Correct Answer - Option 3 : 5

**Concept:**

Number of ways to select 3 points out of the n collinear points = \({\;^n}{C_3}\)

\({\;^n}{C_r}\; = \;\frac{{n!}}{{r!\left( {n\; - \;r} \right)!}}\)

**Calculation:**

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)

⇒ Number of ways to select 3 non-collinear points = ^{15}c_{3} - ^{n}c_{3}

⇒ Number of triangles that can be formed = ^{15}c_{3} - ^{n}c_{3}

⇒ 445 = ^{15}c_{3} - ^{n}c_{3}

⇒ ^{n}c_{3} = ^{15}c_{3} – 445 = 455 – 445 = 10

\(\Rightarrow \frac{{n!}}{{\left( {n - 3} \right)!\; \times 3!}} = 10\)

\(\Rightarrow \frac{{n\left( {n - 1} \right)\left( {n - 2} \right)}}{6} = 10\)

⇒ n (n – 1) (n – 2) = 60

∴ n = 5