It turns out that this 3x3x3 cube, or tri-cube, assembly problem was discovered and described a long time ago. I found -- or rather, my daughter Pamela did -- an account of the history and some variations of it on a web page by Jürgen Köller, Math Basteleien - Soma Cube .

A subset of RUMIS Solitaire, namely the 3I-Q cube (where the 3I and the 4Q stones are not used) was analyzed under the name of the Soma Cube by H.J. Conway (of Life Game fame) and a colleague; they deduced that 240 solutions are possible.
On web page on the Soma Cube (see the URL above) one can also find a statement that the 3C-T cube (where the 3C and the T stones are not used) has no solution. This comes as part of the solution to the problem of assembling 3x3x3 cubes from all possible 'stones' containing 3 and 4 unit-cube elements, a collection known as Rehm's Cube Set.
Also listed are the number of solutions for all the other 3x3x3 cubes, giving the same two series of 7 cubes each that I used in my first post. There is no reason given why the 3C-T cube cannot be assembled.
However, a discussion of the possible solutions to the Soma Cube (3I-Q) is found on another web page, see Soma Cube Secrets .

Here the number of vertices in the assembled cube each stone can occupy is analyzed and used to derive a criterion for the existence of particular solutions, that is, whether a particular stone can be in a specific place within the cube, namely in 'normal', 'central', or 'deficient' position. This notion is introduced as the Hidden Secrets of Soma.
To see whether this secret could be helpful in finding the reason for the impossibility of the 3C-T solution, the maximal vertex counts for the 14 numerically possible 3x3x3 cubes from the Rehm Cube Set were determined, see the table below.

The table gives the sum of the maximal vertex count from all stones in any particular type of cube. There is no difference in the counts between the 3C-T and the 3C-L cubes from which follows that the concept of deficiency in the position of the L stone alone cannot explain the impossibility of assembling the 3C-T cube. In other words, the 'cornerness', namely counting the number of vertices the stones can contribute to the tri-cube, is not sufficient to decide on the possibility of a configuration.
The additional concept needed is the 'parity' as explained in Thiele's Impossibility Proof for the 3C-T tri-cube, see Thorsten Sillke's entries in the link to Math Basteleien. Parity essentially decides whether a stone can occupy a corner position or not, based on the number of 'black' and 'white' cubes in a stone if the tri-cube is colored in a checkered pattern. Then the difference between the L and the T piece becomes obvious in that parity forces the 3I stone, in the 3C-T tri-cube, into a position in which it cannot contribute to a vertex thus making a solution impossible. Conversely, for the 3C-L tricube, parity allows the 3I piece to occupy an edge where it contributes to two vertices thus making solutions possible.

While the web site which will depict known solutions to the cubes is still under construction, I want to describe some examples of known solutions. A solution depicts the visible stones of each face of a cube in a fold-out map that represents the six faces in the form of a cross with the left, top, and right face going across, and the back, top, front and bottom face going down. In each face, the stones are identified by their letter symbols with the shapes distinguished from each other by different colors. 

For example, the Q cube from the 3C series (which is the cube in which the stones 3C and Q are missing) is described by the grid system above. It is known as Rosmari's solution. 

The rules for the game Rumis, by Stefan Kögl, suggest several variations. One is Solitaire which consists of building cubes of various sizes with the building blocks (stones) contained in the game box. The stones come in four different colors, and there are eleven stones of each color. Each stone has a unique shape that comprises between 2 and 4 small unit cubes that are fused together in various geometric forms.
The smallest of the assembled cubes, measuring 3x3x3 units, can be built in a single color; for the larger cubes, stones from more than one color are needed.

In regard to the monochrome 3x3x3 cube task, one can ask several questions:
(i) What are the different classes of cubes that are possible?
(ii) Can an example for each class be constructed?
(iii) Are there different cubes in each class with the same stones but connected in different ways?

Illustrations and details of the analysis are provided below; here are some answers:

a) Of the 11 stones, two cannot be used for assembling a 3x3x3 cube.
b) Of the remaining nine stones, two main series of cubes can be built by leaving either one or the other of two specific stones out. That stone defines the class of cubes. The one not left out is used in all cubes of that series.
c) From the remaining seven stones there are then seven 3x3x3 cubes that could, in principle, be built where each is different by one of the seven stones being left over.
d) I have found at least one example for each cube in one of the two main series of cubes.
e) In the other series, I have found only six examples with the last cube so far having resisted successful assembly. It is my conjecture that that cube cannot be built.
f) For many of the other examples in both classes, there are several different assemblies possible.

Details of the analysis:

The stones consist of all possible arrangements of 2, 3, and 4 unit cubes such that they cannot be transformed into each other by rotations, see the picture; the single unit cube itself is not part of the stone set. Using alphabetic letter analogies to describe their shapes, the possible stones in a set are:

2I -- forming the letter I;
3I, 3C -- forming a larger letter I, and the letter C (or a corner in plane);
4I, 4L, 4T, 4R, 4Q -- these are planar stones where the letter designations I, L, and T are obvious, R resembles the lower-case r while Q stands for 'quad' as it forms a 2x2 square;
4C, 4S, 4D -- these stones extend into the 3rd dimension, C forming a corner, and D and S signifying the right-handed (dexter) and left-handed (sinister) versions of the two remaining 4-unit stone configurations (which are mirror images of each other).

An image with projections of the stones into the three spatial planes is also shown (to be added) to relate the stone shapes to the solution drawings below.

The total unit count of all the stones is 40. To form a 3x3x3 cube one needs 27 units. The 4I stone cannot be used since it is too large. That leaves 36 units and, therefore, stones with a total unit count of 9 will have to remain unused when a 3x3x3 cube is assembled. The only way to achieve this with stones consisting of 2, 3, or 4 units is to leave out one of each, that is, always stone 2I, and then either the 3I or the 3C stone plus one of the remaining 7 4-unit stones. This suggests that two series of 3x3x3 cubes can possibly be assembled; they are designated as the 3I and the 3C Main Series. Within each series, one of the 7 stones also left over identifies the specific 3x3x3 cube.

The two main series with different assemblies of monochrome 3x3x3 Rumis cubes are:

3I-C, -D, -L, -Q, -R, -S, -T
3C-C, -D, -L, -Q, -R, -S, -T

So far, at least one example was found for each of those cubes except for the 3C-T cube. The analysis of several attempts at the 3C-T cube indicates that it is not possible to assemble this 3x3x3 monochrome cube; that is currently my conjecture.

Any thoughts or suggestions?

Udo Pernisz