Question

The number of straight lines that can be formed by joining $20$ points no three of which are in the same straight line except $4$ of them which are in the same line?

• A. $183$
• B. $186$
• C. $197$
• D. $185$

Answer 1

The correct option is D

$185$

Explanation for correct option:

Find the number of straight lines that can be formed.

Consider the set of $20$ points out of which $4$ are in the same line.

We know that,

${}^{n}C_r=\frac{n!}{r!\left(n-1\right)!}$

Here, ${}^{n}C_r$ is the number of combinations, $n$ is the total number of objects in the set, and $r$ is the number of choosing objects from the set.

Now, out of $16$ points we need to select two points to make a straight line that can be expressed as,

$$\begin{gathered} {}^{16}C_2=\frac{16!}{2!\left(16-2\right)!} \\ {}^{16}C_2=\frac{16\times 15\times 14!}{2\times 1\times 14!} \\ {}^{16}C_2=120 \end{gathered}$$

If we select $1$ point from $16$ points and $1$ from $4$ points, then the number of possible straight lines are,

$$\begin{gathered} {}^{16}C_1\times {}^{4}C_1=\frac{16!}{1!\times \left(16-1\right)!}\times \frac{4!}{1!\times \left(4-1\right)!} \\ {}^{16}C_1\times {}^{4}C_1=\frac{16\times 15!}{1\times 15!}\times \frac{4\times 3!}{1\times 3!} \\ {}^{16}C_1\times {}^{4}C_1=16\times 4 \\ {}^{16}C_1\times {}^{4}C_1=64 \end{gathered}$$

Selecting both the points from $4$ points that are collinear and we get a single line.

Hence, the total number of straight lines formed are $120+64+1=185$

Therefore, the correct answer is Option D.