Question

Out of $15$ points in a plane, n points are in the same straight line, $445$ triangles can be formed by joining these points. What is the value of n?

• A. $3$
• B. $4$
• C. $5$
• D. $6$

Answer 1

The correct option is C $5$

Number of triangles that can be formed is equal to the number of ways to select 3 non-collinear points.

Number of ways to select 3 points from 15 points ${=}^{15}{C}_3$

Let n points be collinear.

Number of ways to select 3 points out of the n collinear points${=}^n{C}_3$

So, Number of ways to select 3 non-collinear points = Number of ways to select 3 points using all the points - Number of ways to select 3 points using the collinear points

So, Number of ways to select 3 non-collinear points ${=}^{15}{C}_3{-}^n{C}_3$

So, Number of triangles that can be formed ${=}^{15}{C}_3{-}^n{C}_3$

$\therefore 445{=}^{15}{C}_3{-}^n{C}_3$

$\therefore 445=455{-}^n{C}_3$

${\therefore }^n{C}_3=10$

$\therefore \frac{n!}{3!(n-3)!}=10$

$\therefore \frac{n(n-1)(n-2)}{6}=10$

$\therefore n(n-1)(n-2)=60$

Solving this equation we get $n=5$

The answer is option (C)