Locker Combo uses 3 numbers w/ consec. numbers different; find least number of....

Kas Joy

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There are 1200 students at a school. Each student has a locker. A lock combination uses three numbers, such that consecutive numbers must be different. What is the least number of values that must be on the lock in order for all students to have a unique combination?

I understand that the answer would be along the lines of...
P(least # of locker combos)=(12!)+(11!)+(11!)
=558835200
But I need more explanation as to why this is? It's sort of confusing to me
 
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There are 1200 students at a school. Each student has a locker. A lock combination uses three numbers, such that consecutive numbers must be different. What is the least number of values that must be on the lock in order for all students to have a unique combination?

I understand that the answer would be along the lines of...
P(least # of locker combos)=(12!)+(11!)+(11!)
=558835200
But I need more explanation as to why this is? It's sort of confusing to me
By what reasoning did you obtain this value? ;)
 
Well, we know that if every student has their own locker combination, there will be only 1200 unique combinations. That strongly rules out your proposed answer of 558 million and change. Instead, let's think about some smaller answers. If there's three numbers, there will be exactly one unique combination. What if there's four numbers on the lock? How many combinations then? With five numbers? We can see that it's quickly going to get very difficult to manually count out the number of unique combinations, so we need a formula to help us out. In your class leading up to the assignment of this exercise, you should have learned about the binomial coefficient formula. It's sometimes called "n choose k" as well. It looks like this: \(\displaystyle \begin{pmatrix}n\\ k\end{pmatrix}\). Does this formula ring a bell? What happens if you try to apply it to this exercise?
 
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There are 1200 students at a school. Each student has a locker. A lock combination uses three numbers, such that consecutive numbers must be different. What is the least number of values that must be on the lock in order for all students to have a unique combination?
Suppose that we take three distinct digits at random from the ten digits zero to nine (order does not matter). How many ways can that be done is no two selected digits are consecutive.
Well here a model.
| | |_____ __0__0__0__0__0__0__0__
there three selected digits and seven non-selected. How ways can we put the three
into the eight blanks,
__0__0__0__0__0__0__0__ ?

Here is one way
__0_|_0__0__0_|_0__0__0_|_
and that selects the digits \(\displaystyle 1,~5~\&~9\). You want to be sure that you understand the model.

What is the model for selecting \(\displaystyle 0,~6~\&~8~?\) Is it
_|_0__0__0__0__0_|_0_|_0__ ?

The answer to the above is \(\displaystyle \dbinom{8}{3}=56\). Now your question is about locks where order does matter.
Thus there are \(\displaystyle (3!)\dbinom{8}{3}=336\), three digit combinations with no two consecutive digits.

That is far too few. So how many?
Well there three selected numbers and N-3 non-selected. The non-selected create N-2 blanks. As in the model:
\(\displaystyle (3!)\dbinom{N-2}{3}=?\). So how do you find \(\displaystyle N\) so that \(\displaystyle (3!)\dbinom{N-2}{3}\ge 1200\)
 
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