TITLE: A combinatorial solution to the tool maker problem
QUESTION [6 upvotes]: I'm trying to verify the claim presented in this question. 
Let's assume that we are interested in the following probability in both cases (i.e. sorted piles(SP) and messy heap(MH)):
The probability of choosing $k$ stones from $N$ stones with which the tool maker can make $k$ different tools.

For the (SP) case:
The probability is actually one. Because we have distinct piles of each stone's type. So, one can pick one stone from each pile (Assume we have $k$ types of stone with the frequency of $n_k$ such that $\sum_{k} n_k = N$):
$P_{SP}(X) = \dfrac{{{n_1}\choose{1}}{{n_2}\choose{1}}...{{n_k}\choose{1}}}   {{{n_1}\choose{1}}{{n_2}\choose{1}}...{{n_k}\choose{1}}} = 1$
For the (MH) case:
The selection space now is all of the stones, as there is no labelled type to any arbitrary stone. Hence:
$P_{MH}(X) = \dfrac{{{n_1}\choose{1}}{{n_2}\choose{1}}...{{n_k}\choose{1}}}   {{N}\choose{k}}$

The claim is:
$\lim_{N\to\infty} \dfrac{P_{SP}(X)}{P_{MH}(X)} = k$
Now, one can use Stirling's Approximation leading to:
$\lim_{N\to\infty} \dfrac{P_{SP}(X)}{P_{MH}(X)} = \lim_{N\to\infty} \dfrac{{N}\choose{k}}{{{n_1}\choose{1}}{{n_2}\choose{1}}...{{n_k}\choose{1}}} = \lim_{N\to\infty} \dfrac{\dfrac{N!}{k!(N-k)!}}{n_{1}*n_{2}*...*n_{k}}$
How should I proceed now?!

Update:
I myself solved the question:
Let's assume that all frequencies are the same (for the sake of simplicity):
$n_{1} = n_{2} = ... n_{k} = \psi$,
therefore, $k\psi = N \Rightarrow \psi = \dfrac{N}{k}$
$\lim_{N\to\infty} \dfrac{{N}\choose{k}}{n_{1}*n_{2}*...*n_{k}} = \lim_{N\to\infty} \dfrac{{N}\choose{k}}{(\dfrac{N}{k})^{k}}$
Now by the following approximation, we have:

${{N}\choose{k}} \approx \dfrac{N^k}{k!}$ if $N\gg k$

$\lim_{N\to\infty} \dfrac{{N}\choose{k}}{(\dfrac{N}{k})^{k}} = \lim_{N\to\infty} \dfrac{\dfrac{N^k}{k!}}{\dfrac{N^k}{k^k}} = \dfrac{k^k}{k!}$  
Let $\phi = \dfrac{k^k}{k!}$
We can use the following approximation here:

$\log(k!) \approx k\log k - k$,

So,
$\log \phi = \log k^{k} - \log k! = k\log k - [k\log k -k] = k$
Finally,
$\phi = exp(k)$
I should have reached to $k$, not $exp(k)$.
What's wrong with my argument?!

REPLY [2 votes]: the sorted architecture is roughly $k$-times more efficient than no
    architecture

means that if the sorted architecture expected time to a solution is $t$, then the messy heap can expect to solve in $kt$.
Which I'm fairly sure is nonsense, and there is nothing wrong with your argument.