Number of straight lines can be drawn 
How many different straight lines can be drawn using $9$ points on the triangle of the figure below?

My try: Considering points other than $A$, $B$ and $C$ we got $2\times3 + 2\times1 + 3\times1 = 11$. For those three points,  number of lines are $1 \times1 + 2 \times 1 + 3\times1 = 6$. So the total number is $6 + 11 + 3 = 20$ counting sides as possible lines but the solution gives $24$. What are the other lines which I'm not considering here? Also I think there is a combinatorial solution but I didn't find that.
 A: We have $9$ points and a straight line is determined by two distinct points.
We can use the binomial coefficients.

In how many ways can we choose pairs of points?
  $$\binom{9}{2}=\frac{9\cdot 8}{2}=36$$

But, $A,D,E, B$ are collinear as well as $A,F,G,H,C$ and $B,I,C$.
So, we have $\binom{4}{2}+\binom{5}{2}+\binom{3}{2}$ non-distinct lines.
$$\binom{4}{2}+\binom{5}{2}+\binom{3}{2}=\frac{4\cdot 3}{2}+\frac{5\cdot 4}{2}+\frac{3\cdot 2}{2}=19$$
Now, we have to take $AB, BC$ and $AC$ into account, so the final result is:
$$36-19+3=20$$
A: Note: we have the same count, but here is my solution, anyway.
Considering that the set of points $S$ consists of the set $V$ of points that are vertices and the set $E$ of non-vertex points that lie on an edge, we can count the lines $V \to V,$ $V \to E,$ and $E \to E.$ We note that we need not find the lines $E \to V$ because these are precisely the lines $V \to E.$
Given a line $V \to V,$ our only choices are the three sides of the triangle, hence there are 3 such lines.
Given a line $V \to E,$ we can reconstruct it by first choosing a vertex point in one of three ways. Based on this choice, we obtain a different number of choices for the remaining edge point. Observe that if we choose $A,$ the only edge point that will give a new line is $I.$ Indeed, choosing any of the other edge points would give us a line $V \to V.$ Further, if we choose $B,$ the other edge points could be $F,$ $G,$ or $H.$ Last, if we choose $C,$ the other edge points could be $D$ or $E.$ Ultimately, there are $1 + 3 + 2 = 6$ such lines.
Given a line $E \to E,$ we can reconstruct it by first choosing an edge point in one of six ways. Based on this choice, we obtain a different number of choices for the remaining edge point. Explicitly, if we choose $D$ or $E,$ then we must choose $I,$ $F,$ $G,$ or $H,$ and if we subsequently choose $I,$ then we can choose $F,$ $G,$ or $H.$ Order does not matter here, so these are all the lines $E \to E;$ there are a total of $2 \cdot 4 + 1 \cdot 3 = 11$ such lines.
Overall, there are $3 + 6 + 11 = 20$ distinct straight lines we can obtain from these points.
