Probability of having exactly $v$ different letters in password Consider a password with $t$ characters, with a character set of length $n^2$. What is the probability there are $v$ distinct letters in the password.
So if the password was "abccdadde" $v$ would be $5$ and $t=9$.
So far I have deduced if $t=n=v$ the probability is given by: $$\frac{n^{2}!}{\left(n^{2}-n\right)!n^{2n}}$$
And when $t=v+1=n+1$ the probability is given by:$$\frac{n\left(n+1\right)!n^{2}!}{2n!\left(n^{2}-n\right)!n^{2\left(n+1\right)}}$$
 A: Denote $n^2=N$ for ease of notation, then the number of ways to assign $v$ "dummy" characters (to be replaced by actual characters later under a bijective map) is the definition of the second-kind Stirling number $S(t,v)$. Then there are ${}^NP_v=\binom Nvv!$ ways to assign the actual characters, so the number of admissible passwords is
$$\binom Nvv!S(t,v)$$
and the probability is this over the total password count $N^t$.
A: This is a collection of hints, but too long for a comment.
First, one approach which I feel often helps in these cases is to use the relation
$$\Pr(\text{exactly} \ v) = \Pr(\text{at most} \ v) - \Pr(\text{at most} \ v-1).$$
Perhaps you can try to work out an estimate on $\Pr(\text{at most} \ v)$. Here are some starting points.
Define $p_{k,t,v}$ to be the probability that at most $v-k$ new letters are used in $t$ draws when $k$ letters have already been used. Now "take one step and see what happens", as my old Markov chains teacher used to say. Taking one step reduces $t$ by $1$, as there is now one fewer draw required. If the new letter drawn has yet to be seen, then $k$ increases by $1$; otherwise, it remains fixed.
Might be quite rough computationally to do this recurrence relation. There might be some easier way; I'm not sure. Certainly some combinatorial argument, counting the ways of drawing things, seems plausible, but I was trying to avoid that as it would likely be pretty unwieldy!
