Probability of identical numbers in two sets $l$ and $n$ where $n$ has been drawn without replacement from $N$? The problem
I am trying to produce a general formula for the problem below. This is not homework.
$N$ different balls are inside an urn. The balls have numbers on it and are labelled $1\ldots N$. Then, $n$ balls are drawn without replacement from the urn. A candidate player is asked to choose $l$ numbers from the set $\{1, \ldots, N\}$ without having seen the $n$ balls that were drawn (but the candidate knows $n$, i.e. how many balls there were drawn).
What is the probability of $z$ (with $z \in \{0,\ldots, n\})$ identical numbers in the candidate's set $l$ and the $n$ drawn balls?
What I have tried
I figured out that the probability of exactly $z$ matching numbers in sets of $n$ is
$$
\frac{\binom{N - z}{n - z}}{\binom{N}{n}}
$$
So let's say we have $N=5$ and $n = 3$, there are obviously $10$ ways of drawing $3$ out of $5$ numbers. Then, the probability that exactly $z=2$ prespecified numbers (say, $9$ and $10$) occur is $0.3$.
But don't know how to proceed from here.
Any help is appreciated. Thank you.

I obtain $$p = \frac{\binom{l}{z}\binom{N-l}{n-z}}{\binom{N}{n}} $$ We have $z$ identical numbers if and only if among the balls drawn $z$ have numbers from the set chosen by the candidate, and the other $n-z$ have numbers from the $N-l$ numbers not chosen by the candidate
