#### On one straight line $\mathrm{AB}$, there are $\mathrm{m}$ non-A points, and on another straight line $\mathrm{AC}$, there are $\mathrm{n}$ non-A points. With these places serving as vertices, how many triangles may be constructed? How many possibilities are there if point $\mathrm{A}$ is also considered?Option: 1 $\frac{n m}{2}$Option: 2 $\frac{n(m+n)}{2}$Option: 3 $\frac{m n(m+n)}{2}}$Option: 4 $\mathrm{\frac{2 m+3 n}{2}}$

A triangle has three vertices, hence picking two points on one line, one on the other, and vice versa.

Consider the scenario where one of the triangle's points is the intersection of two lines.

Two $\mathrm{AB}$ points and one $\mathrm{AC}$ point can be combined to form a triangle, as can one $\mathrm{AB}$ point and two $\mathrm{AC}$ points.

The outcome is that there are now more triangles.

$\mathrm{\left({ }^{m} C_{2}\right)\left({ }^{n} C_{1}\right)+\left({ }^{m} C_{1}\right)\left({ }^{n} C_{2}\right)=\frac{m(m-1)}{2} n+\frac{n(n-1)}{2} m}$
$\mathrm{=\frac{1}{2} m n(m-1+n-1)}$
$\mathrm{=\frac{1}{2} m n(m+n-2)}$

More triangles can be obtained if point $\mathrm{A}$ is included. As a result, in this case,

$\mathrm{\frac{m n}{2}(m+n-2)+m n=\frac{m n(m+n)}{2} \text { triangles. }}$

Option c) is correct.