Two approaches to a probability/combinatorics problem The statement:
A city with $6$ districts has $6$ robberies in a particular week. Assume the robberies are located randomly, with all possibilities for which robbery occurred where equally likely. What is the probability that some district had more than $1$ robbery?
The answer:
There are $6^6$ possible configurations for which robbery occurred where. There are $6!$ configurations where each district had exactly $1$ of the $6$, so the probability of the complement of the desired event is $6!/6^6$. So the probability of some district having more than $1$ robbery is $1 - 6!/6^6$.
The problem:
I was thinking about this problem and found another approach. Of course, I know this approach is incorrect, but I can't find why.
The number of ways to distribute k balls into n boxes is: $\binom{n + k - 1}{k}$. So I thought that the boxes could be the districts, and the balls could be the robberies. So there is $\binom{6 + 6 - 1}{6} = \binom{11}{6}$ different configuration of robberies, out of which, just one is the configuration in which every bank has exactly one robbery.
The problem is that $1 - 1 / \binom{11}{6}$ is not equal to $1 - 6! / 6^6$.
What is wrong with this reasoning?
 A: Why does your approach not work?
Robberies are distinguishable.  The formula you used counts the number of ways of placing $n$ indistinguishable objects in $k$ distinguishable boxes. 
Moreover, we would like the simple events in our sample space to be equally likely.  Each of the six robberies can occur in one of six districts, which gives us a sample space of $6^6$ possible ways the robberies could be distributed.  
The $\binom{11}{6}$ events you counted are not equally likely.  There is one way to get the outcome $(6, 0, 0, 0, 0, 0)$.  However, the number of ways of obtaining the outcome $(0, 2, 3, 0, 0, 1)$ is 
$$\binom{6}{2}\binom{4}{3}\binom{1}{1} = 15 \cdot 4 \cdot 1 = 60$$
and the number of ways of obtaining the outcome $(1, 1, 1, 1, 1, 1)$ is 
$$6! = 720$$
You counted the event $(1, 1, 1, 1, 1, 1)$ just once. 
A: The formula that you have used is the number of ways of distributing $k$ indistinguishable balls into $n$ distinguishable boxes. This is quite different from the question at hand.
Suppose you want to distribute $k$ distinguishable balls into $n$ distinguishable boxes. Consider the first ball. It can be placed into any of the $n$ boxes. Number of ways = $n$. Similarly for the second, third,..., $kth$ ball. Hence the total number of ways of doing that is $n^k$. This approach will yield the same solution as the answer mentioned.
