What is the number of non-increasing 4 digit numbers? This problem, though quite simple, has stumped my teenage mind. How many 4-digit numbers are there whose digits are non-increasing? This seemed quite simple at first, meaning I have the digits [9 8 7 6 5 4 3 2 1 0], and I just pick four numbers from that which gives me the answer of 10C4, 210, but this fails to include the fact that a possible solution could be 9999, or 9988, or 6332, a number with repeating digits. How would I go about including this in an equation? What would that equation look like? I presume the solution would be something relating to the casework method, but I'm not very familiar with it in the first place.

Alternative approach that reaches the same answer as that of Bulbasaur.
Actually, his Combination with Repetition article can be construed to be an alternative presentation of Stars and Bars which is also discussed here.
Consider any specific solution to
$$x_1 + x_2 + \cdots + x_{10} = 4, \tag1 $$
where $x_1, x_2, \cdots, x_{10}$ are all required to be 
non negative integers,

and $x_k$ refers to the number of occurrences of the digit
$k$ 
in the 4 digit number, when $k \in \{1,2,\cdots, 9\}$.
You can then let $x_{10}$ refer to the number of zeroes in the 4 digit number.
For each such solution, there is exactly one way that the 4 digits that correspond to the solution may be permuted into a non-increasing order.
Therefore, the enumeration of the possible satisfying number of 4 digit numbers is the same as the enumeration of the number of solutions to the equation in (1) above.
Per the linked articles, the number of solutions to 
$x_1 + x_2 + \cdots + x_k = n$, 
where $x_1, \cdots, x_k$ are required to be non-negative integers is 
$\displaystyle \binom{n + [k-1]}{k-1}.$
In the present problem, $n = 4$ and $k = 10$.
Further, as suggested in the answer by Bulbasaur, since having the leftmost digit equal $0$ is normally prohibited, the specific 4 digit string of $0-0-0-0$ may or may not be required to be deducted from the overall count.
