# Difficult Permutation-Combination Solved QuestionAptitude Discussion

 Q. Ten points are marked on a straight line and eleven points are marked on another straight line. How many triangles can be constructed with vertices from among the above points?
 ✖ A. 495 ✖ B. 550 ✔ C. 1045 ✖ D. 2475

Solution:
Option(C) is correct

We can get the triangles in two different ways.

Taking two points from the line having 10 points

(in $^{10}C_2$ ways, i.e., 45 ways) and one point from the line consisting of 11 points (in 11 ways).

So, the number of triangles here is $45 \times 11 = 495$.

Taking two points from the line having 11 points (in $^{11}C_2$, i.e., 55 ways) and one point from the line consisting of 10 points (in 10 ways), the number of triangles here is $55 \times 10 = 550$

Total number of triangles,

$= 495 + 550$

$= \textbf{1,045 triangles}$