Permutation-Combination
Aptitude

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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

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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,

$= 495 + 550$

$= \textbf{1,045 triangles}$


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