How many ways are there to assign 5 points into three scores with a range of -1 to 4? You have 5 points to assign between 3 scores; that is to say, the three scores must total to 5 exactly. A score may go as high as 4 and as low as -1. How many different ways are there to assign these points?
My attempt at brute force got 27 answers:


*

*[-1,2,4] which can be assigned 6 ways.

*[-1,3,3] which can be assigned 3 ways. 

*[0,1,4] which can be assigned 6 ways.

*[0,2,3] which can be assigned 6 ways.

*[1,1,3] which can be assigned 3 ways.

*[1,2,2] which can be assigned 3 ways.


However, I fear I missed something and am leaving out some answers.

Your answer of $27$ is correct.
For a non-brute-force solution, add one to each score, so that you obtain equivalently three scores in the range $\{0,1,2,3,4,5\}$ adding to $8$.  Then the answer is the coefficient of $x^8$ in
\begin{align*}
(1 + x + x^2 + \cdots + x^5)^3
&= \left( \frac{1}{1-x} - \frac{x^6}{1-x} \right)^3 \\
&= \frac{1}{(1-x)^3} - \frac{3x^6}{(1-x)^3} + \frac{3x^{12}}{(1-x)^3} - \frac{x^{18}}{(1-x)^3}
\end{align*}
The $x^k$ coefficient of $\frac{1}{(1-x)^3}$ is $k + 2 \choose 2$, so the $x^8$ coefficient in the above expression is
$$
{8 + 2 \choose 2} - 3 {2 + 2 \choose 2} + 0 + 0
= 45 - 18 = 27.
$$
