Suppose I have $n$ balls numbered $1$ through $n$, and I take a simple random sample with replacement of size $n$ of these balls. What is the chance that Ball $1$ occurs 5 times or more in my sample?
The total number of samples is $n^n$. We only need to count how many of these have Ball 1 occurring at least 5 times. The 5 places can be picked in ${n\choose 5}$ ways, and I place Ball 1 in these 5 places. The remaining $(n-5)$ places can contain anything, so I believe the total number of such samples should be ${n\choose 5}n^{n-5}$. Hence the chance is ${n\choose 5}/n^5\approx 1/120\approx 0.008$ if $n$ is large.
However, another way of looking at this problem is the following. Let $N$ be the number of times Ball 1 appears in the sample. Since the occurrences are independent and the probability that Ball 1 comes up in any given position is $1/n$, we have $N\sim Bin(n,1/n)$. Using the Poisson approximation to Binomial, $P(N\geq 5)\approx P(Poi(1)\geq 5)\approx 0.004$.
Maybe the first argument I presented is incorrect, but I am not immediately able to see why that might be incorrect.
The first argument is not correct, because you are double counting configurations containing more than 5 copies of Ball 1.
EDIT: See comments below for a better discussion of why this argument overcounts.