Time complexity of variation on Coupon's collector problem I need to know the complexity of the following algorithm:
Draw a set of n numbers from a larger set of m numbers, one by one, randomly, with replacement. The result may be any set of numbers as long as the size is n and the elements are different.
This is a variation of the Coupon collector's problem: we do not have to draw all m elements, any set of different numbers of size n will do. 
Example: 
n = 3, m = 5, result = {1,5,2} produced by successive draws 1,1,5,1,5,2 from pool {1,2,3,4,5}
What is the average time complexity of this problem? 
 A: By way of enrichment here  is the complexity using Stirling numbers
of   the   second   kind.   Using   the  notation   from   this   MSE
link  we  have  $n$
coupons, and ask  about the expected time until  a multiset containing
instances of $j$ different coupons has been drawn.

First let  us verify  that we indeed  have a  probability distribution
here. We have for the number $T$ of coupons being $m$ draws that
$$P[T=m] = \frac{1}{n^m} \times 
{n\choose j-1} \times {m-1\brace j-1} \times (j-1)!
\times (n+1-j).$$
What  happens here  is that  for a  run of  $m$ samples  to  produce a
multiset containing  instances of $j$ different coupons  for the first
time on  the last sample we have  two parts, a prefix  of length $m-1$
and a terminal sample that completes the set. Therefore we must choose
the $j-1$  values excluding  the one that  occurs last for  the prefix
from the $n$ possibilities which gives the first binomial coefficient.
Next we partition  the first $m-1$ slots into  $j-1$ non-empty sets in
an  ordered  set partition.   (Stirling  number  and factorial).   The
smallest value chosen gets the slots listed in the first set, the next
one  those  in   the  second  set  etc.   Finally   we  get  $n-(j-1)$
possibilities ($j-1$  values from the  prefix have already  been used)
for the  terminal sample that  completes the selection.   Combine with
$n^m$ possible choices. 
Recall the OGF  of the Stirling numbers of the  second kind which says
that
$${n\brace k} = [z^n] \prod_{q=1}^k \frac{z}{1-qz}.$$
This gives for the sum of the probabilities
$$\sum_{m\ge 1} P[T=m]
= {n\choose j-1} (j-1)! (n+1-j) 
\frac{1}{n} \sum_{m\ge 1} \frac{1}{n^{m-1}} {m-1\brace j-1}.$$
Focusing on the sum we obtain
$$\sum_{m\ge 1} \frac{1}{n^{m-1}}
[z^{m-1}] \prod_{q=1}^{j-1} \frac{z}{1-qz}
= \prod_{q=1}^{j-1} \frac{1/n}{1-q/n}
\\ = \prod_{q=1}^{j-1} \frac{1}{n-q}
= \frac{(n-j)!}{(n-1)!}.$$
Combining this with the outer factor we get
$${n\choose j-1} (j-1)! (n+1-j) 
\frac{1}{n} \frac{(n-j)!}{(n-1)!}
\\ = {n\choose j-1} (j-1)! \frac{(n+1-j)!}{n!} = 1$$
This confirms it being a probability distribution.
We then get for the expectation that
$$\sum_{m\ge 1} m\times P[T=m]
\\ = {n\choose j-1} (j-1)! (n+1-j) 
\frac{1}{n} \sum_{m\ge 1} \frac{m}{n^{m-1}} {m-1\brace j-1}.$$
We once more focus on the sum to get
$$\sum_{m\ge 1} \frac{m}{n^{m-1}}
[z^{m-1}] \prod_{q=1}^{j-1} \frac{z}{1-qz}
= \sum_{m\ge 1} \frac{m}{n^{m-1}}
[z^{m}] z \prod_{q=1}^{j-1} \frac{z}{1-qz}
\\ = \left.\left( \prod_{q=0}^{j-1} 
\frac{z}{1-qz} \right)'\right|_{z=1/n}
\\ = \left.\left( \prod_{q=0}^{j-1} 
\frac{z}{1-qz} \sum_{p=0}^{j-1} \frac{1-pz}{z}
\frac{1}{(1-pz)^2}
\right)\right|_{z=1/n}
\\ = \left.\left( \prod_{q=0}^{j-1} 
\frac{z}{1-qz} \sum_{p=0}^{j-1} \frac{1}{z}
\frac{1}{1-pz}
\right)\right|_{z=1/n}
\\ = \prod_{q=0}^{j-1} \frac{1/n}{1-q/n} 
\sum_{p=0}^{j-1} \frac{1}{1/n}
\frac{1}{1-p/n}
\\ = \prod_{q=0}^{j-1} \frac{1}{n-q} 
\sum_{p=0}^{j-1}
\frac{n^2}{n-p}
= n \prod_{q=1}^{j-1} \frac{1}{n-q} 
\sum_{p=0}^{j-1}
\frac{1}{n-p}.$$
Retrieving the outer factor we have
$${n\choose j-1} (j-1)! (n+1-j) 
\frac{1}{n} \frac{(n-j)!}{(n-1)!} \times n 
\sum_{p=0}^{j-1} \frac{1}{n-p}.$$
The front simplifies to one as before and we are left with
$$n\sum_{p=0}^{j-1} \frac{1}{n-p}
= n \left(\sum_{p=0}^{n-1} \frac{1}{n-p}
- \sum_{p=j}^{n-1} \frac{1}{n-p}\right).$$
This is
$$\bbox[5px,border:2px solid #00A000]{\Large
 n \times \left( H_n - H_{n-j} \right)}$$
This yields  $n H_n$ when $j  = n$ and  $1$ when $j=1$ which  are both
correct. Using  $H_n \sim \log n  + \gamma$ we  get for $j =  n/2$ the
expectation $n\log 2.$
A: The expected 
number of trials until your collection has $n$ distinct elements is:
$${m\over m}+{m\over m-1}+{m\over m-2}+\cdots +{m\over m-n+1}. $$
