Solving for possible orientations of 3 objects on a 3x3 grid Say you have a 3x3 grid, and 3 objects to work with. Each occupies one space. How would I go about solving the amount of ways they can lay on the grid.
Example: (pardon my bad ASCII art)
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[x][x][x]
Any help would be much appreciated, thanks!

To reiterate what was said in the comments, imagine temporarily that your grid has its positions labeled.
$\begin{bmatrix}1&2&3\\4&5&6\\7&8&9\end{bmatrix}$
A specific arrangement of the objects will then correspond to a selection of three numbers from $1$ to $9$.
$\begin{bmatrix}\circ&\times&\circ\\\circ&\circ&\times\\\circ&\circ&\times\end{bmatrix}$ would correspond to the set $\{2,6,9\}$
Assuming you consider $\begin{bmatrix}\circ&\times&\circ\\\circ&\circ&\times\\\circ&\circ&\times\end{bmatrix}$ to be a different arrangement than $\begin{bmatrix}\circ&\times&\times\\\times&\circ&\circ\\\circ&\circ&\circ\end{bmatrix}$, one can move freely between the set of desired arrangements and the set of subsets of size three of $\{1,2,\dots,9\}$ via a natural bijection.
The fact that there is a bijection between the sets in question implies that if we know how to count how many arrangements one of these questions has, we know how to count both.
We have thus boiled the question down to asking the question, "How many size three subsets exist of the set $\{1,2,\dots,9\}$?"
This is one of the many problems that binomial coefficients were designed to answer.  The answer being $\binom{9}{3}=\frac{9!}{3!6!}=\frac{9\cdot 8\cdot 7}{3\cdot 2\cdot 1} = 84$ possible arrangements.

If you consider rotations and/or reflections of your grids to be considered "the same", for example the two arrangements pictured above, the question becomes harder since we will have over counted.  Approach instead by Burnside's lemma.
