In how many ways we can rearrange 2 people A and B where their distance is at least 4 seats apart from each other in 8 seats? I tried to follow the round table permutation problem but I got lost because I know that there are 10 possible places where they have to appear inside permutation but how to take the number of cases where they are not at this distance.
2026-05-05 15:57:45.1777996665
Permutation , specific distance between elements
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1
For a linear arrangement:
For the moment, suppose that A is to the left of B.
If they are four seats apart, A must be in one of the first four seats.
If they are five seats apart, A must be in one of the first three seats.
If they are six seats apart, A must be in one of the first two seats.
If they are seven seats apart, A must be in the first seat.
They cannot be more than seven seats apart, so there are $4 + 3 + 2 + 1 = 10$ seating arrangements in which A is to the left of B and they are at least four seats apart. By symmetry, there are ten seating arrangements in which B is to the left of A and they are at least four seats apart. Consequently, there are $20$ such seating arrangements.
For a circular arrangement:
Given that there are eight seats, they can only be at least four seats apart if they are opposite each other. If the seats are distinguishable, then there are ten such seating arrangements, one for each of the places A can sit. If seating arrangements are considered to be invariant under rotation, then there is only one possible seating arrangement, namely that A and B sit in opposite seats.