Number of small cuboids inside a larger cube 
How many small cuboids of dimension $2$ m × $3$ m × $4$ m can be
  accommodated in a cube of side $22$ m?

(This is similar to the question from this site. But probably more complex)
This question is available in the following sources.
Source 1: quora.com
Source 2: careerbless.com
Source 3: m4maths.com
As clear from  Quora.com and careerbless.com, answer cannot be $\dfrac{22^3}{2 \times 3 \times 4}=443$, but $443$ is an upper bound. This is clear because there is some waste of material.
Both these sources give answer as $385$ . This is also straight forward, because $22/2=11\ $, $\ 22/3 =7$ (avoiding fraction), $\ 22/4 =5$ (avoiding fraction) and total number of small cuboids possible is then $11 \times 7 \times 5= 385$
But in, careerbless.com, one more option is given by showing the different orientation of the small cuboids. See the table given in my link. Thus they show $442$ smalll cuboids can be fit inside the actual cube, if different orientation and rotation is taken into account.

I want to know the right answer for this problem. Whether it is $385$
  or $442$ ? If $442$ is also a possible answer, how this can be generalized (i.e., is there any generalized approach or formula), because that process appears too complex for me. Please help.


Here is a solution for the 443 tiles of 2×3×4 cells each, in a 22×22×22 volume:

To "build" the solution:


*

*Use 11 layers of tiles as shown above in the leftmost picture, to fill in the volume except for a centered 2×2×16 volume, and 2×2×22 on top of it. 
Since each layer takes 5 + 4×6 + 5 = 34 tiles, this step uses a total of 11 × 34 = 374 tiles.

*Rotate the volume so that you face the void.

*Fill in deeper 2×2×16 volume, as shown in the middle figure. This will leave a 4×1×2 hole. This uses 29 additional tiles.

*Fill in the top two layers as shown in the rightmost figure. This will leave four 1×1×2 holes. This uses 40 additional tiles.
The total number of tiles used is 11×34+29+40 = 443.
The entire volume consists of 22×22×22 = 10,648 cells. Each tile occupies 2×3×4 = 24 cells. The 443 tiles occupy 443×24 = 10,632 cells. We left 1×4×2 + 1×1×2 + 1×1×2 + 1×1×2 + 1×1×2 = 16 cells unoccupied. (Each black box signifies two unoccupied cells, since each layer in the figures above is two cells thick.)
The sum of occupied and unoccupied cells, 10,632 + 16 = 10,648, matches the number of cells in the volume.
Note that there are many other equivalent solutions, as especially in step 3 you can choose where the 1×4×2 void ends up.

These types of packing problems are generally considered hard, in the sense that there are no simple generic algorithms that yield optimal solutions. Often, optimal solutions are found through various types of searches, sometimes even brute-force solutions.
I stumbled onto this solution by accident.
