.
Given seven equal circles on a hexagonal grid, how many permutations are possible where circles
are always adjacent (repetitions such as inversions and reflections excluded)?
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I have another logic/reasoning, and my solution and my answer are different from that by @CPhill.
But before to start, I'd like to discuss the problem's formulation.
The problem says
Given seven equal circles on a hexagonal grid, how many permutations are possible where circles
are always adjacent (repetitions such as inversions and reflections excluded)?
I would re-formulate the problem in mathematically more appropriate way:
Given seven equal circles on a hexagonal grid, how many arrangements are possible where circles
are always adjacent (arrangements that differ by rotations, inversions, and reflections
are considered as indistinguishable).
Below is my solution for this modified formulation.
S O L U T I O N
We consider 7 circles as numbered from 1 to 7 - so the circles are distinguishable.
To start, let assume that the circle '7' is in the center.
Then, accounting for circular permutations (rotations), we have (6-1)! = 5! = 120 different arrangements.
There are 3 different axes of symmetry.
To account for reflections, we divide the number of 120 arrangements by 2 three times.
We get then 120 : 8 = 15 distinguishable arrangements.
We should not make an additional correction for inversion, since the inversion
is just accounted as the 180-degree rotation.
So, now we only need to multiply 15 by 7 to account for the fact that any of 7 circles can be placed in the center.
ANSWER. Under the given conditions (and with my modification) there are 15*7 = 105 different distinguished arrangements.
Solved.