Using the technique from the following MSE
link we have for
the probability from first principles (exponential generating
functions and generalized Stirling numbers) that it is given by
$$\frac{1}{M!} {N\choose M}^{-1} \times
M! [z^M] \prod_{j=1}^C
\sum_{k=1}^{c_j} \frac{c_j!}{(c_j-k)!} \frac{z^k}{k!}
= {N\choose M}^{-1}
[z^M] \prod_{j=1}^C
\sum_{k=1}^{c_j} {c_j\choose k} z^k.$$
This is
$$\bbox[5px,border:2px solid #00A000]{
{N\choose M}^{-1}
[z^M] \prod_{j=1}^C (-1+(1+z)^{c_j}).}$$
For the special case of all clusters having the same size $j$
we get
$${N\choose M}^{-1} [z^M] (-1+(1+z)^j)^C
= {N\choose M}^{-1} [z^M]
\sum_{q=0}^C {C\choose q} (-1)^{C-q} (1+z)^{qj}.$$
This is
$$\bbox[5px,border:2px solid #00A000]{
{N\choose M}^{-1}
\sum_{q=0}^C {C\choose q} (-1)^{C-q} {qj\choose M}.}$$
We can use this to compute the expected number of draws until a
representative from every cluster has been seen. Note that the
complementary probability counts draws where at least one type of
cluster is missing, i.e. the number of draws until having seen all is
more than $M.$ Hence we get for the expectation
$$\bbox[5px,border:2px solid #00A000]{
\mathrm{E}[T] = N-j+1
- \sum_{M=0}^{N-j} {N\choose M}^{-1}
\sum_{q=0}^C {C\choose q} (-1)^{C-q} {qj\choose M}.}$$
As a sanity check when $j=1$ the expectation should be $C.$
We obtain
$$C
- \sum_{M=0}^{C-1} {C\choose M}^{-1}
\sum_{q=M}^C {C\choose q} (-1)^{C-q} {q\choose M}.$$
Now we have
$${C\choose q} {q\choose M} =
\frac{C!}{(C-q)! \times M! \times (q-M)!}
= {C\choose M} {C-M\choose C-q}.$$
Substituting we find
$$C
- \sum_{M=0}^{C-1} {C\choose M}^{-1} {C\choose M}
\sum_{q=M}^C {C-M\choose C-q} (-1)^{C-q}
\\ = C
- \sum_{M=0}^{C-1}
\sum_{q=0}^{C-M} {C-M\choose C-M-q} (-1)^{C-M-q}
= C
- \sum_{M=0}^{C-1}
\sum_{q=0}^{C-M} {C-M\choose q} (-1)^{q}
\\ = C - \sum_{M=0}^{C-1} 0 = C,$$
as claimed. Here we have used that $C-1\ge M$ or $C\ge M+1\gt M$.
