Question

Number of line segments in the given figure is?

(a) 5
(b) 10
(c) 15
(d) 20

Answer 1

Hint: First understand the meaning of a line segment and the number of endpoints it has. Now, use the formula of combinations given as ${}^{n}{{C}_{r}}$ to find the numbers of possible ways to select r things out of a total of n things. Consider n as the total number of points present on the line and r as the number of points to be selected to form a line segment. Now, use the formula $^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ to simplify the expression and get the answer in number.

Complete step-by-step answer:
Here we have been provided with the figure of a line on which several points (A, B, C, D, E) are present. We are asked to determine the total number of line segments that can be formed. First we need to understand the definition of a line segment.
In mathematics, a line segment is a 2 – D figure which has two end points. That means in the given figure we have to select any two points to form a line segment. Now, we know that if we have to select r number of things from a total of n number of things then we use the formula of combination given as ${}^{n}{{C}_{r}}$. So let us consider the total number points present as n and the number of points to be selected to form a line segment as r, so we have n = 5 and r = 2, substituting these values in the formula we get,
$\Rightarrow $ Number of line segments = ${}^{n}{{C}_{r}}={}^{5}{{C}_{2}}$
Using the formula $^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ we get,
$\Rightarrow $ Number of line segments = $\dfrac{5!}{2!\left( 5-2 \right)!}$
$\Rightarrow $ Number of line segments = $\dfrac{5!}{2!\times 3!}$
$\Rightarrow $ Number of line segments = $\dfrac{5\times 4\times 3\times 2\times 1}{2\times 1\times 3\times 2\times 1}$
$\Rightarrow $ Number of line segments = 10

So, the correct answer is “Option b”.

Note: You must know the difference between the two terms ‘permutation’ and ‘combination’. In general if we have to select r things from n things then we apply the combinations formula and if after selection we need to arrange those things also then we apply permutation formula given as ${}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}$. Remember the definition of a line, a ray and a line segment and the number of end points present in each.