Question

If a line segment be cut at n points, then the number of line segments formed is
A. $n\left( {n + 3} \right)$
B. $\dfrac{{n\left( {n - 3} \right)}}{2}$
C. $\dfrac{{\left( {n + 2} \right)\left( {n + 1} \right)}}{2}$
D.n

Answer 1

Since, we know that a line segment has two endpoints.
In this question, we will use the basic knowledge of line segment and combination.
If a line segment is cut at n parts then to find the number of the line segment, first we will find the total number of points which are lying on that line segment. Then we will use the concept of combination.

Complete step-by-step answer:
We know that a line segment has two endpoints.
Let us take a line segment AB where A and B are two end points.
Now, let us assume that this line segment AB cut at n points.
First, we will find the total number of points which are lying on the line segment AB.
If a line segment AB cut at n points. Then, we can say that there are total (n + starting point A + ending point B) points lying on a line segment.
Therefore, there are total $n + 2$ points lying on the line segment when a line segment is cut at n parts.
Now, we will use the concept of combination.
Number of ways of selecting 2 points out of $n + 2$ points is given by ${}^{n + 2}{C_2}$ . Therefore, the total number of line segments formed is ${}^{n + 2}{C_2}$ .
Now, we are going to simplify ${}^{n + 2}{C_2}$ by using the formula,
${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$ .
${}^{n + 2}{C_2} = \dfrac{{\left( {n + 2} \right)!}}{{\left( {2!} \right)\left( {n!} \right)}}$
As we know that $n! = 1 \times 2 \times 3 \times 4 \times ......... \times \left( {n - 1} \right) \times n$
now, we can write,
$\left( {n + 2} \right)! = 1 \times 2 \times 3 \times 4 \times ......... \times \left( {n - 1} \right) \times n \times \left( {n + 1} \right) + \left( {n + 2} \right)$
Therefore,
${}^{n + 2}{C_2} = \dfrac{{1 \times 2 \times 3 \times 4 \times ......... \times \left( {n - 1} \right) \times n \times \left( {n + 1} \right) + \left( {n + 2} \right)}}{{\left( {2!} \right)\left( {1 \times 2 \times 3 \times 4 \times ......... \times \left( {n - 1} \right) \times n} \right)}}$
Now, on cancellation of $1 \times 2 \times 3 \times 4 \times ......... \times \left( {n - 1} \right) \times n$ ,we get,
${}^{n + 2}{C_2} = \dfrac{{\left( {n + 1} \right) \times \left( {n + 2} \right)}}{2}$
Therefore, the total number of line segments formed is $\dfrac{{\left( {n + 1} \right) \times \left( {n + 2} \right)}}{2}$ .
Correct answer is option C

Note: Since, a line is extended in both directions but a line segment has two fixed end points.
We can measure the length of a line segment but we cannot measure the length of a line.
When we need to find the number of ways of selecting r objects out of n objects then we will use the formula ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$ .