Variation of Coupon Collectors Problem for the case where $k\le n$. I was working on this problem which says,
Suppose one draws balls with replacement from an urn containing $n$ unique balls and records its number. Then what is the expected number of draws required for getting 
(a) the set of balls numbered $\{1,2,\cdots,k\}$ recorded? Where $k\leq n$ a fixed number.
(b) the set of any $k$ distinct balls? 
N.B I have solved the part (a) which comes out to be $nH_k$; $H_k=\sum_{j=1}^{k}\frac{1}{j}$. But I'm stuck at the second part. Any help is much appreciated.

By way  of enrichment  here is a  generating function approach  to the
question of when the first $k$  coupons where $k\le n$ have been seen.
Using      the     notation      from      the     following      MSE
link  we  get  from
first principles for the probability of $m$ draws that
$$P[T=m] = \frac{1}{n^m} \times {k\choose k-1} \times
\sum_{q=0}^{n-k} {n-k\choose q} {m-1\brace q+k-1}
(q+k-1)!.$$
What happens here  is that we choose the $k-1$ of  the $k$ values that
go into the prefix, which also determines the value that will complete
the set  with the last draw.  We then choose  a set of $q$  values not
from the $k$ initial ones and partition the first $m-1$ draws or slots
into $q+k-1$ sets, one for each value.
We verify that this is a probability distribution, getting
$$\sum_{m\ge 1} P[T=m] \\ = 
\sum_{m\ge 1} \frac{1}{n^m} \times {k\choose k-1} \times
\sum_{q=0}^{n-k} {n-k\choose q} (m-1)! [z^{m-1}] 
(\exp(z)-1)^{q+k-1}
\\ = k \sum_{m\ge 1} \frac{1}{n^m} \times 
(m-1)! [z^{m-1}] \sum_{q=0}^{n-k} {n-k\choose q} 
(\exp(z)-1)^{q+k-1}
\\ = k \sum_{m\ge 1} \frac{1}{n^m} \times 
(m-1)! [z^{m-1}] (\exp(z)-1)^{k-1}\exp(z(n-k))
\\ = k! \sum_{m\ge 1} \frac{1}{n^m}
\sum_{q=0}^{m-1} {m-1\choose q} {q\brace k-1} (n-k)^{m-1-q}
\\ = k! \sum_{q\ge 0} {q\brace k-1} 
\sum_{m\ge q+1} {m-1\choose q} \frac{1}{n^m} (n-k)^{m-1-q}
\\ = k! \sum_{q\ge 0} {q\brace k-1} \frac{1}{n^{q+1}}
\sum_{m\ge 0} {m+q\choose q} \frac{1}{n^m} (n-k)^m
\\ = k! \sum_{q\ge 0} {q\brace k-1} \frac{1}{n^{q+1}}
\frac{1}{(1-(n-k)/n)^{q+1}}
 = k! \sum_{q\ge 0} {q\brace k-1} \frac{1}{k^{q+1}}.$$
Recall the OGF  of the Stirling numbers of the  second kind which says
that
$${n\brace k} = [z^n] \prod_{p=1}^k \frac{z}{1-pz}.$$
We obtain
$$k! \sum_{q\ge 0} \frac{1}{k^{q+1}}
[z^q] \prod_{p=1}^{k-1} \frac{z}{1-pz}
= (k-1)! \prod_{p=1}^{k-1} \frac{1/k}{1-p/k}
\\ = (k-1)! \prod_{p=1}^{k-1} \frac{1}{k-p} = 1$$
and the  sanity check  goes through. For  the expectation of  when the
first $k$ have  been seen we recycle the above,  inserting a factor of
$m$, starting from
$$k! \sum_{q\ge 0} {q\brace k-1} 
\sum_{m\ge q+1} {m-1\choose q} \frac{m}{n^m} (n-k)^{m-1-q}
\\ = k! \sum_{q\ge 0} {q\brace k-1} 
\sum_{m\ge q+1} {m\choose q+1} \frac{q+1}{m} \frac{m}{n^m} (n-k)^{m-1-q}
\\ = k! \sum_{q\ge 0} {q\brace k-1} (q+1)
\sum_{m\ge q+1} {m\choose q+1} \frac{1}{n^m} (n-k)^{m-1-q}
\\ = k! \sum_{q\ge 0} {q\brace k-1} (q+1) \frac{1}{n^{q+1}}
\sum_{m\ge 0} {m+q+1\choose q+1} \frac{1}{n^m} (n-k)^{m}
\\ = k! \sum_{q\ge 0} {q\brace k-1} (q+1) \frac{1}{n^{q+1}}
\frac{1}{(1-(n-k)/n)^{q+2}}
\\ = n \times k! \sum_{q\ge 0} {q\brace k-1} (q+1) \frac{1}{k^{q+2}}
 = \frac{n}{k^2} \times k! 
\sum_{q\ge 0} {q\brace k-1} (q+1) \frac{1}{k^{q}}.$$
Activating the OGF produces
$$\frac{n}{k^2} \times k! 
\sum_{q\ge 0}  \frac{1}{k^{q}}
[z^q] \left(\prod_{p=0}^{k-1} \frac{z}{1-pz}\right)'
\\ = \frac{n}{k^2} \times k! 
\sum_{q\ge 0}  \frac{1}{k^{q}}
[z^q] \prod_{p=0}^{k-1} \frac{z}{1-pz}
\sum_{p=0}^{k-1} \frac{1}{z(1-pz)}
\\ = \frac{n}{k^2} \times k! 
\prod_{p=0}^{k-1} \frac{1/k}{1-p/k}
\sum_{p=0}^{k-1} \frac{1}{1/k(1-p/k)}
\\ = n \times k! 
\prod_{p=0}^{k-1} \frac{1}{k-p}
\sum_{p=0}^{k-1} \frac{1}{k-p}
= n \times k! \times \frac{1}{k!} \times H_k.$$
This yields the answer
$$\bbox[5px,border:2px solid #00A000]{ n H_k.}$$
What we have here are in fact two annihilated coefficient extractors
(ACE)    more    of   which    may    be    found    at   this    MSE
link.
