Yes, the events $(6,0,0,0,0,0)$ and $(1,2,1,1,1,0)$ are both considered by sticks and stones and by the "all are distinct" method.
Nobody is contradicting that statement when they say
"cases yielded by [sticks and stones] not equi-probable".
What they mean is precisely and simply that the probability of $(6,0,0,0,0,0)$
is not equal to the probability of $(1,2,1,1,1,0)$. That's all. And that's a fatal flaw with the application of sticks and stones to your problem.
It isn't good enough to merely consider both outcomes;
you also have to consider them with the correct probabilities assigned to them.
The technique where you count number of "successful" outcomes divided by the total number of outcomes is valid when all outcomes have the exact same probability,
probability $1/n$ where $n$ is the total number of outcomes.
If there are any outcomes that are more likely than others, you have to do something more sophisticated than just counting.
Since some outcomes counted by sticks and stones are more likely than others
(that is, "cases yielded by [sticks and stones] not equi-probable"),
you can't use sticks and stones as a counting method for probability in this example.
That's why sticks and stones leads to a wrong result.
Here's why we get a correct result by considering $6^6$ equally likely outcomes instead.
The problem statement says,
... all possibilities for which robbery occurred where [are] equally likely.
In order to even talk about "which" robbery occurred where, we have to be able to distinguish one robbery from another. So let's number the robberies $1, 2, 3, 4, 5, 6$ in the sequence in which they occurred during the week.
Further, number the districts $1, 2, 3, 4, 5, 6$ and let $d(n)$ be the number of the district where robbery number $n$ occurred.
Then if you write out all the values of $d(n)$, you will know exactly which robbery occurred where. For example, if it turns out that
$$
d(1) = 1, \ d(2) = 1, \ d(3) = 1, \ d(4) = 1, \ d(5) = 1, \ d(6) = 1
$$
then all robberies occurred in district $1$ and you have the event that you denoted by $(6,0,0,0,0,0)$ when using sticks and stones.
And this is the only way to define $d(n)$ that produces the event $(6,0,0,0,0,0)$; if we write anything but exactly these values of $d(n)$ for each $n$, we get at least one robbery in a district other than district $1$, which means $(6,0,0,0,0,0)$ did not happen.
But some other possibilities for the definition of $d(n)$ are
$$
d(1) = 1, \ d(2) = 2, \ d(3) = 2, \ d(4) = 3, \ d(5) = 4, \ d(6) = 5
$$
and
$$
d(1) = 5, \ d(2) = 4, \ d(3) = 3, \ d(4) = 2, \ d(5) = 2, \ d(6) = 1,
$$
each of which matches the outcome $(1,2,1,1,1,0)$ in the sticks and stones method. And we've still barely begun to list all the possible different ways we could define the function $d(n)$ and still end up with $(1,2,1,1,1,0)$.
But if you know where each robbery occurred, you know exactly how the function $d(n)$ is defined -- you know every value of $d(n)$ in a list like the ones written above. And if you have a full definition of $d(n)$, showing all its values, you know exactly where each robbery occurred.
This gives us a particular one-to-one correspondence between the set of possible definitions of $d(n)$ and the set of possible cases of which robbery occurred where. And remember, each of those cases has the same probability as every other case because the problem explicitly statement says it does.
Now we just need to count the number of cases to determine what the probability of each case is. The number of cases is the same as the number of possible ways to define $d(n)$.
Every possible $d(n)$ is a function from the set $\{1,2,3,4,5,6\}$ to the set $\{1,2,3,4,5,6\}$, and every function from $\{1,2,3,4,5,6\}$ to $\{1,2,3,4,5,6\}$ could be $d(n)$, so the set of cases, which has the same number of elements as the set of possible definitions of $d(n)$,
also has the same number of elements as the set of all functions from $\{1,2,3,4,5,6\}$ to $\{1,2,3,4,5,6\}$.
The number of functions from a set $A$ to a set $B$ is
$\lvert A\rvert^{\lvert B\rvert}$. (The notation $\lvert S\rvert$ is defined as the number of elements in $S$.)
In our problem, $A = B = \{1,2,3,4,5,6\}$ and
$\lvert A\rvert = \lvert B\rvert = 6$, so the number of possible definitions of $d(n)$ is $6^6$.
That's how we get $6^6$ possible outcomes, and how we know they are equally likely (because in this question the problem statement says they are).
The likelihood of each of these outcomes is $$\dfrac{1}{6^6}.$$
With sticks and stones, on the other hand, you have only $\binom{11}{6}$ possible outcomes (each of which says only how many robberies occurred in each district, not which robberies occurred in each district).
One of these outcomes is denoted $(6,0,0,0,0,0)$, which as we have found can only occur when the function $d(n)$ is defined in one particular way,
which is something that happens with probability
${1}/{6^6}.$
So unless you come up with some way to get the sticks and stones method to produce the result that the probability of $(6,0,0,0,0,0)$ is
${1}/{6^6},$
you can't get correct probabilities by this method.
Best Answer
Yes, that’s the problem. If the children are $A$, $B$, $C$, and $D$, there are actually $\binom52\cdot 3!=60$ different ways to distribute the balls so that $A$ gets $2$ balls and $B$, $C$, and $D$ get one each: there are $\binom52=10$ different pairs of balls that can be given to $A$, and the remaining $3$ balls can be permuted amongst $B,C$, and $D$ in $3!=6$ distinguishable orders.