Probability that each number obtained by throwing a die is no smaller than the preceding number 
A fair die is thrown 4 times. Find the probability that the each number obtained is no smaller than the preceding number.



*

*If all numbers obtained are same, number of such outcomes $$=\dbinom{6}{1}=6$$

*If 3 numbers obtained are same, number of ways is equal  to the number of ways of choosing any two numbers $=\dbinom{6}{2}$. But these two number can be arranged as $x,y,y,y$ or $x,x,x,y$ where $x<y$. So $$2\times\dbinom{6}{2}$$

*If 2 numbers are same, the cases are $a_1<a_2<a_3=a_4$, $a_1<a_2=a_3<a_4$, $a_1=a_2<a_3<a_4$, $a_1=a_2<a_3=a_4$. Number of ways $$=3\times\dbinom{6}{3}+\dbinom{6}{2}$$

*if all 4 numbers are different, $$\dbinom{6}{4}$$


Total number of possibilities $=6^4$.
Even though I got the correct answer, is there any shorter method?

Nope, that seems likely to be the most concise way to do it.
Count ways to pick $n\in\{1,2,3,4\}$ unique numbers, and to arrange them with $n-1$ "$>$" signs, to meet the criteria.
$$\begin{array}{|l:l|} \hline
\rm a{=}a{=}a{=}a & \dbinom 3 0 \dbinom 6 1
\\\hdashline\rm a{=}a{=}a{>}b , a{=}a{>}b{=}b, a{>}b{=}b{=}b & \dbinom 3 1\dbinom 6 2
\\\hdashline\rm a{=}a{>}b{>}c, a{>}b{=}b{>}c, a{>}b{>}c{=}c & \dbinom 3 2 \dbinom 6 3
\\\hdashline\rm a{>}b{>}c{>}d & \dbinom 3 3 \dbinom 6 4
\\ \hline\end{array}$$
$$\dfrac{\sum_{n=1}^4 \dbinom{3}{n-1}\dbinom{6}{n}}{6^4} = \dfrac{\dbinom 6 1 + 3\dbinom 6 2 + 3 \dbinom 6 3 + \dbinom 6 4}{6^4}$$
