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.
For part b, you have an expected time of $1$ to get the first distinct ball, $\frac n{n-1}$ for the second distinct ball, $\frac n{n-2}$ for the third, and on to $\frac n{n-k+1}$ for the $k^{\text{th}}$. This sums to $n(H_n-H_{n-k})$
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.
