If a fair six-sided die is rolled four times, in how many outcomes is the value of each roll at least as large as the value of the previous roll? Suppose you roll a fair 6-sided die four times.
Let C be the event that the value of each roll is at least as large as the value of the previous roll.
What is the probability of C?
I know that $$\omega = 6^4 = 1296$$
I also know that to get P(C),  I need to divide C by $\omega$
But what is the fastest/simpler way to get C? 
I could write all the sets that would qualify but that would  A: be time consuming.
B: error prone. 
Is there a formula to find C in this case?
A: A non-decreasing sequence is completely determined by the number of times each outcome occurs.  Hence, the number of non-decreasing sequences of four rolls is the number of solutions of the equation
$$x_1 + x_2 + x_3 + x_4 + x_5 + x_6 = 4 \tag{1}$$
in the non-negative integers.  A particular solution of equation corresponds to the placement of five addition signs in a row of four ones.  For instance,
$$+ + 1 1 + + 1 + 1$$
corresponds to $x_1 = x_2 = 0$, $x_3 = 2$, $x_4 = 0$, $x_5 = x_6 = 1$ (that is, to the sequence of rolls $3, 3, 5, 6$).  The number of such solutions is the number of ways five addition signs can be placed in a row of four ones, which is 
$$\binom{4 + 5}{5} = \binom{9}{5}$$
since we must choose which five of the nine symbols (four ones and five addition signs) will be addition signs.
A: Finding $|C|$ is equivalent to the problem that finding the number of functions from $X=\{1,2,3,4\}$ to $Y=\{1,2,3,4,5,6\}$ and $f(1)\le f(2)\le f(3)\le f(4)$. Then, just select elements of $Y$ with repetition. Since $f$ monotonely increases, $f(1),f(2),f(3),f(4)$ are determined automatically. Thus
$$
|C|=\left(\!\binom{6}{4}\!\right) = \binom{6+4-1}{4}=\binom{9}{4}.
$$
