What is the probability exactly one monster does not get a snack?
My approach was to standard "$n$ box $n$ balls approach"
my current answer:
$$\frac{n! \cdot \binom{n}{2}}{n^n}$$
my doubt is in the question there is nothing about if these snacks are distinguishable or not. Or if the snacks were distinguishable would it alter my a answer. Couldn't see it. Thank you.
Your answer is correct:
$n {n\choose 2}$ for the number of ways of choosing the monster with two snacks and which two snacks they are
${n-1\choose n-2} (n-2)!$ for the number of ways of then choosing the remaining monsters with one snack and which remaining snacks they get
$n^n$ total ways of distributing the snacks
so a probability of $\dfrac{n {n\choose 2} {n-1\choose n-2} (n-2)!}{n^n} = \dfrac{n! {n\choose 2} }{n^n}$ exactly one monster does not get a snack, as you found.
If you had supposed that the snacks were indistinguishable but the monsters distinguishable, and that each possible allocation was equally likely, which would be difficult to arrange in reality, then you might have $\frac{n(n-1)}{2n-1\choose n}$.
With $n=2$ this would suggest $\frac23$ as the probability one of the monsters gets both snacks (either the first gets both, or the second gets both, or they get one each). Your answer would suggest a more plausible $\frac12$, rather like flipping a coin twice and looking for the probability the results were the same as getting one each is more probable than each of the other outcomes.
Your final answer is correct for both cases where the snakes are distinguishable or not, but different methods need to be used.
Even if the snakes are not distinguishable, you can artificially number them because this does not change the probability of the event (which only depends on the number of snakes). However, this artificial distinguishing helps us to have a equally likely sample space.
Below, you can see that under both assumptions we get the same probabilities:
$$\frac{\binom{n}{1} (n-1)! \binom{n}{2}}{n^n}=\\ \underbrace {n(n-1)}_{\small \small \text{selecting monsters with 0 and 2 snakes}} \binom{n}{0,2,1,\cdots,1} \left ( \frac{1}{n} \right )^0 \left ( \frac{1}{n} \right )^2 \left ( \frac{1}{n} \right )^1 \cdots \left ( \frac{1}{n} \right )^1.$$
The LHS is obtained by assuming that the snakes are distinguishable; hence, the sample space is equally likely and we can use counting method.
The RHS is obtained by assuming that the snakes are indistinguishable, which results in a sample space that is not equally likely, and we cannot use the counting method. To compute the second part of the RHS, I used the multi-nominal distribution.
To see why the sample space for distinguishable snakes is equally likely but not for indistinguishable snakes. Consider the following example for $n=4.$
When the snakes are distinguishable, the probability of each of $ s1 | s1 | s3 | s4$ and $ - | s1, s2 | s3 | s4$ (and any other outcomes) is the same as:
$$\left ( \frac{1}{4} \right) ^4. $$
When the snakes are indistinguishable, the probabilities of the two outcomes $ 1 | 1 | 1 | 1 $ and $ 0 | 2 | 1 | 1 $ are not the same, given by $$\binom{4}{1,1,1,1}\left ( \frac{1}{4} \right) ^4 $$ and $$\binom{4}{0,2,1,1}\left ( \frac{1}{4} \right) ^4,$$ respectively.
More details on the LHS:
The probability of that exactly $r$ monsters do not get snakes is
$$\frac{\binom{n}{r} (n-r)! S(n,n-r)}{n^n}$$ where $$S(n,k) = \frac{1}{k!} \sum_{i=0}^k(-1)^{i}\binom{k}{i}(k-i)^n$$ is the Stirling number of the second kind. For $r=n-1$, we have $$S(n,n-1)=\binom{n}{2}.$$ Another way, working for $r=n-1$, is to directly count the ways that $n$ distinct items (snakes) can be assigned to $n-1$ distinct receivers (monsters) such that each receiver at least gets one item. This can be done by selecting two items that are given to the same receiver, $\binom{n}{2}$, and finding all the permutations of the $n-2+1$ objects, $(n-1)!$.
