AHSME 1981 #22 - Number of lines that pass through four distinct points 
How many lines in a three dimensional rectangular coordinate system
  pass through four distinct points of the form $(i, j, k)$ where $i$, $j$, and $k$
  are positive integers not exceeding four? 
$\text{(A)} \ 60 \qquad \text{(B)} \ 64 \qquad \text{(C)} \ 72 \qquad \text{(D)} \ 76 \qquad \text{(E)} \ 100$

I am interested in knowing why my solution below is incorrect:
Any line in 3D space is characterized by an initial point and a vector.
Starting with an initial point of form $(x, y, z)$ and a vector of $(\Delta x, \Delta y, \Delta z)$ you can get other points on the line by adding the vector to the initial point. Since we are interested in a line that passes through four points, the fourth point on this line will be $(x + 3\Delta x, y + 3\Delta y, z + 3\Delta z)$
You need each of these points to be between 1 and 4 (inclusive). If the initial and final points are within the boundary then all points inbetween will also be.
Thus if, $$1 \leq x\leq 4$$
and $$1 \leq x + 3\Delta x \leq 4$$ then you have the following choices:
$x = 1$ and $\Delta x = 0, 1$
$x = 2$ and $\Delta x = 0$
$x = 3$ and $\Delta x = 0$
$x = 4$ and $\Delta x = 0$
Resulting in a total of $1\cdot2 + 1\cdot1 + 1\cdot1 + 1\cdot1 = 5$ valid for $x$ and $\Delta x$. The same holds true for the $y$ and $z$ cases, resulting in a total of $5^3$ total choices. 
However, because the points need to be distinct a vector of $(0, 0, 0)$ is not allowed, and thus we must subtract all cases where $\Delta x = 0$, which make up $(1\cdot1 + 1\cdot1 + 1\cdot1 + 1\cdot1)^3$ cases.
The total is thus $5^3 - 4^3 = 61$. However, the correct answer is 76. Why did I not get the right answer?
 A: You’re assuming that $\Delta x,\Delta y,\Delta z\ge 0$ but that doesn’t cover all of the possibilities. The diagonal that runs from $\langle 1,4,4\rangle$ to $\langle 4,1,1\rangle$, for instance, isn’t covered by any of your cases.
A: You can think of it in a physical sense instead of math.
For each pair of opposite faces, there are 16 lines that can go through the $4\times4$ cube perpendicularly.  So that's $3\times 16=48$.
And for each pair of diagonally opposite edges, there are 4 lines that will go through 4 lattice points.  So that's $6\times 4 = 24$
Then for each opposite pair of vertices there is one line.  $4\times 1$
Add them up and you get 76 :)
ETA:
Doesn't really address your question, but I think these are all the point vector solutions.
\begin{array}{|c|c|c|} \hline
(x,y,z)& (\Delta x, \Delta y, \Delta z) & \text{count}\\ \hline
(1,y,z) & (1, 0, 0) & 16\\ \hline
(x,1,z) & (0, 1, 0) & 16\\ \hline
(x,y,1) & (0, 0, 1) & 16\\ \hline
(1,1,z) & (1, 1, 0) & 4\\ \hline
(4,1,z) & (-1, 1, 0) & 4\\ \hline
(1,y,1) & (1, 0, 1) & 4\\ \hline
(4,y,1) & (-1, 0, 1) & 4\\ \hline
(x,1,1) & (0, 1, 1) & 4\\ \hline
(x,4,1) & (0, -1, 1) & 4\\ \hline
(1,1,1) & (1, 1, 1) & 1\\ \hline
(1,4,4) & (1, -1, -1) & 1\\ \hline
(4,1,4) & (-1, 1, -1) & 1\\ \hline
(4,4,1) & (-1, -1, 1) & 1\\ \hline
\end{array}
A: Just reading your solution, when I got to the part about $x = 4$, then $\Delta x = -1$ is a permissible condition and I stopped right there.
Reading a bit further, we can also immediately see that your enumeration is flawed in a second way, which is that the choices of $(x, \Delta x)$, $(y, \Delta y)$, and $(z, \Delta z)$ are not independent of each other, which is what you are assuming when writing $5^3$.
