# What is the probability that, having $N$ numbered balls and performing $n$ extractions, I have $n−m$ non-repeated balls?

What is the probability that, having $N$ numbered balls (from 1 to N) and performing $n$ extractions, I have exactly, for $m$ given, $n-m$ non-repeated balls?

Example: $N=10, n=8, m=4$ I will consider all the sequences like [1,2,3,4,5,5,6,6], [1,2,3,4,6,6,6,6] ... but not [1,2,3,4,5,5,5,6] since it has 5 non-repeated balls and not 4 = $n-m$.

I tried to solve the problem in this way: if we consider the random variable $X =$ "how many repeated balls (counted with their multiplicities) I have in my sequence", I have:

$$\mathbb{P}(X \leq m) = \binom{n}{m}\frac{N}{N}\cdot\frac{N-1}{N}\cdot\dots\cdot\frac{N-(n-m-1)}{N}\cdot \big[ \frac{N-(n-m)}{N}\big]^m$$

And then $\mathbb{P}(X = m) = \mathbb{P}(X \leq m) - \mathbb{P}(X \leq m-1)$ but this formula doesn't give me the exact result (I generated all the possible sequences and counted the "good" ones to get the real probability).