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Open By Majority (Posted on 2004-03-03) Difficulty: 3 of 5
A group of five people want to put a set of locks on a chest and distribute keys to the locks amongst themselves in such a way that all the locks on the chest could be opened only when at least three of them were present to open it.

How many locks would be needed, and how many keys?

See The Solution Submitted by Brian Smith    
Rating: 4.1429 (7 votes)

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2 for all but 1 | Comment 37 of 45 |
No one may possess 3 keys.

As 3 of 5 need access,
then the minimum LOCK limit is 3;
having single keys only can be ruled out.

By combinations  [5!/(2!*3!)= 10] the suggestion is 10 keys.

The following table will suggest otherwise:

Table:        Person
Key    A    B    C    D    E
1        x          x     x    x
2        x    x           x            
3              x    x           x

In the table Key 1 seems to be over-committed.
Consider the removal of Key E1.

This leaves 9 keys that are needed, as per the table, but remove
any one of the Key1's from the top Key 1 row.

My proof is: (having removed A1, C1, D1 OR E1),
cover 2 alpha columns and test.

Solution:  9 keys, 4 people get 2 but one soul get 1 such the table
distribution is obeyed.
  Posted by brianjn on 2004-03-04 17:07:27
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