If the sides A B , B C and C A of a triangle A B C 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

Question

If the sides A B , B C and C A of a triangle A B C 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:

TestHub · Accepted Answer

We know that for making a triangle, we require three points, which are not on a line.

Here, we have total $3 + 5 + 6 = 14$ points out of which $3, 5, 6$ points respectively, on $A B, B C$ and $C A$ are collinear.

Thus, the total Number of triangles formed

$= {}^{14}C_{3} - {}^{3}C_{3} - {}^{5}C_{3} - {}^{6}C_{3}$

$= \frac{14!}{3! \cdot 11!} - \frac{3!}{3! \cdot 0!} - \frac{5!}{3! \cdot 2!} - \frac{6!}{3! \cdot 3!}$

$= \frac{14 \times 13 \times 12 \times 11!}{3 \times 2 \times 1 \times 11!} - 1 - \frac{5 \times 4 \times 3!}{3! \times 2 \times 1} - \frac{6 \times 5 \times 4 \times 3!}{3 \times 2 \times 1 \times 3!}$

$364 - 1 - 10 - 20 = 333 .$

Hence, total number of triangles is $333 .$

Mathematics - Permutation & Combination Question with Solution | TestHub

Options:

Doubts & Discussion