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Possible Combinations on a Push Lock

Date: 4/17/96 at 15:53:5
From: Anonymous
Subject: lock problem

A combination lock consist of a sequence of "pushes" and each push 
is a collection of buttons (1 or more) that are pushed together.  
The digits available are 1 thru 5.  Furthermore, in each 
combination a digit can occur only once.  You need at least two 
pushes to open the lock. How many different combinations are 

This is as far as I got......

	5! = 120 combinations

This seem to be the right answer but I feel that there is more to 
it. I am not sure I understand the problem fully.  Any help to the 
solution of this problem would be most appreciated.

Date: 4/18/96 at 20:2:40
From: Doctor Ken
Subject: Re: lock problem

Hey there -

If I understand this problem correctly, you can't use any number 
twice in a combination, even if they are in different pushes.  So 
the following would be okay:

4, then 
5 and 2, 
then 3 and 1.  

But this wouldn't:

4, then
5 and 4, then
3 and 4, because you re-use 4.

Notice that this limits us to at most 5 pushes, because after that 
you're going to have to start repeating.  Anyway, let's get some 
notation: if a push consists of 5 and 4, for example, let's write 
that as (5,4).  And let's use the + sign to seperate consecutive 
pushes, so that the first combination we had would be written as 
(4) + (5,2) + (3,1).

Anyway, I think this is a harder problem than one that will be 
able to be solved with just one case: we're going to have to break 
it down.

So let's break it down according to how many digits we use in each 
combination.  We know we need to use at least 2, because we need 
to have at least 2 pushes.  So how many different combinations can 
we have that use exactly 2 digits?  Well, we'll have 5 choices for 
the first one and 4 choices for the second (order does matter in 
this case).  So that's 20 ways of choosing our digits.  Then how 
many different combinations can we get with these 2 digits?  It 
seems pretty obvious, in this case, that all we can do is push one 
button, then the other (i.e. we can't push 2 at the same time).  
So we'll just get 20 different combinations that use 2 digits.

Now let's look at 3 digits, which is a little more interesting.  
How many ways can we choose which 3 digits to use?  Well, again, 
it's 5*4*3 = 60.  

But now we have several choices for whether these buttons are 
pushed one at a time, or together with others.  For instance, if 
we chose the digits 1, 2, and 3, in that order, we could have 
(1) + (2) + (3), or we could have (1,2) + (3), or (1) + (2,3).  
It looks like these are the only possible combinations that use 
only 1, 2, and 3 in that order.  So since every selection of 3 
digits will have these 3 different combination orders, there will 
be a total of 60 * 3 = 180 combinations that use 3 digits.

Can you do the cases where we use 4, and then 5 digits yourself?  
They get tougher the more digits you have, but try to get it 
yourself before you write back for more help.  This is a neat 
problem, and it will be worth it if you solve it yourself.

-Doctor Ken,  The Math Forum

Associated Topics:
High School Permutations and Combinations

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