From Y.A. Rozanov. Probability Theory: A Concise Course: Chapter 1. Problem 6.

Suppose $n$ people sit down at random and independently of each other in an auditorium containing $n+k$ seats. What is the probability that m seats specified in advance $(m \le n)$ will be occupied?

My Solution:

First, there are $ n+k \choose n $ ways to choose n seats from n+k seats. That is the denominator.

What I want to find is the number of ways the m seats can be occupied by n people where there are more people than m and n+k total seats.

I start by thinking of the m seats as baskets that can be filled with n balls. What is the number of ways to fill m baskets with n balls? The first basket can be filled n different ways. The second basket can be filled n-1 different ways. The mths basket can be filled n-m+1 different ways. Therefore, by the usual rule for sampling without replacement, all the m baskets can be filled $\frac{n!}{m!*(n-m)!)}$ different ways. But the other n+k-m baskets (these are the "seats" leftover from the n+k total after m have been selected) can be filled also. The question now is, how many ways can these be filled given that that all the m seats are filled? There are n-m left over balls. And there are n+k-m left over baskets. So I believe these can be filled $n+k-m \choose n-m$ ways.

So my answer is:

$\frac{\frac{n!}{m!*(n-m)!)}{n+k-m \choose n-m}}{ n+k \choose n }$

I'm not confidant in this answer. :(

Sorry that I re-expressed it in terms of baskets instead of seats, but that makes it easier for me to think about.


2 Answers 2


The denominator is right. There are $\binom{n+k}{n}$ ways to choose $n$ seats, and all choices are equally likely.

We have $m$ seats specified in advance. Call them the marked seats. How many ways can seats be chosen so that all the $m$ marked seats are chosen?

We have no choice about the $m$ seats, they must be occupied. So our only freedom is in choosing $n-m$ seats from the $n+k-m$ unmarked seats. There are $\binom{n+k-m}{n-m}$ ways to do this. Now divide.

Remark: You identified the denominator correctly, and got a term $\binom{n+k-m}{n-m}$ in the numerator. But you had an additional term that should not be there.

  • $\begingroup$ But, if the n people are all distinct, then shouldn't the different arrangements of people in the m seats be counted as distinct? The first person might sit in the first of the m seats or the last, for example. $\endgroup$
    – student
    Commented Jan 27, 2013 at 3:39
  • $\begingroup$ If we are going to worry about who sits where, then the denominator will be quite different. When we divide "favourable" by "total" each must count the same sort of thing. The number $\binom{n+k}{n}$ counts the number of seat choices. $\endgroup$ Commented Jan 27, 2013 at 3:43
  • $\begingroup$ I think I see now. The number $ n+k \choose n $ does not regard order. It is the number of ways to pick out n things without regard to their order. So I should ignore the order in the numerator as well. Ok. Thank You very much. $\endgroup$
    – student
    Commented Jan 27, 2013 at 3:47
  • $\begingroup$ Well put. We could do it by counting the number of ways to assign the seats to individual people. That's OK as long as we are consistent. But it is a little more messy, so I prefer the way it was done, with denominator $\binom{n+k}{n}$. $\endgroup$ Commented Jan 27, 2013 at 3:51
  • $\begingroup$ Yes. I think it would be done (partly?) by multiplying the denominator by $n!$ because $n!$ is the number of ways to arrange the sub-population of size n. $\endgroup$
    – student
    Commented Jan 27, 2013 at 3:54

Here' how I understand it: The probability of the first of those m seats to be occupied is 1/(n+k), the second 1/(n+k-1), the third 1/(n+k-2), the mth 1/(n+k-m+1). Then the probability of all m seats being occupied is 1 / (n+k)!/m!= m!/(n+k)!


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