Screwy Pirates

Problem: those screwy pirates are at it again. this time there are 13 pirates and they need to protect their treasure chest. they decide that they should only be able to open the chest if the majority (at least 7) agree that it should be opened. they ask a locksmith to come and put a specific number of locks on the safe. every lock must be opened to open the chest. there can be multiple keys for each lock, but each key only opens one lock (i.e. no skeleton keys). the locksmith can give more than one key to each pirate. how many locks should the locksmith use and what strategy should he use to distribute the keys, such that only when a majority of the pirates agree can the chest be opened?

Somewhat technical answer:
Here is solution that works for any number of pirates and any number of pirates needed to open to chest.

Let T be the total number of pirates. Let N be the number of pirates required to open the chest.

The number of locks needed would be L = (T,N-1) = (T!)/[(N-1)! * (T-N-1)!]. The number of keys each pirate would have be K = (T-1,N-1) = (T-1)!/[(N-1)! * (T-N)!].

For this specific problem, the number of locks would be 1716 and the number of keys per pirate would be 924.

I know I haven't provided an explanation of why and how this system works. We'll leave that lengthy, involved explanation to the author.

Note: the notation (x,y) is the combination notation; I can't use the conventional combination notation in plain text. (x,y) basically asks the question "how many ways can you pick y objects from a group of x objects?" (x,y) = (x!)/[y! * (x-y)!]

Another easy way:

Here's a less elegant solution. How many different combinations of 6 pirates are there? 13 choose 6 is: 13!/6!7! That's a hell of a lot of locks. But you could say, put that many locks on the chest. And for each lock, you distribute keys to all the pirates EXCEPT the selection of 6 corresponding to that lock. Then any selection of 7 pirates is guaranteed to be open all the locks. For consider the first 6 of them. There will be exactly ONE lock that those six are unable, between the 6 of them, to open. But the seventh pirate will be able to open that lock. Because keys to that lock went out to everybody except those 6.

Also, any selection of 6 pirates is guaranteed to encounter exactly one lock that they are unable to open. (So any selection of fewer than 6 is guaranteed to encounter 1 or more locks they are unable to open...)

So that will do the trick, and the explanation of why it would do the trick is elegant. But I don't feel that it's an elegant solution, because of there being so many locks... Anyone have something better?