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.



These green blocks are lying flat; there's 38 of them.  Make ten layers of this.  Then, fill in the blue area with six layers of blocks lying on their side; now there is a 2x4x4 space above it.  Similarly, fill in the orange area with five layers of blocks standing up; now there is no space above those.  Now, partially fill the remaining 2x4x4 space with a single 2x3x4 block.
Finally, you need one more 22x22 layer that has four 1x1x2 spots open.  You can use the blue display from Nominal Animal's post, or you can use the one below that I came up with.
Final score: 10 layers of 38, plus five layers of 2, plus six layers of 2, plus 1, plus one layer of 40.  380 + 10 + 12 + 1 + 40 = 443.

