Probability of $10$ random balls filling $7$ bins. Does indistinguishability matter? 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? 
 A: The reason that the formula $${7+3-1 \choose 3} = {9 \choose 3}$$ miscalculates the required probability is not that it confuses the distinguishability or indistinguishability of the balls, but that is treats each distribution of the balls in the pins as equally likely, although this is not true. For a thorough explanation see here. 
A: 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 $\frac {7+42+35} {282,475,249}$
