non-separable assignments of the vertices of a hypercube I have a question regarding exercise 14.17 in An Introduction to Information Retrieval by Manning et al. 

The problem is:


  "Assuming two classes, show that the percentage of non-separable
  assignments of the vertices of a hypercube decreases with
  dimensionality M for M > 1. For example, for M = 1 the proportion of
  non-separable assignments is 0, for M = 2, it is 2/16. Solve the
  exercise either analytically or by simulation."


The total number of assignments of vertices of an N-dimensional hypercube is: $2^{(2^N)}$

And as I found in here the number of separable assignments is O($2^{(N^2)}$).
So the percentage of non-separable assignments is increasing with N (which is the opposite of what is said in the exercise). 

What am I missing here?
You appear to be correct. I believe this is a simple proof of the opposite fact:

Let $s(n)$ denote the number of separable sets in the $n$-cube. Let $Q_0, Q_1$ denote the subcubes of one smaller dimension, with say last coordinates equal to 0, 1 respectively. The key fact is that any separable set of the whole cube intersects with each of $Q_0, Q_1$ in separable sets. (The separating hyperplane intersects the codimension-1 spaces to form separating hyperplanes.) So 
$$s(n)\leq [s(n-1)]^2.$$

Now the proportion $r(n)$ satisfies
$$r(n)=
\frac{s(n)}{2^{2^n}}\leq
\frac{[s(n-1)]^2}{(2^{2^{n-1}})^2}=
[r(n-1)]^2&lt;
r(n-1),$$
where the last inequality holds because $r(n-1)&lt;1$ for $n&gt;2$.