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 ?

Lets name robberies A,B,C,D,E,F and districts 1,2,3,4,5,6. As per assumption: the robberies are located randomly, with all possibilities for which robbery occurred where equally likely means that 
P(A occuring in 1) = P(A occuring in 2) = P(A occuring in 3) ....
So A  has 6 possibilities 
similarly B has 6 ....
by multiplication rule.
Number of ways in which robberies can occur is 6^6
