Maths Question under Permutations and Combinations?

Question

Five people jointly bought a lottery ticket which won the first prize. They decided to keep their ticket in a locker installed with a number of locks. Each person may hold the keys to more than one lock. What is the minimum number if locks required to serve the purpose that any three of the people together will be able to open the locker but any two of them together will not be able to to open it?

Answer 1

Its a great question.

Since, any combination of 3 ppl shud be able to open a lock, every lock's keys shud be given to 3 ppl. The number of ways 3 keys can be given to 5 ppl is 5C3 = 10

so we need 10 locks.

to make the above statement clearer, take person A and person B. they shudnt be able to open all locks without another person. suppose there is a lock (call it lock 1) which they cudnt open. So, neither A nor B has the key to it. Since, any third person shud be able to open it, key 1 is there with C, D and E. this is why every key is present with 3 out of 5 ppl. Hence, the number of locks required will be the number of ways we can get 2 ppl out of those 5 = 5C2 = 10.

Therefore, 10 locks.

Answer 2

This is a good question.

At least 3 people must hold the key to a given lock. Otherwise there would exist a group of 3 people that could not open the locker. We must now cover every combination of distributing 3 keys among 5 people.

Since C(5,3) = 10, that is the answer.

Labeling each key A-J, distribution will look like this:

1. ABCDEF
2. ABCGHI
3. ADEGHJ
4. BDFGIJ
5. CEFHIJ

Answer 3

vafdrbfndshljdkfjekdfjdshgjdvbdshafv?