I saw this solution to a problem in my textbook:
The problem is we have $10$ indistinguishable balls and $7$ distinguishable bins. What is the probability of having at least one ball in each bin?
From what I saw in Probability: Distinguishable vs Indistinguishable, I think it is false to calculate the number of sample space elements like
${7+3-1 \choose 3} = {9 \choose 3}$
since then we would miscalculate the probability.
I think we should consider the balls distinct then solve it by means of Stirling number of second kind. Am I right?
You didn't say how the bins are filled with the balls so I will assume randomly. I am also assuming that each of the $7$ bins can hold all $10$ balls if need be. Using a "brute force" case method, there are $3$ main ways to fill all $7$ bins. We already know each of the $7$ bins must contain one ball, then the remaining $3$ balls can be placed as follows:
case 1: $1$ bin has $4$ balls and all the others have $1$ ball each. Ways of that happening are $7 \choose 1$.
case 2: $1$ bin has $3$ balls, another bin has $2$ balls, and all the others have $1$ ball each. Ways of that happening are $7 \choose 1$ * $6 \choose 1$ which is also $7 \choose 2$ * $2$.
case 3: $3$ bins have $2$ balls each and the others have $1$ ball each. Ways of that happening are $7 \choose 3$.
The numbers of ways to place $10$ balls in $7$ bins is $7^{10}$.
So the probability is the sum of cases $1,2,$ and $3$, divided by # of ways, so we have $\frac {7+42+35} {282,475,249}$