So there is this problem
A carousel has eight seats, each representing a different animal. Eight girls are seated on the carousel facing forward $($each girl looks at another girl's back$)$. In how many ways can the girls change seats so that each has a different girl in front of her? How does the problem change if all the seats are identical?
So I was wondering whether this is a circular permutation problem or just a linear one and how can you tell? Would you have to use Inclusion-Exclusion Principle to solve this problem?
Since each seat represents a different animal, the question is not a circular permutation question rather a linear permutations. We will name the girls by the numbers as they sit first time so that $1$ faces $5, 2$ faces $6, 3$ faces $7$ and $4$ faces $8$. We use inclusion-exclusion principle. Let $S$ be the set of all permutations of $\{1, 2, 3, 4, 5, 6, 7, 8\}$ and
$T_1$ be the set of permutations in S such that $1$ faces $5$;
$T_2$ be the set of permutations in S such that $2$ faces $6$;
$T_3$ be the set of permutations in S such that $3$ faces $7$;
$T_4$ be the set of permutations in S such that $4$ faces $8$.
Note that $|S| = 8!$. To compute $|T_1|$, seat $1$, there are $8$ ways Then sit $5$ to the opposite of $1$ (only one way). Then seat the remaining $6$ girls. Then $|T_1| = 8\cdot 6!$. Similarly, $|T_2| = |T_3| = |T_4| = 8 \cdot 6!$. To compute $|T_1\cap T_2|$, seat $1$ and $5$ first $(8$ ways$)$, then seat $2$ and $6$ by just seating $2 ( 6$ has to face $2)$. There are $6$ choices and $4!$ ways to seat the remaining $4$ girls Thus, $|T_1\cap T_2|=8\cdot4\cdot4!$.
Similarly, $T_i\cap T_j|=8\cdot4\cdot4!\forall i\ne j$. In $T_1\cap T_2\cap T_3$ first seat $1$ and $5 (8$ ways$)$ then seat $2$ and $6$, $(6$ ways) followed by seating $3$ and $7 (4$ ways$)$ then remaining $2$ girls $(2$ ways$)$. So, $T_1\cap T_2\cap T_3$ and any other intersections of three sets have $8\cdot6\cdot4\cdot2!$ ways.
Similarly, $|T_1\cap T_2\cap T_3\cap T_4|=6\cdot4\cdot2$.
Thus by the inclusion-exclusion principle, the number of ways to change the seating so no boy faces the same person is$$8!-\dbinom41\cdot8\cdot6!+\dbinom42\cdot8\cdot6\cdot4!-\dbinom43\cdot6\cdot4\cdot2!+\dbinom44\cdot8\cdot6\cdot4\cdot2$$
If all seats are identical, the question becomes a circular permutation question. Now we let $S$ be the set of all circular permutations of $\{1, 2, 3, 4, 5, 6, 7, 8\}$ and
$T_1$ be the set of permutations in S such that $1$ faces $5$;
$T_2$ be the set of permutations in S such that $2$ faces $6$;
$T_3$ be the set of permutations in S such that $3$ faces $7$;
$T_4$ be the set of permutations in S such that $4$ faces $8$.
Note that $|S|=\dfrac{8!}{8}=7!$. To calculate $|T_1|$, , seat $2$, and $5$ first (only one way), then seat the remaining $6$ girls. Then $|T_1=6!$. Similarly, $|T_2|=|T_3|=|T_4|=6!$. To compute $|T_1\cap T_2$, seat $1$, and $5$ first, then seat $2$ and $6$ by just seating $2 ($then $6$ has to face $2)$, there are $6$ choices and $4!$ ways to seat the remaining $4$ girls. Thus $|T_1\cap T_2|=6\cdot4!$.
Similarly, $|T_i\cap T_j|=6\cdot4!\forall i\ne j$. In $T_1\cap T_2\cap T_3$ first seat $1$ and $5 ($one way$)$ then seat $2$ and $6$, $(6$ ways$)$ followed seating $3$ and $7 (4$ ways$)$ then remaining $2$ girls (two ways). Thus $T_1\cap T_2\cap T_3$ and any other intersections of three sets have $6 · 4 · 2!$ elements. Similarly, $|T_1\cap T_2\cap T_3\cap T_4|=6\cdot4\cdot2$.
Thus by the inclusion-exclusion principle, the number of ways to change the seating so no boy faces the same person is $$7!-\dbinom41 6!+\dbinom42 6\cdot4!-\dbinom 43 6\cdot4\cdot2+6\cdot4\cdot2\cdot1=2880$$
Note that each of the numbers we computed for identical seats are just the numbers of those with different animals divided by $8$. This reflects that the first case (seat with different animals) the seating are linear permutations, while the identical cases are circular permutations and the total numbers differed by a factor of $8$.