Question

Let $n \geqslant 2$ be an integer. Take n distinct points on a circle and join each pair of points by a line segment. Colour the line segment joining every pair of adjacent points by blue and the rest by red. If the number of red and blue line segments are equal, then the value of n is……………….

Answer 1

Here in the question blue lines would we be n because there are n numbers of pairs. As the question mentions every pair is adjacent, then if blue point is n then red is always n-1. The combination here is $^n{C_2}$.

Complete step-by-step answer:
$n \geqslant 2$
In the question blue lines would be = n
Because the line joining every pair of adjacent points
And red lines would be $^n{C_2}$ - n = n
So, $^n{C_2} = 2n$
${}^n{C_2} = \dfrac{{n!}}{{2!(n - 2)!}}$
After solving this function, we get
$\dfrac{{n(n - 1)}}{2} = 2n$
${n^2} - n = 4n$
${n^2} - n - 4n = 0$
Subtract the number that having the same variable and constant
${n^2} - 5n = 0$
Equalising the equation, we get
${n^2} = 5n$
Dividing both side by n
$\dfrac{{{n^2}}}{n} = \dfrac{{5n}}{n}$
cancelled the denominator by numerator
$n = 5$
If the number of red and blue line segment are equal, then the value of n is 5

Hence, we have here n = 5

Note: Note for solving the question that are related to arrangement or adjacent number use the combination method and always remember the combination formula that is $^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}}$. In this question students get confused between the blue point and the red point. While doing the solution keep in mind that is line joining every pair of adjacent points here the most important point is that the pair is adjacent.