Formula for number of lines you can draw through $n$ points So I've got a homework question I'm stuck on. It's asking me to develop a formula that when given $n$ points, it gives the number of straight lines that can be drawn through those points. 
For example, the first two questions were "How many lines can be drawn through 3 points?" Which is 3, and "How many lines can be drawn through 4 points?" Which is 6. Now, it says "Develop a formula that gives the number of lines that can be drawn through $n$ points." I understand the relationship, and i've developed a formula, but it relies on having a predetermined list of answers.
I've found that where $a_n$ is the number of points you have, you can find the number of lines with $a_n = a_{n-1} + (n-1)$ where $a_{n-1}$ is the previous item in the list. However, I don't think that's the formula that my teacher is looking for. Is there a less complicated way to solve this problem?
 A: It says "drawn through $n$ points", but it really means "drawn through any two of $n$ points", where we assume no three points are collinear.  So the number of lines is the same as the number of ways to choose two points out of $n$, where order doesn't count.  Do you know about permutations, combinations, binomial coefficients?
A: We will assume that our $n$ points are in "general position." This means that no three of our points are on the same line. 
Let our points be $P_1, P_2, P_3, \dots, P_n$.
First draw all the lines through $P_1$ and every other point. There are $n-1$ of these.
Now draw all the lines through $P_2$, and every other point, except for the line through $P_1$ and $P_2$, since that has already been drawn. There are $n-2$ of these, through $P_3$, $P_4$, and so on up to $P_n$.
Now draw all the lines through $P_3$ and every other point that have not already been drawn. We have already taken care of the line through $P_3$ and $P_1$, and also of the line through $P_3$ and $P_2$, so there are $n-3$ of these.
Continue. At the end, all we are drawing is the line through $P_{n-1}$ and $P_n$, just $1$ line.
Thus the total number of lines is 
$$(n-1)+(n-2)+(n-3)+\cdots+2+1,$$
which will look more familiar as 
$$1+2+3+\cdots +(n-1).$$
You probably know a formula for the sum of the first $k$ positive integers.
Another way: Sit in turn on every one of our $n$ points. Draw the lines through that point and every other point. So for each of our $n$ points, you draw $n-1$ lines, for a total of $n(n-1)$.
However, this means that you have drawn every line twice. The line through $P_i$ and $P_j$ has been drawn once when you were sitting on $P_i$, and once again when you were sitting on $P_j$. So $n(n-1)$ overcounts our lines by a factor of $2$. That means that the actual number of lines is 
$$\frac{n(n-1)}{2}.$$ 
A: Assuming no three of the points are collinear (which I think is a fair assumption for this question based on the answer), any such line is uniquely determined by a choice of $2$ of the $n$ points. Conversely given any $2$ points there is a unique line passing through them.
Thus there are $\left(\begin{matrix}n\\2\end{matrix}\right) = \frac{n(n-1)}{2}$ possible lines.
