I am solving one simple problem just for fun. The problem is as follows.

$n$ people have to be seated randomly in a cinema hall which has $(n+k)$ seats.
What is the probability that in this process, some fixed $m$ places ($m \leq n$) get occupied/taken?

The book gives this answer:

${{n} \choose {m}} \cdot {{n-k-m}\choose{n-m}} / {{n+k} \choose {n}}$

There are 2 things I don't like here:

1) I think that ${{n-k-m}\choose{n-m}}$ is obviously a typo and should read ${{n + k-m}\choose{n-m}}$ Why? Well to say the least $n-k-m \leq n-m$ so this binomial coefficient does not make much sense here.

2) This answer seems to treat the $n$ people as indistinguishable which is strange. Since these are $n$ people/persons these are $n$ different people (we can never have identical people, let alone $n$ of them, even twins are different persons). Thus it makes more sense to treat the people as distinguishable. So the order in which the people take the places also matters, right?
But in that case I am getting this answer:

${{n} \choose {m}} \cdot m! \cdot {{n + k-m}\choose{n-m}} \cdot (n-m)! / ( {{n+k} \choose {n}} \cdot n! ) $

What is your opinion of these two items? Am I correct in both or at least in one of them?


The term $\binom{n}{m}$ counts the number of ways $m$ of the $n$ people can occupy the specified $m$ seats. That leaves $n + k - m$ seats available. The remaining people can occupy $n - m$ of them. Hence, the number of favorable cases is $$\binom{n}{m}\binom{n + k - m}{n - m}$$ as you suspected. The denominator $\binom{n + k}{n}$ represents the number of ways the $n$ people can select $n$ of the $n + k$ seats in the theater.

The people are distinguishable in the sense that we are selecting which $m$ of the $n$ people sit in the designated seats. What does not matter is which particular person sits in which particular seat, just which seats are occupied.

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  • $\begingroup$ Hm... But say $m=3$ and you pick the persons $p_1,p_2,p_3$ to seat there (in the pre-chosen seats $4,5,6$ let's say). Then you make no distinction between the person orderings $(p_1,p_2,p_3)$, $(p_2,p_1,p_3)$ and so on. And this means either you assume $(p_1,p_2,p_3)$ = $(p_2,p_1,p_3) = ... $ which means you treat the people as indistinguishable or you assume the seats are indistinguishable (which is even more strange probably since we all know seats in cinemas have some identifiers). Isn't it so? Or ... am I completely deluded here?! $\endgroup$ – peter.petrov Mar 24 '18 at 21:11
  • $\begingroup$ Since the only question is occupying m seats, whether or not the people are distinguishable is irrelevant. $\endgroup$ – herb steinberg Mar 24 '18 at 21:13
  • $\begingroup$ Keep in mind that we are selecting which $m$ people sit in the designated seats, so the people are distinguishable. What does not matter is which of the $m$ people sits in which designated seat. $\endgroup$ – N. F. Taussig Mar 24 '18 at 21:14
  • $\begingroup$ OK... well... if the people are distinguishable how come it does not matter who sits where? The problem does not say this explicitly so ... How do we know it? If you ask me apparently it does matter which of the two assumptions (matters or does not matter) the problem means. Apparently we get different answers with the two assumptions. $\endgroup$ – peter.petrov Mar 24 '18 at 21:17
  • $\begingroup$ We want to find the probability that the fixed $m$ places are taken. The problem does not ask us to find the probability of a particular arrangement of seats. $\endgroup$ – N. F. Taussig Mar 24 '18 at 21:18

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