Points in a triangle QUESTION
In a triangle $[ABC]$ thirteen points were marked, one in $[AB]$, two in $[AC]$ and $10$ in $[BC]$. How many different lines, triangles, quadrilaterals and pentagons is it possible to build with these thirteen points?
Possible solution
$\circ$ lines __________________answer = 34
$\binom {13}2 -\binom{10}2+1$
Total, less than $[CB]$ choose 2, plus 1, the line $CB$
$\circ$ triangles _______________ answer = 166
$\binom {13}3 -\binom{10}3$
Total, less than $[CB]$ choose 3
$\circ$ quadrilaterals _______________ answer = 145
$\binom {10}1\times \binom 33 +\binom{10}2\times \binom 32$
Choose 1 from $[CB]$ or choose 2
$\circ$ pentagons _______________ answer = 45
$\binom {10}2\times \binom 33
Choose 2 from $[CB]$
This are correct ? Thanks in advance
 A: Let the point in [AB] be denoted as $x$ and the two points in $[AC]$ as $y_1,y_2$ and the ten points in $[BC]$ as $z_1,\ldots,z_{10}$.
Lines:
Two ways for $x$ to $y_i$. Ten ways for $x$ to $z_j$. Twenty ways from $y_i$ to $z_j$. So 32 lines.
Here I am assuming that picking two points on the same side of the triangle does not create a new line (but if you clarify in the comments I can update).
If points on the same side of the triangle do create a new line then we also have an additional $\binom{10}{2} = 45$ for the $z_j$ and $\binom{2}{2} = 1$ for $y_1$ to $y_2$.
So $32 + 45 + 1 = 78$ lines.
Triangles:
If $x$ is one of the vertices and the other two are $y_1$ and $y_2$ that gives $1$ triangle.
Then $x,z_i,z_j$ as the vertices of a triangle give us another $\binom{10}{2} = 45$ triangles.
$x, y_i, z_j$ as the vertices of a triangles gives us another $2 \cdot 10 = 20$ triangles.
$y_1,y_2,z_j$ gives $10$ triangles.
$y_i, z_j,z_k$ gives $2 \cdots \binom{10}{2} = 90$ triangles.
So $166$ triangles.
Quadrilaterals:
$x, y_1, y_2, z_j$ gives $10$ quadrilaterals.
$x, y_i, z_j, z_k$ gives $2 \cdot \binom{10}{2} = 90$ quadrilaterals.
$y_1, y_2, z_j, z_k$ gives $\binom{10}{2} = 45$ quadrilaterals.
So $145$ quadrilaterals.
Pentagons:
$x, y_1, y_2, z_j, z_k$ gives $\binom{10}{2} = 45$ pentagons.
And there are no other ways to form pentagons so $45$ is how many we can form.
