The problem involves determining the number of ways passwords can be used to open at least one of nine doors, each requiring a password of between 4 and 10 digits. The original poster presents a combinatorial question related to password possibilities and their application to the doors.
There is ongoing clarification regarding the original question, with participants seeking to understand the intent behind the inquiry. Some have provided insights into the mathematical expressions involved, while others express confusion about the phrasing and implications of the problem.
Participants note potential language barriers affecting the clarity of the question, suggesting that the original phrasing may have contributed to misunderstandings in the discussion.
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?Julian102 said: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 .
haruspex said: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.
True.It came in my exam. You know , sub continental English is quite poor. I provided with the sameandrewkirk said: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?
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 - 1andrewkirk said: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?
This is the number of different possible passwords a door could have, and it includes the option of No Password.Total ways =(10^4+10^5+...10^10 +1)
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}.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