A die was thrown three times. Find probability that first throw is less than the second and second is less than the third throw. My proposed solution follows. Is this correct? If yes, I'd like to know if there is a better way to solve this problem.
We can see that the value of the first throw could only be $1, 2, 3$ or $4$.
Throws are labeled as $T_1, T_2, T_3$.
If $T_1$ is $1$ then $T_2$, $T_3 \in \{2,3,4,5,6\}$
If $T_1$ is $2$ then $T_2$, $T_3 \in \{3,4,5,6\}$
If $T_1$ is $3$ then $T_2$, $T_3 \in \{4,5,6\}$
If $T_1$ is $4$ then $T_2$, $T_3 \in \{5,6\}$
Hence the number of favorable outcomes is $\dbinom{2}{2}+\dbinom{3}{2}+\dbinom{4}{2}+\dbinom{5}{2} = 20$
And $P(A) = \frac{20}{6\times6\times6}=\frac{5}{54}$

That method would be hard to scale up to larger numbers.
As an alternative, for an $N$ sided die:  First note that the probability that the rolls are distinct is $$1\times \frac {N-1}N\times \frac {N-2}N$$
Given that the rolls are distinct, there are $6$ possible, equally probable,  orders for them, only one of which is increasing.  Thus the answer to your question is $$\frac 16\times 1\times \frac {N-1}N\times \frac {N-2}N=\boxed {\frac {(N-1)(N-2)}{6N^2}}$$
Note that if $N=6$ this resolves to $$\frac {5\times 4}{6\times 36}=\frac {20}{6^3}$$  as desired.
Should say:  it's a numerical coincidence that, in your case, the number of ways to order $3$ distinct items happens to be the same as the number of faces on a standard die.
