Question

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

Answer 1

In this problem, we will use the basic knowledge of line segment and combination. If a line segment is cut at $n$ points then to find the number of line segments, 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 solution: We know that a line segment has two endpoints. Let us take a line segment $AB$ where $A$ and $B$ are two end points. 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 that line segment $AB$. If a line segment $AB$ is cut at one point say $C$ then we can say that there are three points $A,B$ and $C$ lying on the line segment. If a line segment $AB$ is cut at two points say $C,D$ then we can say that there are four points $A,B,C$ and $D$ lying on the line segment.
Similarly, if a line segment $AB$ be cut at $n$ points say then we can say that there are total $\left( n \right. +$starting point $A +$ending point $\left. B \right)$points lying on the line segment. That is, there are total $n + 2$ points lying on the line segment when a line segment is cut at $n$ points.
Now we need to find the number of line segments formed when a line segment is cut at $n$ points. We know that to form a line segment we need a minimum two points.
We have total $n + 2$ points and we need to select $2$ points out of these $n + 2$ points to find the number of line segments. Therefore, 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 ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$. Therefore, we get
${}^{n + 2}{C_2} = \dfrac{{\left( {n + 2} \right)!}}{{2!\left( {n + 2 - 2} \right)!}}$
$\Rightarrow {}^{n + 2}{C_2} = \dfrac{{\left( {n + 2} \right)!}}{{\left( {2!} \right)\left( {n!} \right)}}$
As we know that $n! = 1 \times 2 \times 3 \times 4 \times ... \times \left( {n - 1} \right) \times n$, we can write $\left( {n + 2} \right)! = 1 \times 2 \times 3 \times 4 \times ... \times \left( {n - 1} \right) \times n \times \left( {n + 1} \right) \times \left( {n + 2} \right)$.
Therefore, ${}^{n + 2}{C_2} = \dfrac{{1 \times 2 \times 3 \times 4 \times ... \times \left( {n - 1} \right) \times n \times \left( {n + 1} \right) \times \left( {n + 2} \right)}}{{\left( {2!} \right)\left( {1 \times 2 \times 3 \times 4 \times ... \times \left( {n - 1} \right) \times n} \right)}}$
On cancellation of $1 \times 2 \times 3 \times 4 \times ... \times \left( {n - 1} \right) \times n$, we get
$\Rightarrow {}^{n + 2}{C_2} = \dfrac{{\left( {n + 1} \right) \times \left( {n + 2} \right)}}{2}$
Therefore, the total number of line segments formed is $\dfrac{{\left( {n + 1} \right)\left( {n + 2} \right)}}{2}$.
Therefore, option C is correct.

Note: 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 ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$.