Can this probability question be solved using $3$D cube volume? 
A dice is rolled $3$ times. Find probability of getting a larger number than the previous number each time.

The answer in the book is : $^6C_3/6^3=20/216$. 
But I am thinking of solving it using geometry. Had the question been of $2$ dice rolls, I would have drawn a $6\times6$ grid of possible outcomes and found area of upper right triangle. Since it is for $3$ trials, can we draw a $6\times6\times6$ cube and then find the volume to find the probability?
 A: 
Kindly, consider the cube above. It's a $6\times6\times6$ cube.
$\qquad\qquad(661)(662)\dots(666)$
$\qquad\qquad\dots$
$\quad\quad(631)(632)\dots(636)$
$\quad(621)(622)\dots(626)$
$(611)(612)\dots(616)$
$\vdots\qquad\qquad\qquad\vdots$
$\vdots\qquad\qquad\qquad\vdots$
$(211)(212)\dots(216)$
$(111)(112)\dots(116)$
Suppose you number the cube in above fashion.
Now you'll definitely observe a pattern out of it. 
Since you are interested in points of form $(a,b,c)$ where $a<b<c$, following pattern/figure will emerge:

You'll note the following:

*

*It was fun but never again!

*Isn't there an easier/smarter method to solve such questions?

Thus, you'll end up earning for and praising the combinatorics method/logic/approach.

There are anyways a lot of limitations with the cube method:

*

*What if 4 dice are rolled!

*Takes a lot of time to create a mental image of the shape that will come about.

*How will you calculate the area!

*What if there's not really any pattern at all!

For the above reasons we don't prefer to solve using geometry.
Every method has its certain application where it suits best.
I'm not saying to not try solving using geometry, you should, for fun's sake, that what makes maths beautiful and for sure you'll end up learning something useful.
To have some fun, you may visit the $3$D cube here: https://www.geogebra.org/calculator/ennsywtt
A: Yes. You're basically listing all possible $6^3$ outcomes and finding which ones result in "success". You can use the geometric structure to make the problem more intuitive, which you are "allowed" to do because all outcomes here are equiprobable, and each die roll is independent of the others.
