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There are n people in line to enter a cinema that own n seats. Each of the n people have an allotted seat in the cinema hall where they are supposed to sit. The first person forgets his/her seat number on the ticket, so he/she decides to take a random seat to watch the movie. Every person after him/her hold the ticket and remember their seat number.(They are entering the cinema hall one by one.) If their seat on ticket is available, they will take this proper seat, otherwise they will choose a random available seat instead.

Now, in the actual question, it is asked what is the probability of the last person entering the cinema hall sitting in his allotted seat? But what I want to ask is that what are the total number of seating arrangements possible in the cinema hall?

Main Question: what are the total number of seating arrangements possible in the cinema hall?

I proceeded like this: If 0 persons were sitting in wrong places (not in their allotted seats), then only 1 case is possible because it is only possible if the first person sits in his allotted place. It is not possible that 1 person is sitting in a wrong place if he is sitting in a wrong place, then he is sitting in someone else's allotted seat. If 2 people are sitting in a wrong place, then the first person has n-1 choices of seats because he cannot sit in his allotted seat and the second person has to sit in the first person's allotted seat, so there are n-1 such cases. If three persons are sitting in the wrong place, then there are only (n-1)(n-2) possible cases. This is how I tried to proceed to find my answer. Am I proceeding in the correct direction? If not please do point out my mistake. Also, please show me an easier method to find the answer if you find any.

Thanks in advance.

WizardGamer44
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  • The question OP is not asking is usually stated as seating on an airplane, and in that guise has been seen on this website before. – Gerry Myerson May 22 '24 at 13:07
  • "Am I proceeding in the correct direction" Frankly, no. This direction leads only to frustration, tedium, and case-hell. You should use the approaches outlined elsewhere and not approach by direct counting as you are. – JMoravitz May 22 '24 at 13:27
  • @GerryMyerson According to your recommendation I checked out the problem and outline of the problem is identical to mine but the question here is the number of seating arrangements possible which makes this different from that one. – WizardGamer44 May 22 '24 at 17:30
  • As I wrote, Wizard, the question you are NOT asking is usually stated as seating on an airplane. I wrote nothing about the question that you ARE asking, but implicit in what I wrote is that it's possible that the question that you ARE asking was also raised in an earlier discussion here, so it might be worth your while to look at ALL previous posts on the airplane seating problem, to see whether any of them also discuss your question. – Gerry Myerson May 22 '24 at 23:03
  • Typing $${\rm number\ of\ airplane\ seating\ arrangements\ math\ puzzle}$$ into Google brings up lots of stuff, some of which MIGHT be relevant to the question you are asking, Wizard. – Gerry Myerson May 22 '24 at 23:08
  • Ok I will try that. – WizardGamer44 May 23 '24 at 04:32
  • This is NOT a duplicate of https://math.stackexchange.com/questions/5595/taking-seats-on-a-plane although some of the answers there give a hint as to how to answer the question here. Please, folks, read this question, and read that question, and see for yourself that they ask for two very different things. The older question asks for the probability that the last entrant gets her assigned seat, the current question asks how many different seating arrangements are possible. Not the same thing at all. – Gerry Myerson May 23 '24 at 06:16
  • @JMoravitz Could you show me a better method of solving this problem? – WizardGamer44 May 23 '24 at 09:53
  • Could anyone please tell me why this question has been closed?? – WizardGamer44 May 23 '24 at 10:34
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    @GerryMyerson I initially interpreted this as trying to find an alternate way of calculating the probability through means of direct counting. In such an event it is strongly recommended to abandon this approach as it is tedious and does not even lead to a solution as the different arrangements might not even be equally likely to occur. – JMoravitz May 23 '24 at 12:06
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    Regardless, the commentary and approaches in the linked question do lead to an elegant result here. Notice that every seating when treated as a permutation consists of fixed points and at most a single cycle of variable length that involves our drunkard who lost their ticket. As the patrons enter in some canonical order, you merely need to choose some (possibly empty) subset of them to fill out the cycle. As such, by my count there are $2^{n-1}$ arrangements. I still do not find this different enough for reopening. – JMoravitz May 23 '24 at 12:10
  • OK, Wizard, you have your answer (both as to why the question has been closed, and as to how to answer the question). – Gerry Myerson May 23 '24 at 12:17
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    Ok got it. Thanks a lot. – WizardGamer44 May 24 '24 at 12:56

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