When to treat things as distinct in combinatorics I made a mistake on the following combinatorics problem:



*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?


(from here)
My mistake was essentially that I treated robberies as indistinguisable, so I thought the answer was
$$
1 - \frac{1}{\binom{11}{6}}
$$
But the answer turns out to be:
$$ 
1 - \frac{6!}{6^6}
$$
which treats robberies as distinguishable.
The problem is, it's not that I did not consider whether robberies are distinguishable or not; it's that I did consider it and thought that here they were indistinguishable. In some cases, it seems that I can get the right answer anyway as long as I am consistent in my definition, but that is not the case here.
I read this answer, where the OP made the same mistake as mine, but it did not help me much, mostly because it simply says that robberies are distinguishable (which does not come naturally for me).  The additional reason seems to rest on this as well: e.g., we only think that (1, 1, 1, 1, 1, 1) can occur in 6! ways if we think the permutations are meaningful, i.e., if we think that robberies are distinguishable in the first place.
For cases of people, I can tell that they are distinguishable. However, I usually have trouble with objects (like balls, dice, etc.) and events (like robberies). Is there a good way to tell when to treat things as distinguishable?
 A: Now let's analyse what the question wants from us,
The question says that robbery can take place in any of the 6 places and in total 6 robberies take place that week, I hope it's all clear till now,
Now we need to find the probability of any district having more than one robberies that week, we can simply find it by eliminating the cases of having exactly 1 robbery in each district, which is calculated as follows:
Total no of cases where robbery happens (N⁰)= 6⁶,
this is because let's assume first robber comes and then he has 6 options to choose from, he either goes to city-1 or city-2 or ... city-6. Same is the case with rest 5 robbers so all in all we get $6\times6\times6\times6\times6\times6$ cases,
N⁰=6^6
The cases where one robbery happens in exactly one city, then we have to divide 6 cities in 6 people which is simply achieved by 6!, So the
No of cases where one robbery happens in each city(N¹)= 6!
Finally the probability of one robbery per district = 6!/6⁶
And now the
Probability of a district having more than one robbery =1-(6!/6⁶)
Now coming to your doubt, we have different cities as different cases, the way each robber is different the same way each district is different. The cases where u can differentiate two bodies in a way where we can assign them as unique identities are the ones where you count them as different groups.
Example:
You have 5 bags of red colour of identical build and quality=> all are same,
But if you have 5 bags of red colour even of identical build and quality but if each have a small wear(even if it's not intentional) by which each can be identified differently then they are counted as different
I hope this clears your doubt, if it still isn't clear then please feel free to ping me, this is my first answer so please ignore any silly mistakes and make sure to provide an honest feedback.
Signing off,
Aditya
