If the sides AB, BC and CA of a triangle ABC | vedclass

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(NONE) The number of interior points on the sides $AB, BC$ and $CA$ are $3, 5$ and $6$ respectively.
Including the $3$ vertices of the triangle $A, B

Vedclass · Accepted Answer

$579$

Vedclass · Answer

(NONE) The number of interior points on the sides $AB, BC$ and $CA$ are $3, 5$ and $6$ respectively.
Including the $3$ vertices of the triangle $A, B, C$,the total number of points available is $n = 3 + 5 + 6 + 3 = 17$.
To form a triangle,we need to select $3$ points out of these $17$ points.
The total number of ways to select $3$ points is $^{17}C_{3} = \frac{17 	imes 16 	imes 15}{3 	imes 2 	imes 1} = 680$.
However,points lying on the same side are collinear and cannot form a triangle.
Number of points on side $AB$ is $3 + 2 = 5$ (including vertices $A$ and $B$).
Number of points on side $BC$ is $5 + 2 = 7$ (including vertices $B$ and $C$).
Number of points on side $CA$ is $6 + 2 = 8$ (including vertices $C$ and $A$).
Number of triangles to be subtracted = $^{5}C_{3} + ^{7}C_{3} + ^{8}C_{3} = 10 + 35 + 56 = 101$.
Total number of triangles = $680 - 101 = 579$.

If the sides $AB, BC$ and $CA$ of a triangle $ABC$ have $3, 5$ and $6$ interior points respectively,then the total number of triangles that can be constructed using these points as vertices is equal to ....... .

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