Locks for a four-of-seven safe

HardCombinatorics Brainteasers~10m

A firm's $7$ employees want a safe that opens only when at least $4$ of them agree. They install locks, distributing keys so that every group of $4$ can open all locks but no group of $3$ can. What is the minimum number of locks required?

Show hints (2)+

Each group of $3$ must be locked out by some lock whose key they all lack.
One lock per $3$ -subset (keys to the other $4$ ): $\binom{7}{3}=35$ .

Answer

Reveal answer →

Create a free account to see the full worked solution and track your progress.

Create a free account

Interview format: Market Maker, Systematic & Data-Driven Trading

Locks for a four-of-seven safe

Answer

Related questions