Simple random Sampling with replacement (SRSWR) If a simple random sample (with replacement) of size $n$ is drawn from a population of size $N$, where $N \leq n$. Then what is the probability that all the population units (items) are present in the sample?
 A: This problem is a special case of the classical occupancy problem.  If you conduct simple random sampling with replacement, choosing $n$ objects from a population of $N$, then the number $K$ of sampled items has a classical occupancy distribution with probability mass function:
$$\mathbb{P}(K=k|N,n) = \frac{(N)_k \cdot S(n,k)}{N^n}
\quad \quad \quad \text{for all } 1 \leqslant k \leqslant \min(N,n),$$
where $(N)_k = N(N-1)(N-2) \cdots (N-k+1)$ are the falling factorials and $S(n,k)$ are the Stirling numbers of the second kind.  The properties of this distribution are well-known (see e.g., O'Neill 2020).  In your particular problem you are dealing with the case where $N \leqslant n$ seeking the probability that all population objects are sampled, which is:
$$\mathbb{P}(K=N|N,n) = \frac{N! \cdot S(n,N)}{N^n}.$$
These values can easily be calculated for values of $n$ and $N$ that are not too large.  For large values the occupancy distribution can be approximated by the normal density, with accuracy shown in the cited paper.
A: Comment:  You may want to check your analytic answer (approximately) 
against a simulation to make sure you have done the combinatorial analysis correctly.
Here is a simulation for $n = 12$ and $N = 25.$ You might think of it as a kind of 'reversal' of the famous birthday problem. Birth months equally likely. Room of 25 students. What is the probability all 12 months are represented? The answer is about $18\%.$
set.seed(2019);  pop = 1:12;  N = 25
x = replicate( 10^6, length(unique(sample(pop, N, repl=T))) )
mean(x);  mean(x == 12);  2*sd(x == 12)/1000
[1] 10.63755       # aprx E(X)
[1] 0.181771       # aprx P(X = 12)
[1] 0.0007713117   # 95% margin of sim error for P(X = 12)

table(x)/10^6 
n
       6        7        8        9       10       11       12 
0.000029 0.000944 0.014487 0.097603 0.301636 0.403530 0.181771 

With a million iterations, you can expect two, maybe three, place accuracy.
Also, if you consider the the number $N$ required on average
to see all $n$ objects in the population, then you have
the much studied 'coupon collecting' problem.
