8,$ let $A_i$ be the circular permutations of girls such that girl $i$ looks at the back of girl $i-1$, and $A_1$ be the circular permutations of girls such that girl $1$ looks at the back of girl $8$. We need to subtract from the 8! of all possible seatings of girls the number $|A_1 \cup ...\cup A_8|$. Observe that $A_i = 8 \times 6!$ for each $i$. Then $$|A_i \cup A_j|= 8 \times 5!$$ $$|A_i \cup A_j \cup A_k|= 8 \times 4!$$ $$\vdots$$Thus, the answer would be $$8! - 818 \times 6! + 828 \times 5!-\dots -878 \times 1! + 8 = 13000.$$
However, my textbook says the answer is $1625$ (but gives no solution). Any ideas how that answer was arrived at?
As per @dEmigOd 's comment, you need to divide by 8 afterwards.
Think about it this way, $8!$ is the number of combinations seating the girls; 8 options for the "first place", 7 options for the "second place", and so on. However in a circle there is no ordering and so the first place could be any of the 8 places and it would be the same answer. We could shift our answer by one place and have the same order. We can shift to 7 other "linear" orders. Thus we have counted each ordering 8 times when in fact they are the same ordering on the circle.