There are $n$ numbers, and each number is different. Randomly select a number with equal probability each time and record it, and repeat this process $t$ times. Note that the number you have selected before can continue to be selected. That is to say, for each selection, the probability of each number being selected is $\frac{1}{n}$. Finally, the amount of digits obtained is $t$. There may be cases where some digits are equal among these $t$ numbers. Repeated digits are deleted and a set of unique elements is obtained. What is the probability $p_{k}$ that the set size is $k$? The value range of $k$ is of course $1\leq k\leq t$. $n$ is much larger than $t$, so there is no need to consider whether $k$ needs to be smaller than $n$. For example, there are $n=1000$ numbers, select $t=40$ times, and finally get $k=35$ non-repeated numbers. What is the probability formula $p_{k}$ for $k$?
For $k=1$, I think $p_1={(\frac{1}{n})}^{t-1}$. And for $k=t$, I think $p_t=\frac{(t-1)\cdot P_n^{t-1}}{n^t}.$ (The $P_x^y$ is number of permutations).
I'm not sure whether these two cases are correct. For $2\leq k\leq t-1$, I didn't find the right way to calculate.
