#### There are 1320 triangles formed by connecting 21 points in a plane where n points are in the same straight line. What is the numerical value of n?Option: 1 10Option: 2 5Option: 3 15Option: 4 20

The number of triangles that can be formed is equal to the number of 3 non-collinear points that can be chosen.

The number of ways to choose three points from a possible 21 points is given by,

\mathrm{\begin{aligned} &{ }^{21} C_3=\frac{21 !}{3 ! 18 !}\\ &{ }^{21} C_3=\frac{21 \times 20 \times 19}{3 \times 2}\\ &{ }^{21} C_3=1330 \end{aligned}}

Assume that n points are collinear.

The number of ways to choose three points from a set of n collinear points is $\mathrm{^{n}C_{3}}$

?So, the number of ways to select 3 non-collinear points = The number of ways to select 3 points using all the points - The number of ways to select 3 points using the collinear points

Thus, the number of triangles formed is given by,

\mathrm{\begin{aligned} & N={ }^{21} C_3-{ }^n C_3 \\ & 1330=1320-{ }^n C_3 \\ & { }^n C_3=10 \end{aligned}}

Solving the above term to get,

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

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

$\mathrm{ n(n-1)(n-2)=60 }$

$\mathrm{ n=5 }$

Therefore, the value of n is 5.