Questions & Answers

Question

Answers

A. $n(n + 3)$

B. $\dfrac{{n(n + 3)}}{2}$

C. $\dfrac{{(n + 2)(n + 1)}}{2}$

D. $n$

Answer
Verified

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.