Need explanation for a combinatorics riddle (full answer provided) 
11 people in a certain company have access to a safe. The company
  owner wants any group of six people out of the 11 to open the safe,
  But no five-person group can open the safe itself. To achieve this
  goal he decided to put more than one lock on The safe, and give each
  person keys only to some of the locks.
How many locks he has to put on the safe and how many keys each person
  will have to achieve his goal (the company owner wants to reduce the
  number as much as possible The locks, and as much as possible reduce
  the number of keys each person receives)?

Answer:
Each subgroup of 5 people will not be able to open the safe, so each subgroup should have a lock so that the members of the group do not have a key for it.
On the other hand, a key for the same lock is shared for all but 5 members of the subgroup. We achieved two goals in this: each sub-group of 5 people could not open the safe and any subset of 6 you can.
so we need $\binom{11}{5}$ Locks and $\binom{10}{5}$ keys
My question: 
Can I get more elaboration on the answer?
 A: Let’s rephrase this as a question about sets. Let $K$ be the set of all keys (that unlock one of the locks on the safe) and $K_i$ be the set of keys held by person $i$, $1\le i\le 11$. Then we have that $K_i\subset K$. The other conditions of the problem require that
$$K_u\cup K_v\cup K_w\cup K_x\cup K_y\cup K_z=K$$
for all distinct $u,v,w,x,y,z$, and
$$K_v\cup K_w\cup K_x\cup K_y\cup K_z \ne K$$
for all distinct $v,w,x,y,z$.
Solution: The pigeonhole principle will help us out a lot here.
If, for any particular $k\in K$, at least $6$ of the $11$ people hold the key (that is, $k\in K_i$ for exactly $5$ distinct values of $i$), then it is guaranteed by the pigeonhole principle that any group of $6$ people will contain at least one person holding $k$. Conversely, if there exists a key $k\in K$ such that fewer than $6$ of the $11$ people hold that key, then it will be possible to select $5$ people none of whom hold the key, making them unable to open the safe and violating a necessary condition. Therefore, we may conclude that at least $6$ people hold every key.
We may use similar reasoning to show that at most $6$ people hold every key $k\in K$.
Finally, we reach the conclusion that there is a one-to-one mapping between keys $k\in K$ and five-person groups (i.e. for every five-person group, there is exactly one key $k$ that is not held by any person in that group, and is held by all $6$ people not in that group). This entails that there must be $\binom{11}{5}$ keys, as desired.
However, I believe that $\binom{10}{5}$ is not the correct number of keys. Since each key is held by $6$ people as explained above, there should be $6\cdot \binom{11}{5}$ keys.
A: Each group of $5$ people should have a lock they cannot open.  There are $11 \choose 5$ groups of $5$ people, so put $11 \choose 5$ locks on the safe, one for each group of five people and make sure that none of those five have the key to the one that corresponds to that group.  
I disagree with the number of keys.  We have ${11 \choose 5}$ locks.  Each lock has six keys which are distributed to the people who are part of the group of $5$ assigned to not be able to open the lock, so there are $6{11 \choose 5}=2772$ keys.  This does not equal ${10 \choose 5}=252$  Each person does get $252$ keys because they are not a part of ${10 \choose 5}=252$ groups.  There are $10$ other people and $10 \choose 5$ ways to choose a group that they are not part of.
A proper solution would also show that there is no other solution with fewer locks and keys.
A: 
Each subgroup of 5 people will not be able to open the safe, so each subgroup should have a lock so that the members of the group do not have a key for it.

$11 \choose 5$ is the number of different five person subgroups in a population of eleven employees. You need this many locks to cover every possible instance of five people being unable to open the door. The same lock can't be used to lock out two different five person subgroups, else it would lock out at least six people.

On the other hand, a key for the same lock is shared for all but 5 members of the subgroup.

Again, since we can't lock out a six person group, everyone not in a five person subgroup must have a key to the lock that locks out that group. Everyone gets the number of keys equal to the number of five person subgroups they’re not in. $10 \choose 5$ is the number of five person subgroups that don't include a particular person. From the ten people who arent person X, choose five of them.

We achieved two goals in this: each sub-group of 5 people could not open the safe and any subset of 6 you can.

This solution satisfies the stated goal.
The text hints that this is the minimum solution without laying out the above logic. In any case, the answer is correct and the proposed solution is wildly impractical.
