Combinatorics Homework Problem: $n$ People Sit Down. What is the Probability that $m$ Seats are Occupied? 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.
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.
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)!