Combinatorics: n people visit k exhibitions

Question

Edit: Excuse me, the numerator of the given solution should've been a rising factorial. That said, I still don't understand where it comes from?

The question is as follows :

n friends visit k exhibitions, each person visits only 1 exhibition. Find the number of possibilities if

b) Only the number of friends who goes to each exhibition matters. Neither the order nor who goes where matters.

Now, I have the solution which would be $$\frac{[n+1]^{k-1}}{(k-1)!} $$

But I don't really understand why that is so. If I understand that correctly, that would be a stars and bars problem, but I don't seem to grasp the concept of it.

Any help would be greatly appreciated.

Answer 1

Yes, it's your basic stars and bars: You have $n$ 'stars' (the friends), and $k-1$ bars to divide the friends into $k$ groups.

So, you have $n+k-1$ objects that you line up, and there are

$${n+k-1} \choose n$$

possible different ways to place the $n$ 'stars' in those $n+k-1$ slots.

This is not at all your formula though ...

Answer 2

Arranging $n$ stars and $k-1$ bars is a way of modeling this situation. Each arrangement of those $n+k-1$ symbols gives us a set of numbers of people to go to each exhibition. However, the result of that counting problem is:

$$\binom{n+k-1}{k-1}=\frac{(n+k-1)!}{n!(k-1)!}$$

Do we have a condition that each exhibition has to be attended by at least one friend? Do we even know that $n\geq k$?