Selecting 2 non-adjacent objects from N objects in a line I tried thinking of an intuitive way by only considering the objects not at the ends, but then it seems to fail due to overcounting, is there a simpler way to approach the problem?
 A: It's the number of any two objects, minus the number of any two objects next to each other. So:
$${{N}\choose{2}} - (N -1)$$
A: By way of example let's say $N=10$ then list our elements and pick the two we want using an up-arrow, one such selection
$$\begin{array}{cccccccccc}
1&2&3&4&5&6&7&8&9&10\\
\uparrow&\bullet&\bullet&\bullet&\uparrow&\bullet&\bullet&\bullet&\bullet&\bullet
\end{array}$$
where bullets indicate numbers not selected.
We can see that we are just arranging $2$ identical up-arrows and $N-2$ bullets so that the up-arrows are not adjacent. So instead count placements of up-arrows in the $N-1$ spaces between bullets, for our example
$$\overbrace{\square\bullet\square\bullet\square\bullet\square\bullet\square\bullet\square\bullet\square\bullet\square\bullet\square}^{\text{8 }\bullet\text{s, 9 spaces}}$$
The $2$ up-arrows may be arranged in the $N-1$ remaining spaces so that there is at most $1$ arrow per space in

$$\text{Answer}=\frac{(N-1)!}{2!(N-3)!}$$

ways. 
This is because we are essentially arranging $2$ identical up-arrows with the remaining $N-3$ spaces e.g. the selection shown at the top is
$$\uparrow\bullet\square\bullet\square\bullet\square\bullet\uparrow\bullet\square\bullet\square\bullet\square\bullet\square$$
but just looking at arrangements of $2$ identical up-arrows and $9$ identical spaces
$$\uparrow\square\square\square\uparrow\square\square\square\square$$
so in this example of $N=10$ case we count arrangements $7$ remaining spaces and $2$ up-arrows:
$$\frac{9!}{2!7!}$$

Note that this is the binomial coefficient:
$$\frac{(N-1)!}{2!(N-3)!}=\binom{N-1}{2}$$
Hopefully you can see how to generalise this argument for selecting $3,4,5,\ldots$ non-adjacent numbers/objects.
