# Critical combinatorics problem

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1. Dec 27, 2015

### Julian102

There are 9 doors in a house.Each door needs a password to open.Passwords can be of at least 4 digits and 10 at most . In how many ways password can be used to open at least one door .

2. Dec 27, 2015

### andrewkirk

I'm afraid your question is unclear.
Are you asking how many different possible passwords there are? If so then the number of doors is irrelevant.
Are you asking for the probability that a random sequence of between 4 and 10 digits will open at least one door? If so then 'how many ways' is not the way to ask that question.
Or are you asking something else? If so, what?

3. Dec 28, 2015

### haruspex

As Andrew posted, this is most unclear. Do you mean, perhaps, that given some set of (presumably distinct) passwords that has been assigned to the doors, how many different passwords would be able to open at least one?
If so, it rather depends what lengths of passwords were actually used. It also assumes that a match can arise anywhere in the sequence, e.g. if a given door has the password abcdef then the key abcabcdefx will open it (as well as opening door with password defx).
So, please clarify the question and post an attempt.

4. Dec 31, 2015

### Julian102

True.It came in my exam. You know , sub continental English is quite poor. I provided with the same
There can be (10^4+10^5+............10^10) passwords. Total ways =(10^4+10^5+............10^10 +1) Hence these are the total ways .But we add 1 because there is a way where no password can be used(I forgot to mention in the question.....sorry for that). Hence we can open the doors so that at least 1 door is opened in every case. Let x=(10^4+10^5+............10^10 +1)C1+(10^4+10^5+............10^10 +1)C2 + ...................+(10^4+10^5+............10^10 +1)C9 = (10^4+10^5+............10^10 +1)^9 - 1

5. Dec 31, 2015

### andrewkirk

Julian, here are some comments just to help you with how these things would be expressed in conventional mathematical writing.
This is the number of different possible passwords a door could have, and it includes the option of No Password.
Let S be the unordered set of all passwords used by the nine doors, where we denote a null password by the word NULL. Note that S has at most nine elements, but may have less if some doors have the same password. S must have at least one element because even if all doors have no password, we have S={NULL}.
Then the expression you have labelled as x is the number of different possible values that S can have.

I can't see any way of relating this to something about 'how many ways are there to open at least one door' though.

Was the question originally put to the students in English, or in an Indian language? If it was originally put in English, and was similar to what you have written, it sounds like you have the misfortune of having a lecturer that is teaching in a language that she/he does not understand well enough to do the job. I think it would work better if they taught in their native language instead.