Possible Combinations on a Push LockDate: 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 possible? 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 |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/