Suppose $k$ people de-board an airplane and get into a hall where they are assigned at most $n$ queues. The number of ways in which this can be done is $k! n^k$ or $n(n+1)\cdots(n+k-1)$?
From the discussion here, the second answer makes sense, but to me, the first one does.
Here's how I am arriving at the first answer: Suppose the people come down the airstair in one of the $k!$ ways; let that order be $p_1 \to p_2 \to \cdots \to p_n$ where $p_i$ is the $i$th person to de-board the plane. Now each $p_i$ has $n$ options (queues) to choose; he either goes to one of the empty queues (if one exists) or stands behinds someone. In total, are $k! n^k$ ways.
The other answer also seems to be correct. I think both the answers are correct; the only difference is in the assumption of people being similar objects (according to that answer) and dissimilar objects (according to my solution). Is my understanding correct?
There are four interpretations to the problem. There is one interpretation where $n(n+1)\cdots(n+k-1)$ is correct, and there is another where $k!n^k$ is correct. The problem is ambiguous on these two issues:
Can the people choose the order they deplane?
Can people cut in a queue, or are they required to join a queue at the back?
If the people can choose the order they deplane, but they must get in the back of any queue, then the answer is $k!n^k$, for exactly the reason you stated.
If they people cannot change the order they deplane, but they can enter a queue at any point, then the answer is $n(n+1)\cdots (n+k-1)$.
The first person to deplane has $n$ choices, entering any empty queue.
The second person to deplane has $n+1$ choices. They can enter at the back of any queue, or cut in from of the first person.
The third person has $n+2$ choices. They can enter at the back of any queue, or they can cut in front of one of the previous two people.
$\vdots$
The $k^\text{th}$ person has $n+k-1$ choices. They can enter at the back of any queue, or they can cut in front of any of the previous $k-1$ people.