Correct approach to evaluate probability? 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?
For the above problem, the solution provided is this:
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 $\frac{6!}{6^6}$
. So the probability of some district having
more than $1$ robbery is
$1 − \frac{6!}{6^6} ≈ 0.9846$.
Note that this also says that if a fair die is rolled $6$ times, there’s over a $98\%$ chance
that some value is repeated!
But this is my approach: 
I thought that this is equivalent of dividing 6 dots into 6 boxes, there are $11\choose6$ ways of distributing, out of which there is only $1$ way in which $1$ dot (equivalent to robbery) is there in each box (equivalent to district). Hence the probability of more than one robbery in any district is $1 - \frac{1}{11\choose6}$
I am unable to digest how the above situation is equivalent to the dice situation. Can anyone explain me, what is wrong with my approach, and how the dice thing is equivalent ?

A model of tossing a die six times is appropriate here, since it is stated that "the robberies are located randomly, with all possibilities for which robbery occurred where equally likely."  
Cases yielded by stars and bars are not  equi-probable
You would expect a much lower frequency for all robberies  taking place in one district, $(6-0-0-0-0-0)$ compared to a more even distribution of robberies, e.g.  $(1-2-1-1-0-0)\;\;$ The first has $6!\over 5!$ $= 6$ permutations, the second has $6!\over{3!2!}$ $= 60$
Btw, note that if you write down all possible stars and bars solutions, and add up the permutations of each such solution, you will get $6^6$
