Sorry for the long silence. Believe it or not, it was impossible to find even 10 minutes until now, not a full hour I need to explain everything in a decent way. Even now I'm starting but I'm not at all sure I'll finish. I'll try to do it in as few shots as possible but I apologize in advance if I will need to bump this thread a few times.

*Part 1. Generating functions and counters.*

The key idea is that if $A_j$ are some sets of non-negative integers and $N_a$ is the number of ordered representations $a=a_1+a_2+\dots$ with $a_j\in A_j$, then
$$
\sum_{a\ge 0} N_az^a=\prod_j\left(\sum_{a_j\in A_j} z^{a_j}\right)\\,.
$$
Unfortunately, this simplest form is not quite suitable for the weighted letter word counting. However, we can tweak this formula a bit.

Suppose we have an alphabet with letter weights $w_1,w_2,\dots$ and can use any letter any number of times. The number of words of weight $W$ we can create is
$$
\sum_{\sum_j \ell_jw_j=W}\frac{\left(\sum_j\ell_j\right)!}{\prod_j\ell_j!}
$$
(the standard combination with repetitions formula).
Thus, taking into account that $\sum_{\ell\ge 0}\frac{Z^\ell}{\ell!}=e^Z$, we get almost what we want for the coefficient at $z^W$ if we consider the product
$$
\prod_j \exp(z^{w_j})
$$

The only problem is that each word is counted not with the weight $1$, as it should in the uniform sampling, but with the weight that is the inverse factorial of its length, which skews the uniform distribution quite a bit. The way to compensate for that is to guess the typical length $L$ and to change the function to
$$
\prod_j \exp(Lz^{w_j})
$$
Now, the coefficient is multiplied by $L^{\sum_j\ell_j}$, which is approximately proportional to $\left(\sum_j\ell_j\right)!$ as long as $\sum_j\ell_j\approx L$. As a matter of fact, this weighing emphasizes the words of the length $L$ a bit stronger than it should because $\frac{L^\ell}{\ell!}$ is maximized at $L$. However, as long as $\ell-L=o(\sqrt L)$, the skewing it introduces is negligible. This is what I called the *local* counting function: the weights are way too much suppressed outside a small window but we hope that outside that window we have only a small portion of words anyway, so suppressing them even further changes nothing in the picture.

If we have a clear idea of what the typical length is, we can introduce all other kinds of counters. To count the number of distinct letters, we need a variable that appears only once if the letter is used at all no matter how many times the letter is used after that. The factor $1+s(e^{Lz^{w_j}}-1)$ does exactly that if you look at the "typical" power of $s$ in the expansion (recall that $\frac{z^{\ell w_j}}{\ell!}$ corresponds to using the $j$-th letter $\ell$ times, so $s$ should appear once if $\ell>0$ and not appear if $\ell=0$). The unique letter counter should appear only if $\ell=1$, so $e^{Lz^{w_j}}-(1-s)Lz^{w^j}$ does the job adding the $s$-factor to the "linear term" in the expansion of the exponent but not anywhere else. You can now play a bit setting various counters yourself to see what generating functions to consider in various cases.

*Part 2. The central term extraction.*

Suppose now that we have a function $F(s,r)=\sum_{\ell,w}N(\ell,w)s^\ell r^w$ of two variables (you can trivially generalize this to more than two variables as well but I do not want to do the one-variable case because some games you can play with several variables would be invisible there). Suppose that we want to estimate $N(L,W)$. The obvious upper bound (the whole is larger than its part) is
$$
N(W,L)\le s^{-L}r^{-W}F(s,r)
$$
where we are free to choose $s$ and $r$. Of course, we are going to choose them so that the right hand side is as small as possible. This leads to the minimization problem
$$
G(s,r)-L\log s-W\log r\to\min\\,.
$$
Suppose that $(1,r)$ is a stationary point of the objective function, i.e., the differential vanishes there. Suppose also that, after switching to the variables $\log s,\log r$ (which makes the subtracted terms linear), the second differential is bounded by
$A^2(ds)^2+B^2(dr)^2$ near this stationary point. Then we can easily control the sum of all terms that correspond to pairs $\ell,w$ that differ a lot from the pair $(L,W)$ in $F(s,r)$ by looking at $F(se^\sigma,re^\rho)$. We still have
$$
N(\ell,w)s^\ell r^w\le e^{-\sigma\ell-\rho w}F(se^\sigma,re^\rho)=
e^{-\sigma(\ell-L)-\rho (w-W)} e^{-\sigma L-\rho W}F(se^\sigma,re^\rho)\le
e^{-\sigma(\ell-L)-\rho (w-W)}e^{A^2\sigma^2+B^2\rho^2}F(s,r)
$$
due to the stationarity and the second derivative estimate.

Now, choosing $|\sigma|=A^{-1},|\rho|=B^{-1}$, we see that
$$
N(\ell,w)s^\ell r^w\le \exp(-A^{-1}|\ell-L|+B^{-1}|w-W|)F(s,r)\\,,
$$
so the terms with $|\ell-L|>A'$ or $|w-W|>B'$ can contribute at most $[Be^{-A'/A}+Ae^{-B'/B}]$ to $F(s,r)$. This crude bound is often enough to show that the main contribution comes from the terms with $w\approx W$, $\ell\approx L$, which sort of allows us to say that the pair $(L,W)$ is typical in the weighted counting where the pair $(\ell,w)$ has the weight $N(\ell,w)s^\ell r^w$. The key idea is that if $s=1$, then this counting is uniform in $\ell$ for each fixed $w$, so getting a typical pair in such weighted counting is essentially the same as getting a typical $L$ for fixed $W$.

*Part 3. Circle method and mountain pass.*

To be continued...